Expanding the inverse tangent in logarithms, writing $\frac{x}{1+x^2}=\Re\frac1{x-i}$, and expanding $\log(1-x^2)=\log(1-x)+\log(1+x)$, each of the resulting four indefinite integrals has a closed form. Each term is amenable to automatic integration, (an example), which means that after taking limits, slogging through simplifications and special values, such as those found here, the closed form can be computed.
For example, for the term above,
$$ \int_0^1\frac{\log(1-ix)\log(1-x)}{x-i}\,dx =
-\frac{K\pi }{4}-\frac{17 i \pi ^3}{384}-\frac{1}{2} i K \log2+\frac{13}{192} \pi ^2 \log2+\frac{3}{32} i \pi (\log2)^2-\frac{(\log2)^3}{48}+3 \,\text{Li}_3({\textstyle\frac{1+i}{2}})-\frac{45 \zeta(3)}{32}.
$$
Now, the integrand of the integral in question is the real part of the sum
$$ \frac i2 \frac{\log(1-ix)\log(1-x)}{x-i} - \frac i2\frac{\log(1+i x)\log(1-x)}{x-i}+\frac i2\frac{\log(1-ix)\log(1+x)}{x-i}-\frac i2\frac{\log(1+ix)\log(1+x)}{x-i},
$$
where each term has a closed form for its integral, as above, in terms of $\pi$, $K$, $\log 2$ and $\text{Li}_3$.
After sufficient simplification, the integral of that sum is
$$\begin{aligned} &\int_0^1 \frac{\arctan x\log(1-x^2)}{x-i}\,dx = \\
&-\frac{1}{4} i K\pi -\frac{\pi ^3}{48}+\frac{1}{32} i \pi ^2 \log2-\frac{1}{8} \pi (\log2)^2+K \log2+\frac{7}{32} i \zeta(3), \end{aligned}$$
of which the real part gives the answer
$$ -\frac{\pi ^3}{48}-\frac{1}{8} \pi (\log2)^2+ K \log2$$
We have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=\sum_{k\geq1}\frac{1}{k^{2}}\int_{0}^{\infty}e^{-\pi kx}\arctan\left(x\right)dx=\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\int_{0}^{\infty}\frac{e^{-\pi kx}}{1+x^{2}}dx
$$ and this is the Laplace transform of $\frac{1}{1+x^{2}}
$ at $s=\pi k
$. This can be calculated (see for example here for $s=1
$) $$\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\left(\textrm{Ci}\left(\pi k\right)\sin\left(\pi k\right)+\frac{\pi\cos\left(\pi k\right)}{2}-\textrm{Si}\left(\pi k\right)\cos\left(\pi k\right)\right)=$$ $$=-\frac{3}{8}\zeta\left(3\right)-\frac{1}{\pi}\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right)
$$ where $\textrm{Ci}\left(x\right)
$ and $\textrm{Si}\left(x\right)
$ are the cosine and the sine integral. Now note, using the power series of $\textrm{Si}\left(x\right)
$ $$\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right)=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\left(\pi k\right)^{2n-1}}{\left(2n-1\right)\left(2n-1\right)!}=$$ $$=\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\pi^{2n-1}\left(2^{2n-3}-1\right)\zeta\left(4-2n\right)}{\left(2n-1\right)\left(2n-1\right)!}
=\pi\left(2^{-1}-1\right)\zeta\left(2\right)-\frac{\pi^{3}\zeta\left(0\right)}{18}=$$ $$=-\frac{\pi^{3}}{18}
$$ because $\zeta\left(-2n\right)=0,\,\forall n\geq1
$. So we have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=-\frac{3}{8}\zeta\left(3\right)+\frac{\pi^{2}}{18}
$$ as conjectured by dbanet.
Best Answer
Integrating by parts twice,
$$ \begin{align} \int_{0}^{1} (\arctan x)^{2} \ dx &= x (\arctan x)^{2} \Big|^{1}_{0} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \arctan(x) \ln(1+x^{2}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 + \int_{0}^{1} \frac{\ln(1+x^{2})}{1+x^{2}} \ dx \end{align}$$
Let $x = \tan t $.
Then
$$\begin{align}\int_{0}^{1} (\arctan x)^{2} \ dx &=\frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \int_{0}^{\pi /4} \ln (\cos t) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 -2 \int_{0}^{\pi /4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nt) - \ln 2 \right) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_{0}^{\pi /4} \cos (2nt) \ dt + \frac{\pi}{2} \ln 2 \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2} \right)}{n^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - C \end{align}$$
Fourier series of Log sine and Log cos