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
$$ -\frac{\pi ^3}{48}-\frac{1}{8} \pi (\log2)^2+ K \log2$$
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
Best Answer
I finally get a solution (i swear i didn't know it when i have posted the question)
Define for $x\in [0,1]$ the function $F$:
$\displaystyle F(x)=\int_0^x \dfrac{\ln t}{1+t}dt$
Notice that $F(1)=-\dfrac{\pi^2}{12}$
(use Taylor's development)
and, after performing the change of variable $y=\dfrac{t}{x}$,
$\displaystyle F(x)=\int_0^1 \dfrac{x\ln(xy)}{1+xy}dy$
Since that:
$\Big[F(x)\arctan x\Big]_0^1=-\dfrac{\pi^3}{48}$
then,
$\displaystyle -\dfrac{\pi^3}{48}=\int_0^1 \dfrac{F(x)}{1+x^2}dx+\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(xy)}{(1+xy)(1+x^2)}dxdy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(x)}{(1+xy)(1+x^2)}dxdy+\int_0^1\int_0^1 \dfrac{x\ln(y)}{(1+xy)(1+x^2)}dxdy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\left[\dfrac{\ln x\ln(1+xy)}{1+x^2}\right]_{y=0}^{y=1} dx+ \displaystyle \int_0^1 \left[-\dfrac{\ln y\ln(1+xy)}{1+y^2}+\dfrac{\ln y\ln(1+x^2)}{2(1+y^2)}+\dfrac{y\ln y\arctan x}{1+y^2}\right]_{x=0}^{x=1}dy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx= \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx-\int_0^1\dfrac{\ln y\ln(1+y)}{1+y^2}dy+\dfrac{\ln 2}{2}\int_0^1 \dfrac{\ln y}{1+y^2}dy+ \dfrac{\pi}{4}\times \int_0^1 \dfrac{y\ln y}{1+y^2}dy$
Using Taylor's development,
$\displaystyle \int_0^1 \dfrac{y\ln y}{1+y^2}dy=-\dfrac{\pi^2}{48}$
And it's well known that, $\displaystyle -G=\int_0^1\dfrac{\ln y}{1+y^2}dy$
Therefore,
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=-\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{192}$
And finally,
$\displaystyle \int_0^1 \dfrac{\arctan x \ln x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$
(I hope there is no mistake, this proof is too wonderful to be true )
NB:
Added, July 2, 2019.
The above computation is the result of "reverse engineering". I was searching for a way to express $\pi^3$ as in integral. If you introduce the function, for $x\in [0;1]$, \begin{align}\displaystyle F(x)&=\int_0^x \dfrac{\ln t}{1+t}dt\\ &=\int_0^1 \dfrac{x\ln(tx)}{1+tx}dt \end{align} Observe that, \begin{align}\frac{\partial F(x)}{\partial x}&=\dfrac{\ln x}{1+x}\\ F(1)&=-\frac{\pi^2}{12} \end{align}
Then, \begin{align}-\frac{\pi^3}{48}&=\Big[F(x)\arctan x\Big]_0^1\\ \end{align} And, \begin{align}\frac{\partial F(x)}{\partial x}\arctan x=\frac{\arctan x\ln x}{1+x}\end{align}
Thus, one can apply integration by parts, \begin{align}\int_0^1 \frac{\arctan x\ln x}{1+x}\,dx&=\int_0^1 \frac{\partial F(x)}{\partial x}\arctan x\,dx\end{align} and so on,