$\underline{\bf Question}$
$\displaystyle\sum_{n = 1}^ \infty \frac{ \coth(\pi nx) + x \coth \bigg( \dfrac{\pi n}{x} \bigg) }{ {n}^{3} }$
$\underline{\bf My Try}$ By Mittag-Leffler expansion for the hyperbolic cotangent,
${\displaystyle{\coth(\pi z) = -\frac1{\pi z} + \frac{2z^2}\pi \sum_{m=1}^\infty\frac1{z^2+m^2}coth(πz)}}$
where $A = \Bbb Z \setminus \{0\}$ .
By symmetry again, the given sum is
$\displaystyle S = \sum_{n=1}^\infty \frac{\coth(\pi n x) + x \coth\left(\frac{\pi n}x\right)}{n^3} = \frac12 \sum_{n\in A} \frac{\coth(\pi n x) + x \coth\left(\frac{\pi n}x\right)}{n^3}$ Now rewrite the summand using the M-L expansion.
$\begin{gathered}\displaystyle S = \frac12 \sum_{n\in A} \frac1{n^3} \left(-\frac1{\pi nx} + \frac{n^2x^2}\pi \sum_{m\in A} \frac1{n^2x^2+m^2} - \frac x{\pi n} + \frac{n^2}{\pi x^2} \sum_{m\in A} \frac1{\frac{n^2}{x^2} + m^2}\right) \\\\ = -\frac1{2\pi x} \left[(x^3+1)\sum_{n\in A}\frac1{n^4} - x^3 \sum_{(m,n)\in A^2} \frac1n\cdot\frac1{n^2x^2+m^2} - x^2 \sum_{(m,n)\in A^2} \frac1n\cdot\frac1{m^2x^2+n^2}\right]\end{gathered}$ By symmetry,
$\displaystyle \sum_{(m,n)\in A^2} \frac1n \cdot \frac1{m^2x^2+n^2} = \sum_{(m,n)\in A^2} \frac1m\cdot\frac1{n^2x^2+m^2}$
and recalling the definition of the Riemann zeta function, this brings us to
$\displaystyle S = -\frac{\zeta(4)}\pi \cdot \frac{x^3+1}x + \frac1{2\pi x} \sum_{(m,n)\in A^2} \left(\frac{x^3}n + \frac{x^2}m\right) \frac1{n^2x^2+m^2}$
and that's unfortunately as far as I've gotten