Evaluation of a tough double integral This is an integral coming from personal research, and very important to me, but it does not
seem
an easy job to do. If a solution is not possible then I'd be glad with a closed form only.
$$\int_{[0,1]^2} \frac{(1-x-y+x y+x \log(x)-x y\log(x)+y \log(y)- x y\log(y)+x y\log(x)\log(y))\log(1+x y)}{x y (1-x) (1-y)\log(x)\log(y)} \ dx \ dy$$
Hopefully this will be seen by Cleo too and maybe we'll find its closed form.
 A: Write the integral as
$$ \int_{[0,1]^2}h(x)h(y)\log(1+xy)\,dx\,dy =
\sum_{n\geq1}(-1)^{n+1}\frac{a_n^2}{n}, \qquad h(x) = \frac{1-x+x\log
  x}{x(1-x)\log x}, $$
where the coefficients $a_n$ are given by
$$ a_n = \int_0^1 h(x)x^n\,dx = \gamma - H_n + \log n. $$
Using the sums
$$ \sum_{n\geq1} \frac{(-1)^{n+1}}{n} = \log2 $$
$$ \sum_{n\geq1} \frac{H_n(-1)^{n+1}}{n} =
\tfrac12\big(\zeta(2)-\log^22\big), $$
$$ \sum_{n\geq1}\frac{H_n^2(-1)^{n+1}}{n} =
-\tfrac12\zeta(2)\log2+\tfrac13\log^22+\tfrac34\zeta(3), $$
$$ \sum_{n\geq1}\frac{(-1)^{n+1}\log n}{n} =
-\gamma\log2+\tfrac12\log^22, $$
$$ \sum\frac{(-1)^{n+1}\log^2n}{n} =
-\gamma\log^22+\tfrac13\log^32-2\gamma_1\log2, $$
where $\gamma_1$ is a Stieltjes gamma constant, and noting that the
sum
$$ \sum_{n\geq 1}(-1)^{n+1}\frac{H_n\log n}{n} = \int_0^1
\frac{du}{u}\big(\lambda_1(u-1)-\lambda_1(-1)\big), \qquad \lambda_s(t) =
\frac{\partial \mathrm{Li}_s}{\partial s}(t)$$
has no closed form at all, the integral can be written as
$$
-\gamma\zeta(2)-\gamma^2\log2-\tfrac12\zeta(2)\log2+\gamma\log^22+\tfrac23\log^32-2\gamma_1\log2\\+\tfrac34\zeta(3)
- 2\sum_{n\geq1}\frac{(-1)^{n+1}H_n\log n}{n}. $$
A: I can simplify the integrand into;
\begin{equation}
\frac{(y\log y + (1-y))(x \log x +(1-x)}{(1-x)(1-y)}
\end{equation}
The computation of 
\begin{equation}
\int_{0}^{1} \int_{0}^{1} \frac{(y\log y + (1-y))(x \log x +(1-x))}{(1-x)(1-y)}
 dx dy
\end{equation}
is proving to be tougher than expected!
