(f, x, y, **kwargs)
Given an expression, f, 3 tests will be done to see what type
    of composite bivariate it might be, options for u(x, y) are::

        x*y
        x+y
        x*y+x
        x*y+y

    If it matches one of these types, ``u(x, y)``, ``P(u)`` and dummy
    variable ``u`` will be returned. Solving ``P(u)`` for ``u`` and
    equating the solutions to ``u(x, y)`` and then solving for ``x`` or
    ``y`` is equivalent to solving the original expression for ``x`` or
    ``y``. If ``x`` and ``y`` represent two functions in the same
    variable, e.g. ``x = g(t)`` and ``y = h(t)``, then if ``u(x, y) - p``
    can be solved for ``t`` then these represent the solutions to
    ``P(u) = 0`` when ``p`` are the solutions of ``P(u) = 0``.

    Only positive values of ``u`` are considered.

    Examples
    ========

    >>> from sympy.solvers.solvers import solve
    >>> from sympy.solvers.bivariate import bivariate_type
    >>> from sympy.abc import x, y
    >>> eq = (x**2 - 3).subs(x, x + y)
    >>> bivariate_type(eq, x, y)
    (x + y, _u**2 - 3, _u)
    >>> uxy, pu, u = _
    >>> usol = solve(pu, u); usol
    [sqrt(3)]
    >>> [solve(uxy - s) for s in solve(pu, u)]
    [[{x: -y + sqrt(3)}]]
    >>> all(eq.subs(s).equals(0) for sol in _ for s in sol)
    True
Given an expression, f, 3 tests will be done to see what type
    of composite bivariate it might be, options for u(x, y) are::
Given an expression f 3 tests will be done to see what type of composite bivariate it might be options for u ( x y ) are ::
def bivariate_type(f, x, y, **kwargs):
    """Given an expression, f, 3 tests will be done to see what type
    of composite bivariate it might be, options for u(x, y) are::

        x*y
        x+y
        x*y+x
        x*y+y

    If it matches one of these types, ``u(x, y)``, ``P(u)`` and dummy
    variable ``u`` will be returned. Solving ``P(u)`` for ``u`` and
    equating the solutions to ``u(x, y)`` and then solving for ``x`` or
    ``y`` is equivalent to solving the original expression for ``x`` or
    ``y``. If ``x`` and ``y`` represent two functions in the same
    variable, e.g. ``x = g(t)`` and ``y = h(t)``, then if ``u(x, y) - p``
    can be solved for ``t`` then these represent the solutions to
    ``P(u) = 0`` when ``p`` are the solutions of ``P(u) = 0``.

    Only positive values of ``u`` are considered.

    Examples
    ========

    >>> from sympy.solvers.solvers import solve
    >>> from sympy.solvers.bivariate import bivariate_type
    >>> from sympy.abc import x, y
    >>> eq = (x**2 - 3).subs(x, x + y)
    >>> bivariate_type(eq, x, y)
    (x + y, _u**2 - 3, _u)
    >>> uxy, pu, u = _
    >>> usol = solve(pu, u); usol
    [sqrt(3)]
    >>> [solve(uxy - s) for s in solve(pu, u)]
    [[{x: -y + sqrt(3)}]]
    >>> all(eq.subs(s).equals(0) for sol in _ for s in sol)
    True

    """

    u = Dummy('u', positive=True)

    if kwargs.pop('first', True):
        p = Poly(f, x, y)
        f = p.as_expr()
        _x = Dummy()
        _y = Dummy()
        rv = bivariate_type(Poly(f.subs({x: _x, y: _y}), _x, _y), _x, _y, first=False)
        if rv:
            reps = {_x: x, _y: y}
            return rv[0].xreplace(reps), rv[1].xreplace(reps), rv[2]
        return

    p = f
    f = p.as_expr()

    # f(x*y)
    args = Add.make_args(p.as_expr())
    new = []
    for a in args:
        a = _mexpand(a.subs(x, u/y))
        free = a.free_symbols
        if x in free or y in free:
            break
        new.append(a)
    else:
        return x*y, Add(*new), u

    def ok(f, v, c):
        new = _mexpand(f.subs(v, c))
        free = new.free_symbols
        return None if (x in free or y in free) else new

    # f(a*x + b*y)
    new = []
    d = p.degree(x)
    if p.degree(y) == d:
        a = root(p.coeff_monomial(x**d), d)
        b = root(p.coeff_monomial(y**d), d)
        new = ok(f, x, (u - b*y)/a)
        if new is not None:
            return a*x + b*y, new, u

    # f(a*x*y + b*y)
    new = []
    d = p.degree(x)
    if p.degree(y) == d:
        for itry in range(2):
            a = root(p.coeff_monomial(x**d*y**d), d)
            b = root(p.coeff_monomial(y**d), d)
            new = ok(f, x, (u - b*y)/a/y)
            if new is not None:
                return a*x*y + b*y, new, u
            x, y = y, x
def bivariate_type ( f , x , y , * * kwargs ) : u = Dummy ( 'u' , positive = True ) if kwargs . pop ( 'first' , True ) : p = Poly ( f , x , y ) f = p . as_expr ( ) _x = Dummy ( ) _y = Dummy ( ) rv = bivariate_type ( Poly ( f . subs ( { x : _x , y : _y } ) , _x , _y ) , _x , _y , first = False ) if rv : reps = { _x : x , _y : y } return rv [ 0 ] . xreplace ( reps ) , rv [ 1 ] . xreplace ( reps ) , rv [ 2 ] return p = f f = p . as_expr ( ) # f(x*y) args = Add . make_args ( p . as_expr ( ) ) new = [ ] for a in args : a = _mexpand ( a . subs ( x , u / y ) ) free = a . free_symbols if x in free or y in free : break new . append ( a ) else : return x * y , Add ( * new ) , u def ok ( f , v , c ) : new = _mexpand ( f . subs ( v , c ) ) free = new . free_symbols return None if ( x in free or y in free ) else new # f(a*x + b*y) new = [ ] d = p . degree ( x ) if p . degree ( y ) == d : a = root ( p . coeff_monomial ( x ** d ) , d ) b = root ( p . coeff_monomial ( y ** d ) , d ) new = ok ( f , x , ( u - b * y ) / a ) if new is not None : return a * x + b * y , new , u # f(a*x*y + b*y) new = [ ] d = p . degree ( x ) if p . degree ( y ) == d : for itry in range ( 2 ) : a = root ( p . coeff_monomial ( x ** d * y ** d ) , d ) b = root ( p . coeff_monomial ( y ** d ) , d ) new = ok ( f , x , ( u - b * y ) / a / y ) if new is not None : return a * x * y + b * y , new , u x , y = y , x