Calculating in closed form $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^4(m^2+n^2)}$

How would you tackle this series by real analysis?

$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^4(m^2+n^2)}$$

Since: $$\sum_{n\geq 1}\frac{1}{m^2+n^2}=\frac{-1+m\pi\coth(m\pi)}{2m^2}\tag{1}$$ our sum equals: $$ -\frac{\zeta(6)}{2}+\frac{\pi}{2}\sum_{m\geq 1}\frac{\coth(m\pi)}{m^5}=-\frac{\pi\zeta(5)-\zeta(6)}{2}+\pi\sum_{m\geq 1}\frac{1}{m^5(e^{2m\pi}-1)}\tag{2}$$ and maybe the last series has a nice closed form. Here there is a related paper.

Following Simon Plouffe, $$ \sum_{m\geq 1}\frac{1}{m^5(e^{2m\pi}-1)}=\sum_{m\geq 1}\text{Li}_5(e^{-2\pi m})=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(z)}{(2\pi)^z}\zeta(z+5)\zeta(z)\,dz.\tag{3}$$ If now we set $f(z)=\frac{\Gamma(z)}{(2\pi)^z}\zeta(z+5)\zeta(z)$, we have: $$\begin{eqnarray*} \text{Res}\left(f(z),z=0\right)&=&-\frac{\zeta(5)}{2},\\\text{Res}\left(f(z),z=-1\right)&=&\frac{\pi^5}{540}\\\text{Res}\left(f(z),z=-3\right)&=&-\frac{\pi^5}{540}\\\text{Res}\left(f(z),z=-4\right)&=&\frac{2\pi^4}{3}\zeta'(-4)\\\text{Res}\left(f(z),z=-5\right)&=&-\frac{\pi^5}{1890}\tag{4}\end{eqnarray*}$$ hence a closed form for $(2)$ just depends on a closed form for $\zeta'(-4)$, or, by the reflection formula, on a closed form for: $$\zeta'(5)=-\sum_{n\geq 1}\frac{\log n}{n^5}.\tag{5}$$