Evaluate the double sum $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}$ As a follow up of this nice question I am interested in 
$$
S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m n\left(m^2-n^2\right)^2}
$$
Furthermore, I would be also very grateful for a solution to
$$
S_2=\sum_{m=1}^{\infty}\sum_{n=m+1}^{\infty}\frac{  1}{m n\left(m^2-n^2\right)^2}
$$
Following my answer in the question mentioned above and the numerical experiments of @Vladimir Reshetnikov it's very likely that at least 
$$
S_1+S_2 = \frac{a}{b}\pi^6
$$
I think both sums may be evaluated by using partial fraction decomposition and the integral representation of the Polygamma function but I don't know how exactly and I guess there could be a much more efficient route.

Numerically, I get
$$
S_1+S_2 = 0.14836252987273216621
$$
which agrees with
$$
\frac{\pi^6}{6480}
$$
Also numerically,
$$
S_1 = 0.074181264936366083104
\\
S_2 = 0.074181264936366083104
$$
are seemingly equal.
