human
Complete the below
@vectorize(0)
def constantsimp(expr, constants):

    ''''Simplifies an expression with arbitrary constants in it.
    This function is written specifically to work with
    :py:meth:`~sympy.solvers.ode.dsolve`, and is not intended for general use.
    Simplification is done by "absorbing" the arbitrary constants into other
    arbitrary constants, numbers, and symbols that they are not independent
    of.
    The symbols must all have the same name with numbers after it, for
    example, ``C1``, ``C2``, ``C3``.  The ``symbolname`` here would be
    \'``C``\', the ``startnumber`` would be 1, and the ``endnumber`` would be 3.
    If the arbitrary constants are independent of the variable ``x``, then the
    independent symbol would be ``x``.  There is no need to specify the
    dependent function, such as ``f(x)``, because it already has the
    independent symbol, ``x``, in it.
    Because terms are "absorbed" into arbitrary constants and because
    constants are renumbered after simplifying, the arbitrary constants in
    expr are not necessarily equal to the ones of the same name in the
    returned result.
    If two or more arbitrary constants are added, multiplied, or raised to the
    power of each other, they are first absorbed together into a single
    arbitrary constant.  Then the new constant is combined into other terms if
    necessary.
    Absorption of constants is done with limited assistance:
    1. terms of :py:class:`~sympy.core.add.Add`\s are collected to try join
    constants so `e^x (C_1 \cos(x) + C_2 \cos(x))` will simplify to `e^x
    C_1 \cos(x)`;
    2. powers with exponents that are :py:class:`~sympy.core.add.Add`\s are
    expanded so `e^{C_1 + x}` will be simplified to `C_1 e^x`.
    Use :py:meth:`~sympy.solvers.ode.constant_renumber` to renumber constants
    after simplification or else arbitrary numbers on constants may appear,
    e.g. `C_1 + C_3 x`.
    In rare cases, a single constant can be "simplified" into two constants.
    Every differential equation solution should have as many arbitrary
    constants as the order of the...'''

Can you provide a the Python program you requested?
assistant
#Here is the Python program you requested.
    Cs = constants
    orig_expr = expr
    constant_subexprs = _get_constant_subexpressions(expr, Cs)
    for xe in constant_subexprs:
        xes = list(xe.free_symbols)
        if (not xes):
            continue
        if all([(expr.count(c) == xe.count(c)) for c in xes]):
            xes.sort(key=str)
            expr = expr.subs(xe, xes[0])
    try:
        (commons, rexpr) = cse(expr)
        commons.reverse()
        rexpr = rexpr[0]
        for s in commons:
            cs = list(s[1].atoms(Symbol))
            if ((len(cs) == 1) and (cs[0] in Cs)):
                rexpr = rexpr.subs(s[0], cs[0])
            else:
                rexpr = rexpr.subs(*s)
        expr = rexpr
    except Exception:
        pass
    expr = __remove_linear_redundancies(expr, Cs)
    def _conditional_term_factoring(expr):
        new_expr = terms_gcd(expr, clear=False, deep=True, expand=False)
        if new_expr.is_Mul:
            infac = False
            asfac = False
            for m in new_expr.args:
                if (m.func is exp):
                    asfac = True
                elif m.is_Add:
                    infac = any(((fi.func is exp) for t in m.args for fi in Mul.make_args(t)))
                if (asfac and infac):
                    new_expr = expr
                    break
        return new_expr
    expr = _conditional_term_factoring(expr)
    if (orig_expr != expr):
        return constantsimp(expr, Cs)
    return expr