Simplifying $\int_0^1 x^{a-1} (1-x)^{b-1} \, _2F_1\left(1,d;c+d+1;2-\frac{1}{x}\right) \, dx$ I have been working on a statistics project that hinges on the following function.
$$
\int_0^1 x^{a-1} (1-x)^{b-1} \, _2F_1\left(1,d;c+d+1;2-\frac{1}{x}\right) \, dx$$
Is it all possible to remove the integral or make this easier to work with/compute? I'm very unfamiliar with hypergeometric functions, so since this popped out of my equations I've been pretty lost.
here 2F1 is the hypergeometric function
 A: Comment.
For $$\int_0^1 (1-y)^{b-1} (2-y)^{-a-b} \, _2F_1(1,d;c+d+1;y) \, dy\tag1$$ I find an "almost" match.
Gradshteyn & Ryzhik 7.512.9 is
\begin{align} &
\int_{0}^{1}\!{x}^{\gamma-1} \left( 1-x \right) ^{\rho-1} \left( 1-zx
 \right) ^{-\sigma}{\mbox{$_2$F$_1$}(\alpha,\beta;\,\gamma;\,x)}
\,{\rm d}x
\\
&={\frac {\Gamma \left( \gamma \right) \Gamma \left( \rho
 \right) \Gamma \left( \gamma+\rho-\alpha-\beta \right)  \left( 1-z
 \right) ^{-\sigma}}{\Gamma \left( \gamma+\rho-\alpha \right) \Gamma
 \left( \gamma+\rho-\beta \right) }
{\mbox{$_3$F$_2$}(\rho,\sigma,\gamma+\rho-\alpha-\beta;\,\gamma+\rho-\beta,\gamma+\rho-\alpha;\,{\frac {z}{z-1}})}
}
\end{align}
This yields
\begin{align} &
\int_{0}^{1}\!{x}^{c+d} \left( 1-x \right) ^{b-1} \left( 2-x \right) 
^{-a-b}{\mbox{$_2$F$_1$}(1,d;\,c+d+1;\,x)}\,{\rm d}x
\\ &
={\frac {\Gamma
 \left( c+d+1 \right) \Gamma \left( b \right) 
}{\Gamma \left( c+d+b
 \right)  \left( c+b \right) }}\;
{\mbox{$_3$F$_2$}(b,a+b,c+b;\,c+1+b,c+d+b;\,-1)}
\end{align}
But because of the $x^{c+d}$ in there, it does not match $(1)$.

Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.
