How to evaluate the infinite sum of $\frac{1}{ (n^2+a)(n^2+b)(n^2+c)}$ Playing around with Mathematica, I found a rather nice result for this summation
\begin{align}
    \sum_{n=0}^\infty \frac{1}{(n^2 + a)(n^2 +b) (n^2 + c)} 
    &= \frac{1}{2 a b c} 
    - \frac{ \pi     \left[ 
        \sqrt{ab}(a-b) \coth(\pi\sqrt{c})
        +
        \sqrt{bc}(b-c) \coth(\pi \sqrt{a})
        +
        \sqrt{ca}(c-a) \coth( \pi \sqrt{b})
    \right]}{2 \sqrt{a b c} \, (a-b)(b-c)(c-a)}.
\end{align}
Does anybody know how to obtain this result analytically?

Partial fraction decomposition gives
$$\frac1{(n^2+a)(n^2+b)(n^2+c)}=\sum_{\mathrm{cyc}}\frac1{(a-c)(b-c)}\cdot\frac1{n^2+c}$$
Now use the well-known sum
$$\sum_{n=0}^\infty\frac1{n^2+k}=\frac{1+\sqrt k\pi\coth\sqrt k\pi}{2k}$$
which should lead you to the final result.
