Integral as a Beta function 
Is there a way to express integrals like $$\int_0^1 x^{-a} (1+cx)^{1-a} \text dx$$ as a Beta function, where $c$ is positive (or negative) constant and $a>1$?

I tried to use the substitution
\begin{equation} u = \left[{\dfrac{\left(c+1\right)^{a-1}x^a}{\left(cx+1\right)^{a-1}}}\right]
\end{equation}
\begin{equation}
\dfrac {du}{dx}=\dfrac{\left(1-a\right)c\left(c+1\right)^{a-1}x^a}{\left(cx+1\right)^a}+a\left(c+1\right)^{a-1}x^{a-1}\left(cx+1\right)^{1-a}
\end{equation}
But, I always get an $x$ term messing around. Another option is  to use q-Beta functions..but it is too restrictive because $0 < q < 1$.
 A: If we let $t=-cx$, $dx=dt/(-c)$, and $x^{-a}=(-c)^at^{-a}$, then
$$\int_0^1 x^{-a} (1+cx)^{1-a} \text dx=(-c)^{a-1}\int_0^{-c}t^{-a}(1-t)^{1-a}dt$$
Recalling the definition of the beta function
$$B(\nu,\mu,x)=\int_0^x t^{\nu-1}(1-t)^{\mu-1}dt,\quad \nu,\mu>0,\  x\le1$$
and blithely ignoring the obvious problem with the limits we can write
$$\int_0^1 x^{-a} (1+cx)^{1-a} \text dx=(-c)^{a-1}\text{B}(-c,1-a,2-a)$$
with the restriction that $a<1$ in order to meet the requirements on the beta function parameters. To the casual observer this may seem absurd because of the argument $-c$ in the function. I ran the original integral in Mathematica and came out with the same result, albeit with a few extra caveats. Namely,
$$
\Re(1/c) \ge 0 \  || \  \Re(1/c) \le -1 \  || \  1/c \not\in \mathbb{R}
$$
Unfortunately, I have no access to an analytically continued incomplete beta function and cannot test this out. In fact, I'm not even sure what the parallel lines mean; in computer code this usually means or, but there are other mathematical meanings, as well.
