# Closed form for $\int\limits_0^\infty \frac{x^\alpha(1-x)^\beta}{(x-c)^\gamma(x-b)^\gamma(x-\bar{b})^\gamma} \mathrm{d}x$

I would like to find a "closed form" for the integral

$$I(\beta) = \int_0^\infty \frac{x^\alpha(1-x)^\beta}{(x-c)^\gamma(x-b)^\gamma(x-\bar{b})^\gamma}\mathrm{d}x$$ where $$\gamma=3/2$$, $$\alpha,\beta>0$$ , $$\{b,\bar{b}\}$$ is a pair of complex conjugates, and $$c$$ is real.

Defining $$a\equiv -\alpha -\beta +7/2$$ and supposing $$a>0$$, we can write $$I$$ as $$[1]$$ $$I(\beta) = (-1)^\beta B(a,\alpha+1) R\left(a;-\beta,\frac32,\frac32,\frac32;-1,-c,-b,-\bar{b}\right)$$

where $$B$$ is the Beta function, and $$R$$ is a generalized Carlson elliptic function. My question is, can this expression be simplified? Is there a way to write this R function in a "more closed form"? For instance, in terms of hypergeometric functions?

In particular, I am interested in taking the derivative with respect to $$\beta$$ and then take the limit $$\beta \rightarrow 0$$ (the original integral has a log term). As noted here,

$$\int_0^{\infty}\!\!\!\!\frac{x^{\alpha-1}\ln^n(cx+d)}{(ax+b)^\sigma(cx+d)^\rho} \! \mathrm dx=(-1)^n\left(\!\frac{d}{c}\!\right)^\alpha\!\! b^{-\sigma}\!\frac{\partial^n}{\partial \rho^n}\!\!\left[d^{-\rho}B(\alpha,\sigma+\rho-\alpha) \ _2F_1\!\!\left(\!\sigma,\alpha;\sigma+\rho;1\!-\!\frac{ad}{bc}\!\right)\!\right]$$ so maybe there is a similar formula for $$I(\beta)$$ in terms of the Beta function and some hypergeometric function.

$$[1]$$: By using equation $$(\text{T}.2)$$ in page $$224$$.