Find the integral $\int_0^\infty \frac{\log(1+x^2)}{e^{2\pi x}+1}dx$ With reference to my previous question : Show that $\int_0^\infty \frac{x\log(1+x^2)}{e^{2\pi x}+1}dx=\frac{19}{24} - \frac{23}{24}\log 2 - \frac12\log A$
The following integral should be manageable :

$$\int_0^\infty \frac{\log(1+x^2)}{e^{2\pi x}+1}dx = ?$$

But it seams that wolframalpha struggles to find a closed form solution. 
Maybe the answer should be in terms Barnes G function or the derivative of the Hurwtiz zeta function.

Update
My attempt so far, consider 
$$F(a,b) = \int^\infty_0 \frac{\log(a+t^2)}{e^{2\pi (t+b)}-1}\,dt$$
We consider the derivative 
$$\frac{\partial }{\partial a}F(a,b) = \int^\infty_0 \frac{dt}{(t^2+a)(e^{2\pi (t+b)}-1)}\,dt$$
Use the following expansion
$$\frac{2t}{e^{2\pi t}-1} =\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2} $$
Hence have 
$$\frac{1}{e^{2\pi (t+b)}-1} =\frac{1}{2(t+b)\pi}-\frac{1}{2}+\frac{t+b}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+(t+b)^2} $$
$$\frac{\partial }{\partial a}F(a,b) = \int^\infty_0 \frac{1}{t^2+a}\left\{ \frac{1}{2(t+b)\pi}-\frac{1}{2}+\frac{t+b}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+(t+b)^2}\right\}dt$$
This reduces to 
$$\frac{\partial }{\partial a}F(a,b)= \frac{1}{2\pi}\int^\infty_0 \frac{1}{(t^2+a)(t+b)} dt-\frac{1}{2}\int^\infty_0 \frac{dt}{t^2+a}+\\\frac{1}{\pi}\sum_{k=1}^\infty\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt$$
The first and second are simple it remains to evaluate 
$$\sum_{k=1}^\infty\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt$$
Then integrate with respect to $a$.
Or evaluate 
$$\sum_{k=1}^\infty\int^\infty_0\frac{\log(t^2+1)(t+b)}{k^2+(t+b)^2}dt$$
Where $b = i+1/2$.
 A: By integration by parts, such integral equals
$$ I=\frac{1}{\pi}\int_{0}^{+\infty}\frac{x\log\left(1+e^{-2\pi x}\right)}{1+x^2}\,dx $$
and since $\mathcal{L}^{-1}\left(\frac{x}{1+x^2}\right)=\cos(s)$ and $\mathcal{L}\left(\frac{(-1)^{k+1} e^{-2\pi k x}}{k}\right) = \frac{(-1)^{k+1}}{k(s+2\pi k)}$, we also have:
$$ I = 2\int_{0}^{+\infty}\cos(s)\sum_{k\geq 1}\frac{(-1)^{k+1}}{2\pi k(s+2\pi k)}\,ds $$
so $I$ is a series of weighted sine and cosine integrals depending on a Lerch transcendent.
That is not surprising since
$$ \int_{0}^{+\infty}\log(1+x^2)e^{-2k\pi x}\,dx = -\frac{\text{Ci}(2\pi k)}{\pi k}=\frac{1}{\pi k}\int_{2\pi k}^{+\infty}\frac{\cos x}{x}\,dx=\frac{1}{\pi k}\int_{0}^{+\infty}\frac{\cos x}{x+2\pi k}\,dx $$
So:
$$\begin{eqnarray*} I &=& \frac{1}{\pi}\int_{0}^{+\infty}\cos(2\pi u)\sum_{k\geq 1}\frac{(-1)^{k+1}}{k (u+k)}\,du\\&=&\int_{0}^{+\infty}\frac{\cos(2\pi u)}{2\pi u}\left[\log(4)+\psi\left(\tfrac{1+u}{2}\right)-\psi\left(\tfrac{2+u}{2}\right)\right]\,du\tag{1}\end{eqnarray*}$$
Additionally, from
$$ \zeta(s)=\frac{1}{\Gamma(s)}\left(1-\frac{2}{2^s}\right)^{-1}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx \tag{2}$$
holding for every $s$ such that $\text{Re}(s)>0$, it is not difficult to derive
$$ \int_{0}^{+\infty}\frac{\log x}{e^{2\pi x}+1}\,dx = -\frac{\log(2)\log(8\pi^2)}{4\pi} \tag{3} $$
through a substitution and differentiation under the integral sign. We also have:
$$\begin{eqnarray*} I &=& \int_{0}^{+\infty}\frac{\log(x+i)}{e^{2\pi x}+1}\,dx + \int_{0}^{+\infty}\frac{\log(x-i)}{e^{2\pi x}+1}\,dx\\
&=& \left(\int_{i}^{+\infty+i}+\int_{-i}^{+\infty-i}\right)\frac{\log x}{e^{2\pi x}+1}\,dx \tag{4}\end{eqnarray*}$$
hence our problem is equivalent to finding a closed form for the principal value of the integral
$$ J = \text{PV}\int_{0}^{i}\text{Re}\left(\frac{\log x}{e^{2\pi x}+1}\right)\,dx \tag{5}$$
by choosing a suitable contour avoiding the singularity at $x=\frac{i}{2}$. The singularity is associated with the term corresponding to $h=0$ in
$$ \frac{1}{e^{2\pi x}+1}=\frac{1}{2}-\sum_{h\geq 0}\frac{4x}{(2h+1)^2 \pi+4\pi x^2}\tag{6}$$
and for every $h>0$ we have
$$ \text{Re}\int_{0}^{i}\log(x)\frac{4x}{(2h+1)^2 \pi+4\pi x^2}\,dx = \frac{1}{4\pi}\,\text{Li}_2\left(\frac{4}{(2h+1)^2}\right)\tag{7}$$
while the singular term associated with $h=0$ gives a contribute of $\frac{1}{2}\text{arctanh}\frac{1}{2}$. It follows that the problem of finding a closed form for $I$ is equivalent to finding a closed form for:
$$ S = \sum_{h\geq 1}\text{Li}_2\left(\frac{1}{(h+1/2)^2}\right) =\sum_{n\geq 1}\frac{\zeta\left(2n,\tfrac{3}{2}\right)}{n^2}.\tag{8}$$
Now we may exploit:
$$ \sum_{n\geq 1}\zeta\left(2n,\tfrac{3}{2}\right)x^{2n} = \frac{4x^2}{4x^2-1}+\frac{\pi x\tan(\pi x)}{2}\tag{9} $$
and check that $S$ just depends on
$$\text{PV}\int_{0}^{1}\frac{x\log x}{4x^2-1}\,dx,\qquad \color{blue}{\text{PV}\int_{0}^{1}\log(x)\tan(\pi x)\,dx}.\tag{10}$$
According to Mathematica, the first integral is a combination of $\pi^2,\log^2(2)$ and $\text{Li}_2\left(\tfrac{1}{4}\right)$.
That is not difficult to prove, and reduces the whole problem to finding a closed form for the blue integral. The computation of the blue integral boils down to the computation of
$$ \int_{0}^{1}\text{arctanh}(x)\cot\left(\tfrac{\pi x}{2}\right)\,dx =-\frac{2}{\pi}\int_{0}^{1}\frac{\log\sin\tfrac{\pi x}{2}}{1-x^2}\,dx=-\frac{2}{\pi}\int_{0}^{1}\frac{\log\cos\tfrac{\pi x}{2}}{x(2-x)}\,dx\tag{11}$$
where
$$ \log\cos\tfrac{\pi x}{2}=\sum_{k\geq 1}(-1)^k\frac{1-\cos(\pi k x)}{k} $$
has a nice Fourier series, but $\frac{1}{x(2-x)}$ does not seem to. However, a decent numerical approximation can be recovered from the fact that over the interval $\left(0,1\right)$ we have:
$$ \text{arctanh}(x)\cot\left(\tfrac{\pi x}{2}\right)\approx\frac{2}{\pi}\sqrt{1-x^2}. \tag{12}$$
