Double summation $\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$ As a follow to this answer I came across the double sum $$\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}.$$
But unfortunately I do not have skills in techniques to handle double summation . 
Help appreciated.
I've made some research in MSE and found several questions which could be helpful: 
1) $\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}=-\frac{17\pi^4}{1440}$
2) $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{\pi^6}{12960}$ 
3) $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}=  \frac{1}{3}\zeta(6)$


The double summation is equal to
  $$\frac{11\zeta(4)}{8}=\frac{11\pi^4}{720}.$$

Note that
$$\frac{m^2+n^2}{m n\left(m^2-n^2\right)^2}=
\frac{1}{2 mn(m+n)^2}+
\frac{1}{2 mn(m-n)^2}.$$
Now consider the Tornheim double sums:
$$T(a,b,c)= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} 
\frac{1}{m^an^b(m+n)^c}.$$
Then
$$ \sum_{m,n=1\, m\neq n}^{\infty} \frac{1}{mn(m+n)^2}=T(1,1,2)-\sum_{m=1}^{\infty}\frac{1}{4m^4}=T(1,1,2)-\frac{\zeta(4)}{4},$$
$$\begin{align} \sum_{m=1}^{\infty} \sum_{n=1}^{m-1} 
\frac{1}{mn(m-n)^2}
&= \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} 
\frac{1}{mn(m-n)^2}\\
&= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} 
\frac{1}{nk^2(n+k)}=T(1,2,1)\end{align}$$
and again
$$ \sum_{m=1}^{\infty} \sum_{n=m+1}^{\infty} 
\frac{1}{mn(m-n)^2}=T(1,2,1).$$
Hence
$$\begin{align}\sum_{m,n=1\, m\neq n}^{\infty}{\frac{m^2+n^2}{mn(m^2-n^2)^2}}
&=\frac{T(1,1,2)-\zeta(4)/4}{2} +T(1,2,1)\\
&=\frac{11\zeta(4)}{8}
\end{align}$$
where we used 
$$T(1,1,2)=\zeta(4)/2\quad,\quad T(1,2,1)=5\zeta(4)/2$$
(see page 31 in The evaluation of Tornheim double sums).
