Does anyone have an idea of solving following integral $$ I =\int_{0}^{\infty}\frac{x\Gamma \left(a,b x\right)}{(x^2+1)}\,dx\,,$$ where $a,b>0$ are positive real values?
Mathematica gives an answer, but I do not know how it can be derived by using integral tables.
\begin{align} I=&\frac{1}{{2 a (a+1)}}\biggl(\pi (a+1) b^a \csc \left(\frac{\pi a}{2}\right) \, _1F_2\left(\frac{a}{2};\frac{1}{2},\frac{a}{2}+1;-\frac{b^2}{4}\right) \\ &+a b \left((a+1) b \Gamma (a-2) \, _2F_3\left(1,1;2,\frac{3}{2}-\frac{a}{2},2-\frac{a}{2};-\frac{b^2}{4}\right)-\pi b^a \sec \left(\frac{\pi a}{2}\right) \, _1F_2\left(\frac{a}{2}+\frac{1}{2};\frac{3}{2},\frac{a}{2}+\frac{3}{2};-\frac{b^2}{4}\right)\right)+2 \Gamma (a+2) (\psi ^{(0)}(a)-\log (b))\biggr) \end{align}
Eq. 2.10.1.3 in Integrals and Series [Vol 2 - Spl Functions] - A. Prudnikov would be one of the closet answer, but I could not match parameters well.
Assuming that $a$ is a positive integer and $b>0$, considering $$I_a =\int_{0}^{\infty}\frac{x}{x^2+1}\,\Gamma \left(a,b x\right)\,dx$$ there is only one which is simple $$I_1=\frac{\pi}{2} \sin (b)-(\text{Ci}(b) \cos (b)+\text{Si}(b) \sin (b))$$ where appear the since and cosine integral functions.
The problem is that, for $a>1$, it seems that the result can only be expressed in terms of the Meijer G function.
The case of $a=2$ is
$$I_2=\frac {1}{\sqrt \pi}\,G_{1,3}^{3,1}\left(\frac{b^2}{4}| \begin{array}{c} 0 \\ 0,0,\frac{3}{2} \end{array} \right)$$
For $a>2$, we have
$$I_a=\frac {2^{a-2}}{\sqrt \pi}\, G_{2,4}^{4,1}\left(\frac{b^2}{4}| \begin{array}{c} 0,1 \\ 0,0,\frac a 2,\frac{a+1}{2} \end{array} \right)$$
Edit
As written in comments, for rational values of $a$ we have quite nasty results. As an example, for $a=\frac 32$, the result given by a CAS is $$2I_{\frac 32}=-\sqrt{\pi } \left(2 b^2 \, _2F_3\left(1,1;\frac{3}{4},\frac{5}{4},2;-\frac{b^2}{4}\right)+\log (4b)+\gamma -2\right)+\pi ^{3/2} \left(C\left( \sqrt{\frac{2b}{\pi }}\right)-S\left( \sqrt{\frac{2b}{\pi }}\right)\right)+ \pi \sqrt{2b} (\sin (b)-\cos (b))$$