r"""
This module contains :py:meth:`~sympy.solvers.ode.riccati.solve_riccati`,
a function which gives all rational particular solutions to first order
Riccati ODEs. A general first order Riccati ODE is given by -

.. math:: y' = b_0(x) + b_1(x)w + b_2(x)w^2

where `b_0, b_1` and `b_2` can be arbitrary rational functions of `x`
with `b_2 \ne 0`. When `b_2 = 0`, the equation is not a Riccati ODE
anymore and becomes a Linear ODE. Similarly, when `b_0 = 0`, the equation
is a Bernoulli ODE. The algorithm presented below can find rational
solution(s) to all ODEs with `b_2 \ne 0` that have a rational solution,
or prove that no rational solution exists for the equation.

Background
==========

A Riccati equation can be transformed to its normal form

.. math:: y' + y^2 = a(x)

using the transformation

.. math:: y = -b_2(x) - \frac{b'_2(x)}{2 b_2(x)} - \frac{b_1(x)}{2}

where `a(x)` is given by

.. math:: a(x) = \frac{1}{4}\left(\frac{b_2'}{b_2} + b_1\right)^2 - \frac{1}{2}\left(\frac{b_2'}{b_2} + b_1\right)' - b_0 b_2

Thus, we can develop an algorithm to solve for the Riccati equation
in its normal form, which would in turn give us the solution for
the original Riccati equation.

Algorithm
=========

The algorithm implemented here is presented in the Ph.D thesis
"Rational and Algebraic Solutions of First-Order Algebraic ODEs"
by N. Thieu Vo. The entire thesis can be found here -
https://www3.risc.jku.at/publications/download/risc_5387/PhDThesisThieu.pdf

We have only implemented the Rational Riccati solver (Algorithm 11,
Pg 78-82 in Thesis). Before we proceed towards the implementation
of the algorithm, a few definitions to understand are -

1. Valuation of a Rational Function at `\infty`:
    The valuation of a rational function `p(x)` at `\infty` is equal
    to the difference between the degree of the denominator and the
    numerator of `p(x)`.

    NOTE: A general definition of valuation of a rational function
    at any value of `x` can be found in Pg 63 of the thesis, but
    is not of any interest for this algorithm.

2. Zeros and Poles of a Rational Function:
    Let `a(x) = \frac{S(x)}{T(x)}, T \ne 0` be a rational function
    of `x`. Then -

    a. The Zeros of `a(x)` are the roots of `S(x)`.
    b. The Poles of `a(x)` are the roots of `T(x)`. However, `\infty`
    can also be a pole of a(x). We say that `a(x)` has a pole at
    `\infty` if `a(\frac{1}{x})` has a pole at 0.

Every pole is associated with an order that is equal to the multiplicity
of its appearance as a root of `T(x)`. A pole is called a simple pole if
it has an order 1. Similarly, a pole is called a multiple pole if it has
an order `\ge` 2.

Necessary Conditions
====================

For a Riccati equation in its normal form,

.. math:: y' + y^2 = a(x)

we can define

a. A pole is called a movable pole if it is a pole of `y(x)` and is not
a pole of `a(x)`.
b. Similarly, a pole is called a non-movable pole if it is a pole of both
`y(x)` and `a(x)`.

Then, the algorithm states that a rational solution exists only if -

a. Every pole of `a(x)` must be either a simple pole or a multiple pole
of even order.
b. The valuation of `a(x)` at `\infty` must be even or be `\ge` 2.

This algorithm finds all possible rational solutions for the Riccati ODE.
If no rational solutions are found, it means that no rational solutions
exist.

The algorithm works for Riccati ODEs where the coefficients are rational
functions in the independent variable `x` with rational number coefficients
i.e. in `Q(x)`. The coefficients in the rational function cannot be floats,
irrational numbers, symbols or any other kind of expression. The reasons
for this are -

1. When using symbols, different symbols could take the same value and this
would affect the multiplicity of poles if symbols are present here.

2. An integer degree bound is required to calculate a polynomial solution
to an auxiliary differential equation, which in turn gives the particular
solution for the original ODE. If symbols/floats/irrational numbers are
present, we cannot determine if the expression for the degree bound is an
integer or not.

Solution
========

With these definitions, we can state a general form for the solution of
the equation. `y(x)` must have the form -

.. math:: y(x) = \sum_{i=1}^{n} \sum_{j=1}^{r_i} \frac{c_{ij}}{(x - x_i)^j} + \sum_{i=1}^{m} \frac{1}{x - \chi_i} + \sum_{i=0}^{N} d_i x^i

where `x_1, x_2, \dots, x_n` are non-movable poles of `a(x)`,
`\chi_1, \chi_2, \dots, \chi_m` are movable poles of `a(x)`, and the values
of `N, n, r_1, r_2, \dots, r_n` can be determined from `a(x)`. The
coefficient vectors `(d_0, d_1, \dots, d_N)` and `(c_{i1}, c_{i2}, \dots, c_{i r_i})`
can be determined from `a(x)`. We will have 2 choices each of these vectors
and part of the procedure is figuring out which of the 2 should be used
to get the solution correctly.

Implementation
==============

In this implementation, we use ``Poly`` to represent a rational function
rather than using ``Expr`` since ``Poly`` is much faster. Since we cannot
represent rational functions directly using ``Poly``, we instead represent
a rational function with 2 ``Poly`` objects - one for its numerator and
the other for its denominator.

The code is written to match the steps given in the thesis (Pg 82)

Step 0 : Match the equation -
Find `b_0, b_1` and `b_2`. If `b_2 = 0` or no such functions exist, raise
an error

Step 1 : Transform the equation to its normal form as explained in the
theory section.

Step 2 : Initialize an empty set of solutions, ``sol``.

Step 3 : If `a(x) = 0`, append `\frac{1}/{(x - C1)}` to ``sol``.

Step 4 : If `a(x)` is a rational non-zero number, append `\pm \sqrt{a}`
to ``sol``.

Step 5 : Find the poles and their multiplicities of `a(x)`. Let
the number of poles be `n`. Also find the valuation of `a(x)` at
`\infty` using ``val_at_inf``.

NOTE: Although the algorithm considers `\infty` as a pole, it is
not mentioned if it a part of the set of finite poles. `\infty`
is NOT a part of the set of finite poles. If a pole exists at
`\infty`, we use its multiplicity to find the laurent series of
`a(x)` about `\infty`.

Step 6 : Find `n` c-vectors (one for each pole) and 1 d-vector using
``construct_c`` and ``construct_d``. Now, determine all the ``2**(n + 1)``
combinations of choosing between 2 choices for each of the `n` c-vectors
and 1 d-vector.

NOTE: The equation for `d_{-1}` in Case 4 (Pg 80) has a printinig
mistake. The term `- d_N` must be replaced with `-N d_N`. The same
has been explained in the code as well.

For each of these above combinations, do

Step 8 : Compute `m` in ``compute_m_ybar``. `m` is the degree bound of
the polynomial solution we must find for the auxiliary equation.

Step 9 : In ``compute_m_ybar``, compute ybar as well where ``ybar`` is
one part of y(x) -

.. math:: \overline{y}(x) = \sum_{i=1}^{n} \sum_{j=1}^{r_i} \frac{c_{ij}}{(x - x_i)^j} + \sum_{i=0}^{N} d_i x^i

Step 10 : If `m` is a non-negative integer -

Step 11: Find a polynomial solution of degree `m` for the auxiliary equation.

There are 2 cases possible -

    a. `m` is a non-negative integer: We can solve for the coefficients
    in `p(x)` using Undetermined Coefficients.

    b. `m` is not a non-negative integer: In this case, we cannot find
    a polynomial solution to the auxiliary equation, and hence, we ignore
    this value of `m`.

Step 12 : For each `p(x)` that exists, append `ybar + \frac{p'(x)}{p(x)}`
to ``sol``.

Step 13 : For each solution in ``sol``, apply an inverse transformation,
so that the solutions of the original equation are found using the
solutions of the equation in its normal form.
"""


from itertools import product
from sympy.core import S
from sympy.core.add import Add
from sympy.core.numbers import oo, Float
from sympy.core.function import count_ops
from sympy.core.relational import Eq
from sympy.core.symbol import symbols, Symbol, Dummy
from sympy.functions import sqrt, exp
from sympy.functions.elementary.complexes import sign
from sympy.integrals.integrals import Integral
from sympy.polys.domains import ZZ
from sympy.polys.polytools import Poly
from sympy.polys.polyroots import roots
from sympy.solvers.solveset import linsolve


def riccati_normal(w, x, b1, b2):
    """
    Given a solution `w(x)` to the equation

    .. math:: w'(x) = b_0(x) + b_1(x)*w(x) + b_2(x)*w(x)^2

    and rational function coefficients `b_1(x)` and
    `b_2(x)`, this function transforms the solution to
    give a solution `y(x)` for its corresponding normal
    Riccati ODE

    .. math:: y'(x) + y(x)^2 = a(x)

    using the transformation

    .. math:: y(x) = -b_2(x)*w(x) - b'_2(x)/(2*b_2(x)) - b_1(x)/2
    """
    return -b2*w - b2.diff(x)/(2*b2) - b1/2


def riccati_inverse_normal(y, x, b1, b2, bp=None):
    """
    Inverse transforming the solution to the normal
    Riccati ODE to get the solution to the Riccati ODE.
    """
    # bp is the expression which is independent of the solution
    # and hence, it need not be computed again
    if bp is None:
        bp = -b2.diff(x)/(2*b2**2) - b1/(2*b2)
    # w(x) = -y(x)/b2(x) - b2'(x)/(2*b2(x)^2) - b1(x)/(2*b2(x))
    return -y/b2 + bp


def riccati_reduced(eq, f, x):
    """
    Convert a Riccati ODE into its corresponding
    normal Riccati ODE.
    """
    match, funcs = match_riccati(eq, f, x)
    # If equation is not a Riccati ODE, exit
    if not match:
        return False
    # Using the rational functions, find the expression for a(x)
    b0, b1, b2 = funcs
    a = -b0*b2 + b1**2/4 - b1.diff(x)/2 + 3*b2.diff(x)**2/(4*b2**2) + b1*b2.diff(x)/(2*b2) - \
        b2.diff(x, 2)/(2*b2)
    # Normal form of Riccati ODE is f'(x) + f(x)^2 = a(x)
    return f(x).diff(x) + f(x)**2 - a

def linsolve_dict(eq, syms):
    """
    Get the output of linsolve as a dict
    """
    # Convert tuple type return value of linsolve
    # to a dictionary for ease of use
    sol = linsolve(eq, syms)
    if not sol:
        return {}
    return {k:v for k, v in zip(syms, list(sol)[0])}


def match_riccati(eq, f, x):
    """
    A function that matches and returns the coefficients
    if an equation is a Riccati ODE

    Parameters
    ==========

    eq: Equation to be matched
    f: Dependent variable
    x: Independent variable

    Returns
    =======

    match: True if equation is a Riccati ODE, False otherwise
    funcs: [b0, b1, b2] if match is True, [] otherwise. Here,
    b0, b1 and b2 are rational functions which match the equation.
    """
    # Group terms based on f(x)
    if isinstance(eq, Eq):
        eq = eq.lhs - eq.rhs
    eq = eq.expand().collect(f(x))
    cf = eq.coeff(f(x).diff(x))

    # There must be an f(x).diff(x) term.
    # eq must be an Add object since we are using the expanded
    # equation and it must have atleast 2 terms (b2 != 0)
    if cf != 0 and isinstance(eq, Add):

        # Divide all coefficients by the coefficient of f(x).diff(x)
        # and add the terms again to get the same equation
        eq = Add(*((x/cf).cancel() for x in eq.args)).collect(f(x))

        # Match the equation with the pattern
        b1 = -eq.coeff(f(x))
        b2 = -eq.coeff(f(x)**2)
        b0 = (f(x).diff(x) - b1*f(x) - b2*f(x)**2 - eq).expand()
        funcs = [b0, b1, b2]

        # Check if coefficients are not symbols and floats
        if any(len(x.atoms(Symbol)) > 1 or len(x.atoms(Float)) for x in funcs):
            return False, []

        # If b_0(x) contains f(x), it is not a Riccati ODE
        if len(b0.atoms(f)) or not all((b2 != 0, b0.is_rational_function(x),
            b1.is_rational_function(x), b2.is_rational_function(x))):
            return False, []
        return True, funcs
    return False, []


def val_at_inf(num, den, x):
    # Valuation of a rational function at oo = deg(denom) - deg(numer)
    return den.degree(x) - num.degree(x)


def check_necessary_conds(val_inf, muls):
    """
    The necessary conditions for a rational solution
    to exist are as follows -

    i) Every pole of a(x) must be either a simple pole
    or a multiple pole of even order.

    ii) The valuation of a(x) at infinity must be even
    or be greater than or equal to 2.

    Here, a simple pole is a pole with multiplicity 1
    and a multiple pole is a pole with multiplicity
    greater than 1.
    """
    return (val_inf >= 2 or (val_inf <= 0 and val_inf%2 == 0)) and \
        all(mul == 1 or (mul%2 == 0 and mul >= 2) for mul in muls)


def inverse_transform_poly(num, den, x):
    """
    A function to make the substitution
    x -> 1/x in a rational function that
    is represented using Poly objects for
    numerator and denominator.
    """
    # Declare for reuse
    one = Poly(1, x)
    xpoly = Poly(x, x)

    # Check if degree of numerator is same as denominator
    pwr = val_at_inf(num, den, x)
    if pwr >= 0:
        # Denominator has greater degree. Substituting x with
        # 1/x would make the extra power go to the numerator
        if num.expr != 0:
            num = num.transform(one, xpoly) * x**pwr
            den = den.transform(one, xpoly)
    else:
        # Numerator has greater degree. Substituting x with
        # 1/x would make the extra power go to the denominator
        num = num.transform(one, xpoly)
        den = den.transform(one, xpoly) * x**(-pwr)
    return num.cancel(den, include=True)


def limit_at_inf(num, den, x):
    """
    Find the limit of a rational function
    at oo
    """
    # pwr = degree(num) - degree(den)
    pwr = -val_at_inf(num, den, x)
    # Numerator has a greater degree than denominator
    # Limit at infinity would depend on the sign of the
    # leading coefficients of numerator and denominator
    if pwr > 0:
        return oo*sign(num.LC()/den.LC())
    # Degree of numerator is equal to that of denominator
    # Limit at infinity is just the ratio of leading coeffs
    elif pwr == 0:
        return num.LC()/den.LC()
    # Degree of numerator is less than that of denominator
    # Limit at infinity is just 0
    else:
        return 0


def construct_c_case_1(num, den, x, pole):
    # Find the coefficient of 1/(x - pole)**2 in the
    # Laurent series expansion of a(x) about pole.
    num1, den1 = (num*Poly((x - pole)**2, x, extension=True)).cancel(den, include=True)
    r = (num1.subs(x, pole))/(den1.subs(x, pole))

    # If multiplicity is 2, the coefficient to be added
    # in the c-vector is c = (1 +- sqrt(1 + 4*r))/2
    if r != -S(1)/4:
        return [[(1 + sqrt(1 + 4*r))/2], [(1 - sqrt(1 + 4*r))/2]]
    return [[S.Half]]


def construct_c_case_2(num, den, x, pole, mul):
    # Generate the coefficients using the recurrence
    # relation mentioned in (5.14) in the thesis (Pg 80)

    # r_i = mul/2
    ri = mul//2

    # Find the Laurent series coefficients about the pole
    ser = rational_laurent_series(num, den, x, pole, mul, 6)

    # Start with an empty memo to store the coefficients
    # This is for the plus case
    cplus = [0 for i in range(ri)]

    # Base Case
    cplus[ri-1] = sqrt(ser[2*ri])

    # Iterate backwards to find all coefficients
    s = ri - 1
    sm = 0
    for s in range(ri-1, 0, -1):
        sm = 0
        for j in range(s+1, ri):
            sm += cplus[j-1]*cplus[ri+s-j-1]
        if s!= 1:
            cplus[s-1] = (ser[ri+s] - sm)/(2*cplus[ri-1])

    # Memo for the minus case
    cminus = [-x for x in cplus]

    # Find the 0th coefficient in the recurrence
    cplus[0] = (ser[ri+s] - sm - ri*cplus[ri-1])/(2*cplus[ri-1])
    cminus[0] = (ser[ri+s] - sm  - ri*cminus[ri-1])/(2*cminus[ri-1])

    # Add both the plus and minus cases' coefficients
    if cplus != cminus:
        return [cplus, cminus]
    return cplus


def construct_c_case_3():
    # If multiplicity is 1, the coefficient to be added
    # in the c-vector is 1 (no choice)
    return [[1]]


def construct_c(num, den, x, poles, muls):
    """
    Helper function to calculate the coefficients
    in the c-vector for each pole.
    """
    c = []
    for pole, mul in zip(poles, muls):
        c.append([])

        # Case 3
        if mul == 1:
            # Add the coefficients from Case 3
            c[-1].extend(construct_c_case_3())

        # Case 1
        elif mul == 2:
            # Add the coefficients from Case 1
            c[-1].extend(construct_c_case_1(num, den, x, pole))

        # Case 2
        else:
            # Add the coefficients from Case 2
            c[-1].extend(construct_c_case_2(num, den, x, pole, mul))

    return c


def construct_d_case_4(ser, N):
    # Initialize an empty vector
    dplus = [0 for i in range(N+2)]
    # d_N = sqrt(a_{2*N})
    dplus[N] = sqrt(ser[2*N])

    # Use the recurrence relations to find
    # the value of d_s
    for s in range(N-1, -2, -1):
        sm = 0
        for j in range(s+1, N):
            sm += dplus[j]*dplus[N+s-j]
        if s != -1:
            dplus[s] = (ser[N+s] - sm)/(2*dplus[N])

    # Coefficients for the case of d_N = -sqrt(a_{2*N})
    dminus = [-x for x in dplus]

    # The third equation in Eq 5.15 of the thesis is WRONG!
    # d_N must be replaced with N*d_N in that equation.
    dplus[-1] = (ser[N+s] - N*dplus[N] - sm)/(2*dplus[N])
    dminus[-1] = (ser[N+s] - N*dminus[N] - sm)/(2*dminus[N])

    if dplus != dminus:
        return [dplus, dminus]
    return dplus


def construct_d_case_5(ser):
    # List to store coefficients for plus case
    dplus = [0, 0]

    # d_0  = sqrt(a_0)
    dplus[0] = sqrt(ser[0])

    # d_(-1) = a_(-1)/(2*d_0)
    dplus[-1] = ser[-1]/(2*dplus[0])

    # Coefficients for the minus case are just the negative
    # of the coefficients for the positive case.
    dminus = [-x for x in dplus]

    if dplus != dminus:
        return [dplus, dminus]
    return dplus


def construct_d_case_6(num, den, x):
    # s_oo = lim x->0 1/x**2 * a(1/x) which is equivalent to
    # s_oo = lim x->oo x**2 * a(x)
    s_inf = limit_at_inf(Poly(x**2, x)*num, den, x)

    # d_(-1) = (1 +- sqrt(1 + 4*s_oo))/2
    if s_inf != -S(1)/4:
        return [[(1 + sqrt(1 + 4*s_inf))/2], [(1 - sqrt(1 + 4*s_inf))/2]]
    return [[S.Half]]


def construct_d(num, den, x, val_inf):
    """
    Helper function to calculate the coefficients
    in the d-vector based on the valuation of the
    function at oo.
    """
    N = -val_inf//2
    # Multiplicity of oo as a pole
    mul = -val_inf if val_inf < 0 else 0
    ser = rational_laurent_series(num, den, x, oo, mul, 1)

    # Case 4
    if val_inf < 0:
        d = construct_d_case_4(ser, N)

    # Case 5
    elif val_inf == 0:
        d = construct_d_case_5(ser)

    # Case 6
    else:
        d = construct_d_case_6(num, den, x)

    return d


def rational_laurent_series(num, den, x, r, m, n):
    r"""
    The function computes the Laurent series coefficients
    of a rational function.

    Parameters
    ==========

    num: A Poly object that is the numerator of `f(x)`.
    den: A Poly object that is the denominator of `f(x)`.
    x: The variable of expansion of the series.
    r: The point of expansion of the series.
    m: Multiplicity of r if r is a pole of `f(x)`. Should
    be zero otherwise.
    n: Order of the term upto which the series is expanded.

    Returns
    =======

    series: A dictionary that has power of the term as key
    and coefficient of that term as value.

    Below is a basic outline of how the Laurent series of a
    rational function `f(x)` about `x_0` is being calculated -

    1. Substitute `x + x_0` in place of `x`. If `x_0`
    is a pole of `f(x)`, multiply the expression by `x^m`
    where `m` is the multiplicity of `x_0`. Denote the
    the resulting expression as g(x). We do this substitution
    so that we can now find the Laurent series of g(x) about
    `x = 0`.

    2. We can then assume that the Laurent series of `g(x)`
    takes the following form -

    .. math:: g(x) = \frac{num(x)}{den(x)} = \sum_{m = 0}^{\infty} a_m x^m

    where `a_m` denotes the Laurent series coefficients.

    3. Multiply the denominator to the RHS of the equation
    and form a recurrence relation for the coefficients `a_m`.
    """
    one = Poly(1, x, extension=True)

    if r == oo:
        # Series at x = oo is equal to first transforming
        # the function from x -> 1/x and finding the
        # series at x = 0
        num, den = inverse_transform_poly(num, den, x)
        r = S(0)

    if r:
        # For an expansion about a non-zero point, a
        # transformation from x -> x + r must be made
        num = num.transform(Poly(x + r, x, extension=True), one)
        den = den.transform(Poly(x + r, x, extension=True), one)

    # Remove the pole from the denominator if the series
    # expansion is about one of the poles
    num, den = (num*x**m).cancel(den, include=True)

    # Equate coefficients for the first terms (base case)
    maxdegree = 1 + max(num.degree(), den.degree())
    syms = symbols(f'a:{maxdegree}', cls=Dummy)
    diff = num - den * Poly(syms[::-1], x)
    coeff_diffs = diff.all_coeffs()[::-1][:maxdegree]
    (coeffs, ) = linsolve(coeff_diffs, syms)

    # Use the recursion relation for the rest
    recursion = den.all_coeffs()[::-1]
    div, rec_rhs = recursion[0], recursion[1:]
    series = list(coeffs)
    while len(series) < n:
        next_coeff = Add(*(c*series[-1-n] for n, c in enumerate(rec_rhs))) / div
        series.append(-next_coeff)
    series = {m - i: val for i, val in enumerate(series)}
    return series

def compute_m_ybar(x, poles, choice, N):
    """
    Helper function to calculate -

    1. m - The degree bound for the polynomial
    solution that must be found for the auxiliary
    differential equation.

    2. ybar - Part of the solution which can be
    computed using the poles, c and d vectors.
    """
    ybar = 0
    m = Poly(choice[-1][-1], x, extension=True)

    # Calculate the first (nested) summation for ybar
    # as given in Step 9 of the Thesis (Pg 82)
    dybar = []
    for i, polei in enumerate(poles):
        for j, cij in enumerate(choice[i]):
            dybar.append(cij/(x - polei)**(j + 1))
        m -=Poly(choice[i][0], x, extension=True)  # can't accumulate Poly and use with Add
    ybar += Add(*dybar)

    # Calculate the second summation for ybar
    for i in range(N+1):
        ybar += choice[-1][i]*x**i
    return (m.expr, ybar)


def solve_aux_eq(numa, dena, numy, deny, x, m):
    """
    Helper function to find a polynomial solution
    of degree m for the auxiliary differential
    equation.
    """
    # Assume that the solution is of the type
    # p(x) = C_0 + C_1*x + ... + C_{m-1}*x**(m-1) + x**m
    psyms = symbols(f'C0:{m}', cls=Dummy)
    K = ZZ[psyms]
    psol = Poly(K.gens, x, domain=K) + Poly(x**m, x, domain=K)

    # Eq (5.16) in Thesis - Pg 81
    auxeq = (dena*(numy.diff(x)*deny - numy*deny.diff(x) + numy**2) - numa*deny**2)*psol
    if m >= 1:
        px = psol.diff(x)
        auxeq += px*(2*numy*deny*dena)
    if m >= 2:
        auxeq += px.diff(x)*(deny**2*dena)
    if m != 0:
        # m is a non-zero integer. Find the constant terms using undetermined coefficients
        return psol, linsolve_dict(auxeq.all_coeffs(), psyms), True
    else:
        # m == 0 . Check if 1 (x**0) is a solution to the auxiliary equation
        return S.One, auxeq, auxeq == 0


def remove_redundant_sols(sol1, sol2, x):
    """
    Helper function to remove redundant
    solutions to the differential equation.
    """
    # If y1 and y2 are redundant solutions, there is
    # some value of the arbitrary constant for which
    # they will be equal

    syms1 = sol1.atoms(Symbol, Dummy)
    syms2 = sol2.atoms(Symbol, Dummy)
    num1, den1 = [Poly(e, x, extension=True) for e in sol1.together().as_numer_denom()]
    num2, den2 = [Poly(e, x, extension=True) for e in sol2.together().as_numer_denom()]
    # Cross multiply
    e = num1*den2 - den1*num2
    # Check if there are any constants
    syms = list(e.atoms(Symbol, Dummy))
    if len(syms):
        # Find values of constants for which solutions are equal
        redn = linsolve(e.all_coeffs(), syms)
        if len(redn):
            # Return the general solution over a particular solution
            if len(syms1) > len(syms2):
                return sol2
            # If both have constants, return the lesser complex solution
            elif len(syms1) == len(syms2):
                return sol1 if count_ops(syms1) >= count_ops(syms2) else sol2
            else:
                return sol1


def get_gen_sol_from_part_sol(part_sols, a, x):
    """"
    Helper function which computes the general
    solution for a Riccati ODE from its particular
    solutions.

    There are 3 cases to find the general solution
    from the particular solutions for a Riccati ODE
    depending on the number of particular solution(s)
    we have - 1, 2 or 3.

    For more information, see Section 6 of
    "Methods of Solution of the Riccati Differential Equation"
    by D. R. Haaheim and F. M. Stein
    """

    # If no particular solutions are found, a general
    # solution cannot be found
    if len(part_sols) == 0:
        return []

    # In case of a single particular solution, the general
    # solution can be found by using the substitution
    # y = y1 + 1/z and solving a Bernoulli ODE to find z.
    elif len(part_sols) == 1:
        y1 = part_sols[0]
        i = exp(Integral(2*y1, x))
        z = i * Integral(a/i, x)
        z = z.doit()
        if a == 0 or z == 0:
            return y1
        return y1 + 1/z

    # In case of 2 particular solutions, the general solution
    # can be found by solving a separable equation. This is
    # the most common case, i.e. most Riccati ODEs have 2
    # rational particular solutions.
    elif len(part_sols) == 2:
        y1, y2 = part_sols
        # One of them already has a constant
        if len(y1.atoms(Dummy)) + len(y2.atoms(Dummy)) > 0:
            u = exp(Integral(y2 - y1, x)).doit()
        # Introduce a constant
        else:
            C1 = Dummy('C1')
            u = C1*exp(Integral(y2 - y1, x)).doit()
        if u == 1:
            return y2
        return (y2*u - y1)/(u - 1)

    # In case of 3 particular solutions, a closed form
    # of the general solution can be obtained directly
    else:
        y1, y2, y3 = part_sols[:3]
        C1 = Dummy('C1')
        return (C1 + 1)*y2*(y1 - y3)/(C1*y1 + y2 - (C1 + 1)*y3)


def solve_riccati(fx, x, b0, b1, b2, gensol=False):
    """
    The main function that gives particular/general
    solutions to Riccati ODEs that have atleast 1
    rational particular solution.
    """
    # Step 1 : Convert to Normal Form
    a = -b0*b2 + b1**2/4 - b1.diff(x)/2 + 3*b2.diff(x)**2/(4*b2**2) + b1*b2.diff(x)/(2*b2) - \
        b2.diff(x, 2)/(2*b2)
    a_t = a.together()
    num, den = [Poly(e, x, extension=True) for e in a_t.as_numer_denom()]
    num, den = num.cancel(den, include=True)

    # Step 2
    presol = []

    # Step 3 : a(x) is 0
    if num == 0:
        presol.append(1/(x + Dummy('C1')))

    # Step 4 : a(x) is a non-zero constant
    elif x not in num.free_symbols.union(den.free_symbols):
        presol.extend([sqrt(a), -sqrt(a)])

    # Step 5 : Find poles and valuation at infinity
    poles = roots(den, x)
    poles, muls = list(poles.keys()), list(poles.values())
    val_inf = val_at_inf(num, den, x)

    if len(poles):
        # Check necessary conditions (outlined in the module docstring)
        if not check_necessary_conds(val_inf, muls):
            raise ValueError("Rational Solution doesn't exist")

        # Step 6
        # Construct c-vectors for each singular point
        c = construct_c(num, den, x, poles, muls)

        # Construct d vectors for each singular point
        d = construct_d(num, den, x, val_inf)

        # Step 7 : Iterate over all possible combinations and return solutions
        # For each possible combination, generate an array of 0's and 1's
        # where 0 means pick 1st choice and 1 means pick the second choice.

        # NOTE: We could exit from the loop if we find 3 particular solutions,
        # but it is not implemented here as -
        #   a. Finding 3 particular solutions is very rare. Most of the time,
        #      only 2 particular solutions are found.
        #   b. In case we exit after finding 3 particular solutions, it might
        #      happen that 1 or 2 of them are redundant solutions. So, instead of
        #      spending some more time in computing the particular solutions,
        #      we will end up computing the general solution from a single
        #      particular solution which is usually slower than computing the
        #      general solution from 2 or 3 particular solutions.
        c.append(d)
        choices = product(*c)
        for choice in choices:
            m, ybar = compute_m_ybar(x, poles, choice, -val_inf//2)
            numy, deny = [Poly(e, x, extension=True) for e in ybar.together().as_numer_denom()]
            # Step 10 : Check if a valid solution exists. If yes, also check
            # if m is a non-negative integer
            if m.is_nonnegative == True and m.is_integer == True:

                # Step 11 : Find polynomial solutions of degree m for the auxiliary equation
                psol, coeffs, exists = solve_aux_eq(num, den, numy, deny, x, m)

                # Step 12 : If valid polynomial solution exists, append solution.
                if exists:
                    # m == 0 case
                    if psol == 1 and coeffs == 0:
                        # p(x) = 1, so p'(x)/p(x) term need not be added
                        presol.append(ybar)
                    # m is a positive integer and there are valid coefficients
                    elif len(coeffs):
                        # Substitute the valid coefficients to get p(x)
                        psol = psol.xreplace(coeffs)
                        # y(x) = ybar(x) + p'(x)/p(x)
                        presol.append(ybar + psol.diff(x)/psol)

    # Remove redundant solutions from the list of existing solutions
    remove = set()
    for i in range(len(presol)):
        for j in range(i+1, len(presol)):
            rem = remove_redundant_sols(presol[i], presol[j], x)
            if rem is not None:
                remove.add(rem)
    sols = [x for x in presol if x not in remove]

    # Step 15 : Inverse transform the solutions of the equation in normal form
    bp = -b2.diff(x)/(2*b2**2) - b1/(2*b2)

    # If general solution is required, compute it from the particular solutions
    if gensol:
        sols = [get_gen_sol_from_part_sol(sols, a, x)]

    # Inverse transform the particular solutions
    presol = [Eq(fx, riccati_inverse_normal(y, x, b1, b2, bp).cancel(extension=True)) for y in sols]
    return presol