I need to evaluate the following integral:

$ $ \int_{1}^\infty dx\left({x+t^2-1}\right)^{-\frac{a}{2}}x^{-\frac{b}{2}}\left({x+r^2-1}\right)^{-\frac{c}{2}}$ $

where, $ a$ , $ b$ , $ c$ are odd and, $ t$ and $ r$ are real. Surprisingly, Mathematica is not able to compute this, nor does return any divergence issue within this interval. But on the other hand, it is computing the integral for $ [{0},\infty]$ and $ [{0},1]$ intervals, respectively, as follows: $ $ \int_{0}^\infty dx\left({x+t^2-1}\right)^{-\frac{a}{2}}x^{-\frac{b}{2}}\left({x+r^2-1}\right)^{-\frac{c}{2}} = \pi Csc[\frac{(b+c)}{2}\pi](t^2-1)^{-\frac{a}{2}}\Biggl(-\frac{(r^2-1)^{1-\frac{b}{2}-\frac{c}{2}}}{\Gamma(c/2)}\Gamma(1-\frac{b}{2})_2F_1(\frac{a}{2},1-\frac{b}{2},\frac{(4-b-c)}{2};\frac{r^2-1}{t^2-1})+{\Gamma(\frac{a+b+c-2}{2})}\frac{1}{\Gamma(\frac{a}{2})}\space_2F_1(\frac{c}{2},\frac{(a+b+c-2)}{2},\frac{(b+c)}{2};\frac{r^2-1}{t^2-1})(t^2-1)^{1-\frac{b}{2}-\frac{c}{2}}\Biggr),$ $ and, $ $ \int_{0}^1 dx\left({x+t^2-1}\right)^{-\frac{a}{2}}x^{-\frac{b}{2}}\left({x+r^2-1}\right)^{-\frac{c}{2}} = \frac{2}{2-b}{(r^2-1)}^{-\frac{c}{2}}{(t^2-1)}^{-\frac{a}{2}}AppellF1(1-\frac{b}{2},\frac{c}{2},\frac{a}{2},2-\frac{b}{2};\frac{1}{1-r^2};\frac{1}{1-t^2})$ $ where, $ _2F_1$ are the regularized Hypergeometric $ _2F_1$ functions.

I’ve tried with turning on the “PrincipalValue” option in order to trace any divergence related issues for the above integral in $ [1,\infty]$ , but my attempt didn’t succeed as Mathematica didn’t return any message or warning. So I am wondering: 1. Is this possibly due to a bug, especially related to manipulating $ AppellF1$ ? 2. Could someone please shed some light on how to get this integral done in $ [1,\infty]$ ?

Any help or suggestions would be much appreciated and thanks in advance for that.