InvalidComparison error from integrate

[oscarbenjamin]

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 integrate(sin(x**n),x)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\SymPyWorkbook\lib\site-packages\sympy\integrals\integrals.py", line 1477, in integrate
    return integral.doit(**doit_flags)
  File "C:\SymPyWorkbook\lib\site-packages\sympy\integrals\integrals.py", line 541, in doit
    function, xab[0], **eval_kwargs)
  File "C:\SymPyWorkbook\lib\site-packages\sympy\integrals\integrals.py", line 1012, in _eval_integral
    h = meijerint_indefinite(g, x)
  File "C:\SymPyWorkbook\lib\site-packages\sympy\integrals\meijerint.py", line 1621, in meijerint_indefinite
    res = _meijerint_indefinite_1(f.subs(x, x + a), x)
  File "C:\SymPyWorkbook\lib\site-packages\sympy\integrals\meijerint.py", line 1684, in _meijerint_indefinite_1
    if b < 0 or f.subs(x, 0).has(nan, zoo):
  File "C:\SymPyWorkbook\lib\site-packages\sympy\core\relational.py", line 304, in __nonzero__
    raise TypeError("cannot determine truth value of Relational")
TypeError: cannot determine truth value of Relational

David: I suppose I expected it to return an unevaluated integral - integral(sin(x**n),x)

@jksuom:

We should have, for example, `if (b < 0) == True or ...` here
  File "C:\SymPyWorkbook\lib\site-packages\sympy\integrals\meijerint.py", line 1684, in _meijerint_indefinite_1
    if b < 0 or f.subs(x, 0).has(nan, zoo):

[oscargus]

I guess b.is_negative is also an alternative?

[jksuom]

I think that b would be a number in typical cases. b.is_negative would then work as well though it would take a little more time.

[oscargus]

I thought I recognized the code...

This is already fixed (but not merged) in #17343.