What about the definite integral $\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy$? I would like to know if it is possible to justify the calculation of next definite integral
$$\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy.$$

Question.(Corrected, see the comments) My reasonings and expriments with Wolfram Alpha online calculator, suggest me that the following identity holds $$\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy=-\frac{1}{32}\left(16C-\pi^2+4\pi\log 2\right),$$
  where $C$ denotes the Catalan's constant. Am I right? Do you know it or is it possible to justify? Many thanks.

My motivation was to compute an example of a reduction formula for integrals that I've known from a preprint by M.L. Glasser (Universidad de Valladolid). I've deduced the conjeture in the Question from my subsequent calculations for the limits of integration to deduce the closed-form of our doble integral  (to me seem that were difficult calculations and the justification should be tedious, this is why I am asking here) with the mentioned CAS.

As already pointed out in the comments, we have
$$ I \equiv \int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y = - \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \, . $$
Using
$$ \int \limits_0^x \operatorname{artanh} (t) \, \mathrm{d} t = x \operatorname{artanh}(x) + \frac{1}{2} \log (1 - x^2) = \frac{1}{2} \left[x \log \left(\frac{1+x}{1-x}\right) +  \log (1 - x^2) \right]  $$
for $x \in (-1,1)$, we can integrate by parts to get
\begin{align}
I &= - \frac{\pi}{4} \log(2) + \int \limits_0^1 \frac{t \operatorname{artanh}(t) + \frac{1}{2} \log(1-t^2)}{1+t^2} \, \mathrm{d} t \\
&= - \frac{\pi}{4} \log(2) + \frac{1}{2} \int \limits_0^1 \frac{2 t \log(1+t) - t \log(1-t^2) + \log(1-t^2)}{1+t^2} \, \mathrm{d} t \\
&\equiv - \frac{\pi}{4} \log(2) + I_1 + I_2 + I_3 \, .
\end{align}
We obtain
$$ I_1 = \int \limits_0^1 \frac{t \log(1+t)}{1+t^2} \, \mathrm{d} t = \frac{\pi^2}{96} + \frac{\log^2 (2)}{8} $$
from this question and we find
$$ I_2 = \frac{1}{2} \int \limits_0^1 \frac{- t \log(1-t^2)}{1+t^2} \, \mathrm{d} t = \frac{1}{4} \int \limits_0^1 \frac{-  \log(1-s)}{1+s} \, \mathrm{d} s = \frac{\pi^2}{48} - \frac{\log^2 (2)}{8}$$
using the substitution $t^2 = s$ and this question. A well-known representation of Catalan's constant and the change of variables $t = \tan(s)$ yield
\begin{align}
I_3 &= \frac{1}{2} \int \limits_0^1 \frac{\log(1-t^2)}{1+t^2} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^1 \frac{\log(t)}{1+t^2} \, \mathrm{d} t + \frac{1}{2} \int \limits_0^1 \frac{\log \left(\frac{1}{t} - t\right)}{1+t^2} \, \mathrm{d} t \\ &= - \frac{\mathrm{C}}{2} + \frac{1}{2} \int \limits_0^{\pi/4} \log \left(\frac{\cos(s)}{\sin(s)} - \frac{\sin(s)}{\cos(s)}\right) \, \mathrm{d} s \\ &= - \frac{\mathrm{C}}{2} + \frac{1}{2} \int \limits_0^{\pi/4} \log \left(\frac{2 \cos(2s)}{\sin(2 s)}\right) \, \mathrm{d} s \\
&= - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) - \frac{1}{4} \int \limits_0^{\pi/2} \log(\tan(r)) \, \mathrm{d} r \\ 
&= - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) \, ,
\end{align}
where the last integral vanishes by symmetry (use $r \rightarrow \frac{\pi}{2} - r$). 
Putting everything together, we confirm the conjectured result:
\begin{align}
I &= - \frac{\pi}{4} \log(2) + \frac{\pi^2}{96} + \frac{\log^2 (2)}{8} + \frac{\pi^2}{48} - \frac{\log^2 (2)}{8} - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) \\
&= - \frac{\mathrm{C}}{2} - \frac{\pi}{8} \log(2) + \frac{\pi^2}{32} \, .
\end{align}
