Does this integral $\int_0^\infty \frac{dx}{(1+e^x)(a+x)}$ have a closed form?

Since: $$\mathcal{L}\left(\frac{1}{1+e^x}\right) = \frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\qquad \mathcal{L}^{-1}\left(\frac{1}{a+x}\right)=e^{-as}$$ the original integral equals: $$ \int_{0}^{+\infty}\frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds=\frac{1}{2}\int_{0}^{+\infty}\int_{\frac{s+1}{2}}^{\frac{s+2}{2}}\sum_{n\geq 0}\frac{e^{-as}}{(n+u)^2}\,du\,ds $$ and by integration by parts, the LHS just depends on: $$ \int_{0}^{+\infty}\left[\log\Gamma\left(\frac{s+2}{2}\right)-\log\Gamma\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds$$ that can be further expanded by exploiting the Weierstrass product for the $\Gamma$ function, leading to a fast-convergent series, whose first term is $-\frac{\gamma}{2a}$.