Integration of hypergeometric function involving power and exponential function I would be grateful if someone can explain how to evaluate the following
$$ 
\int_{0}^{\infty} y^a exp(-y/b)\,_2F_1(1,1;2;-c y^2) dy 
$$
where $a>0$, $b>0$, and $c>2$. I searched on special functions (using Gradestien book and others) but I could not find an answer.
 A: I do not know if a closed form exist.
Let $y=\frac{x}{\sqrt{c}}$ and $k=\frac{1}{b \sqrt{c}}$ to make the integral to be
$$ c^{-\frac{a+1}{2} }\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$ Let
$$J_a=\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$
For $J_2$ it is simple. Now appear  the Meijer G function in all the cases I tried
$$J_1=\frac{G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,0,\frac{1}{2}
\end{array}
\right)}{2 \sqrt{\pi }}\qquad \qquad  J_3=\frac{2 G_{1,3}^{3,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0 \\
 0,0,\frac{3}{2}
\end{array}
\right)}{\sqrt{\pi } k^2}$$
$$J_4=\frac{4 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{3}{2},2
\end{array}
\right)}{\sqrt{\pi } k^3}\qquad \qquad  J_5=\frac{8 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,2,\frac{5}{2}
\end{array}
\right)}{\sqrt{\pi } k^4}$$
$$J_6=\frac{16 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{5}{2},3
\end{array}
\right)}{\sqrt{\pi } k^5}\qquad \qquad J_7=\frac{32 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,3,\frac{7}{2}
\end{array}
\right)}{\sqrt{\pi } k^6}$$
$$J_8=\frac{64 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{7}{2},4
\end{array}
\right)}{\sqrt{\pi } k^7}\qquad \qquad J_9=\frac{128 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,4,\frac{9}{2}
\end{array}
\right)}{\sqrt{\pi } k^8}$$ and, I hope, we could conjecture that for $a >3$
$$\color{blue}{J_a=\frac {2^{a-2}}{k^{a-1}\,\sqrt \pi}G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
 0,1 \\
 0,0,\frac{k-1}2,\frac{k}{2}
\end{array}
\right)}$$
