A closed form indeed exists for this integral:
$$\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=\pi^2\left(\log\frac{\pi\,A^9\sqrt{2}}{\Gamma(\frac{1}{4})^2}-\frac{9}{8}\right)+2\,\pi\,C,$$
where $A$ is the Glaisher-Kinkelin constant and $C$ is the Catalan constant.
A more general result:
$$\int_0^\infty\frac{\log\left(1+\frac{a}{x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=8\,\pi^2\psi^{(-2)}\left(\frac{\sqrt{a}}{2\pi}\right)-\frac{a}2\left(1+\log\frac{4\pi^2}{a}\right)-2\pi\sqrt{a}\left(1+2\log\Gamma\left(\frac{\sqrt{a}}{2\pi}\right)\right),$$
where $\psi^{(-2)}(z)$ is the generalized polygamma function.
The proof is based on Binet's second formula, but I still need to sort out some details.
We first notice that
\begin{align*}
\int_{0}^{\infty}\frac{\operatorname{gd}(x)}{e^x-1}\,dx
&= \int_{0}^{\infty} \frac{1}{e^x-1} \left( \int_{0}^{x} \frac{dy}{\cosh y} \right) \, dx \\
&= \int_{0}^{\infty} \frac{1}{\cosh y} \left( \int_{y}^{\infty} \frac{dx}{e^x - 1} \right) \,d y \\
&= -2 \int_{0}^{\infty} \frac{\log(1 - e^{-y})}{e^y + e^{-y}} \, dy \\
&=-2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta \tag{$e^{-x}=\tan\theta$}.
\end{align*}
The last integral is our starting point. We introduce two tricks to evaluate this.
Step 1. Notice that $\tan(\frac{\pi}{4}-\theta)=\frac{1-\tan\theta}{1+\tan\theta}$. So by the substitution $\theta \mapsto \frac{\pi}{4}-\theta$, it follows that
$$ \int_{0}^{\frac{\pi}{4}}\log(1+\tan\theta)\,d\theta
= \int_{0}^{\frac{\pi}{4}}\log\left(\frac{2}{1+\tan\theta}\right)\,d\theta $$
and hence both integrals have the common value $\frac{\pi}{8}\log 2$. Applying the same idea to our integral, it then follows that
\begin{align*}
-2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta
&= -2\int_{0}^{\frac{\pi}{4}}\log\left(\frac{2\tan\theta}{1+\tan\theta}\right)\,d\theta \\
&= -2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta - \frac{\pi}{4}\log 2.
\end{align*}
Step 2. In order to compute the last integral, we notice that for $\theta\in\mathbb{R}$ with $\cos\theta\neq0$, we have
\begin{align*}
-\log\left|\tan\theta\right|
&= \log\left|\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right|
= \operatorname{Re} \log\left(\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right) \\
&= \operatorname{Re}\left( \sum_{n=1}^{\infty} \frac{1+(-1)^n}{n} e^{2in\theta} \right) \\
&= \sum_{k=0}^{\infty} \frac{2}{2k+1}\cos(4k+2)\theta.
\end{align*}
So by term-wise integration, we obtain
\begin{align*}
-2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta
&= \sum_{k=0}^{\infty} \frac{4}{2k+1} \int_{0}^{\frac{\pi}{4}} \cos(4k+2)\theta \, d\theta \\
&= 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}
= 2K,
\end{align*}
where $K$ is the Catalan's constant.
Best Answer
Maple does this in terms of the Meijer G function. I general, probably nothing simpler can be expected.
If $b$ is rational, then Maple seems to find something where there is a sum over the roots of a certain polynomial... Example
$$ \int_0^\infty \frac{(1-z^2)\;dz}{(1+z^2)(1+az)^{7/4}} = \frac{4a^2-4}{3a(a^2+1)} +\frac{a}{a^2+1} \sum \frac{(\rho^4-2)\log(1-\rho)}{\rho^3(\rho^4-1)} $$ where the $\Sigma$ is over all eight roots $\rho$ of the polynomial $Z^8-2Z^4+(1+a^2)$.