When I use Simplify[] or FullSimplify[], specifying domain reals, I get terms with $0^n$ as the coefficient. Isn't that always zero, and thus be simplified out? A example of input/output is below, with the expressions truncated/replaced since they are extremely long.
Input:
Assuming[Reals && c > 5 , FullSimplify[expression]]
Output:
$$
\int_0^1 i x... +0^{-1+c} \text{Hypergeometric2F1Regularized}\left[\frac{-1+c}{c},1-c i,2-\frac{1}{c},0^c\right]\ F'[x] \, dx
$$
Note, I added the assumption that $c >5$ as overkill to avoid any possible issue with raising 0 to a negative number (or zero), but it didn't help. (Also, note, the $i$ above is a variable, not the imaginary number). Does anyone know why this happens?
Usually, especially in the case of binomial identities, $$ 0^n=\left\{\begin{array}{c}1\text{ if }n=0\\0\text{ if }n>0\end{array}\right. $$ is used.
In your particular example, the exponent of $0$ is assumed to be greater than $0$, so I don't see the need for $0^{-1+c}$ or $0^c$. The answer is still correct, but unnecessarily complicated. If your assumption was $c\ge 1$, then $0^{-1+c}$ would be needed, but not $0^c$.