Give this one a go if you're so inclined. A clever sub will do it.
$\displaystyle \int_{0}^{\infty}\int_{y}^{\infty}\frac{(x-y)^{2}log(\frac{x+y}{x-y})}{xysinh(x+y)}dxdy$
Here's another way.
An alternative sub is to let $\displaystyle x=\frac{u+t}{2}, \;\ y=\frac{u-t}{2}$
Then, we get:
$\displaystyle I=2\int_{0}^{\infty}\frac{1}{sinh(u)}\int_{0}^{u}\ frac{t^{2}log(\frac{u}{t})}{u^{2}-t^{2}}dtdu$
Then, in the inner integral, let t=uw to get the product of two known integrals.
$\displaystyle I=2\int_{0}^{\infty}\frac{udu}{sinh(u)}\int_{0}^{1 }\left(1-\frac{1}{1-w^{2}}\right)log(w)dw$
And you should get:
$\displaystyle \frac{{\pi}^{2}({\pi}^{2}-8)}{16}$