I have been trying to solve a research problem for a while now and in doing so, I stumbled upon the following integral:
$$\int\_0^{\infty } r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)} \, dr.$$
However, I've got no idea how to solve that. Therefore, I'd like to figure out the solution for this integral.



What is the integral of $r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)}$?
An solution for this integral can be found at [Mathmatica.SE](https://mathematica.stackexchange.com/a/225672), which is reproduced next.


After applying the change of variable technique with $x=2^r-1$ we get


$$f=\frac{e^{-\frac{\sqrt{r}}{b}} r^{\frac{d}{2}-1} \log \_2(r+1)}{2 \left(b^d \Gamma (d)\right)} $$


$$\text{Integrate}[f,\{r,0,\infty \},\text{Assumptions}\to d\in \mathbb{R}\land b\in \mathbb{R}\land d>0\land b>0]$$


then, the solution to this integral is


\begin{align}
\frac{1}{\log(4)b^d\Gamma(d)} \left[\frac{2\pi}{d} \csc\left(\frac{\pi d}{2}\right) \, {}\_1F\_2\left(\frac{d}{2};\frac{1}{2},\frac{d}{2}+1;-\frac{1}{4 b^2}\right)\\
+ 
\frac{1}{b^2 (d+1)} \left(2 \left\{-\pi b \sec \left(\frac{\pi d}{2}\right) \, {}\_1F\_2\left(\frac{d}{2}+\frac{1}{2};\frac{3}{2},\frac{d}{2}+\frac{3}{2};-\frac{1}{4 b^2}
\right) \\ +\,(d+1) b^d \Gamma (d-2)\, {}\_2F\_3\left(1,1;2,\frac{3}{2}-\frac{d}{2},2-\frac{d}{2};-\frac{1}{4 b^2}\right) \\
+ 2(d^3-2 d^2-d+2) b^{d+2} \Gamma (d-2) (\log (b)+\psi ^{(0)}(d))
\right\}\right)\right]
\end{align}


