An solution for this integral can be found at https://mathematica.stackexchange.com/questions/225636/integral-of-r-frac2r-1-log-2-e-frac-sqrt2r-1b-left2r-1-ri?noredirect=1#comment574252_225636
Mathmatica.SE, 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
$$\frac{b^{-d} \left(\frac{2 \pi \csc \left(\frac{\pi d}{2}\right) \, _1F_2\left(\frac{d}{2};\frac{1}{2},\frac{d}{2}+1;-\frac{1}{4 b^2}\right)}{d}+\frac{2 \left(-\pi b \sec \left(\frac{\pi d}{2}\right) \, _1F_2\left(\frac{d}{2}+\frac{1}{2};\frac{3}{2},\frac{d}{2}+\frac{3}{2};-\frac{1}{4 b^2}\right)+(d+1) b^d \Gamma (d-2) \, _2F_3\left(1,1;2,\frac{3}{2}-\frac{d}{2},2-\frac{d}{2};-\frac{1}{4 b^2}\right)+2 \left(d^3-2 d^2-d+2\right) b^{d+2} \Gamma (d-2) (\log (b)+\psi ^{(0)}(d))\right)}{b^2 (d+1)}\right)}{\log (4) \Gamma (d)}$$\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}
