Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$
As stated above, this sum may be evaluated using residue theory. I will state the result: for $f$ sufficiently "well-behaved" (meaning that it vanishes sufficiently fast along the vertical sections of the typical rectangular contour used to derive the following relation):
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \operatorname*{Res}_{z=z_k} [\pi \, \cot{\pi z} \, f(z)]$$
where $z_k$ is a non-(real integral) pole of $f$ in the complex plane.
In this case, $f(z)=1/(z^4+a^4)$ and the poles are at $z = a\,e^{i (2 k-1) \pi/4}$, $k \in \{1,2,3,4\}$. So, evaluation of the sum reduces to summing the residues at these poles:
$$-\sum_k \text{Res}_{z=z_k} \pi \, \cot{\pi z} \, f(z) = -\sum_{k=1}^4 \frac{\pi \cot{(\pi e^{i (2 k-1) \pi/4})}}{4 a^3 e^{i 3 (2 k-1) \pi/4}} $$
Now, you may deduce that
$$\cot{(b\, e^{i t})} = \frac{\sin(2 b \cos{t})}{\cosh(2 b \cos{t})-\cos(2 b \sin{t})}-i \frac{ \sinh(2 b \sin{t})}{\cosh(2 b \cos{t})-\cos(2 b \sin{t})}$$
You may either verify this formula in a program like Wolfram Alpha or Mathematica, or you can derive this by using the cosine and sine addition theorems.
The algebra involved can be potentially tedious and error-prone. I will outline here a few tips to get to the correct result. Rewrite the sum over the residues as (negative sign included):
$$\frac{\pi}{4 a^3} \left [ e^{-i 3 \pi/4} \frac{-\sin{(\sqrt{2} \pi a)} + i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} + \\ e^{-i 9 \pi/4} \frac{\sin{(\sqrt{2} \pi a)} + i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} + \\ e^{-i 15 \pi/4} \frac{\sin{(\sqrt{2} \pi a)} - i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} + \\e^{-i 21 \pi/4} \frac{-\sin{(\sqrt{2} \pi a)} - i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} \right ]$$
It should be plain that the exponentials can be reduced, and you end up with two pairs of complex conjugates if everything is done correctly. At this point I will leave the details to the reader and state the final result:
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh{(\sqrt{2} \pi a)}+\sin{(\sqrt{2} \pi a)}}{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}}$$
BONUS
As a check, you can derive the well-known formula
$$\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$$
by considering the behavior of the above result in the limit as $a \to 0$. Note that
$$\sum_{n=1}^{\infty} \frac{1}{n^4+a^4} = \frac12 \left [ \sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} - \frac{1}{a^4}\right]$$
Now Taylor expand the result far enough to get a nonvanishing result:
$$ \frac{\sinh{(\sqrt{2} \pi a)}+\sin{(\sqrt{2} \pi a)}}{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} \sim \frac{2 \sqrt{2} \pi a \left ( 1+ \frac{\pi^4 a^4}{30} \right )}{2 \pi^2 a^2 \left ( 1+ \frac{\pi^4 a^4}{90} \right )} \sim \frac{\sqrt{2}}{\pi a} \left ( 1+ \frac{\pi^4 a^4}{45} \right) $$
Putting this altogether, one easily obtains the desired sum.
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$\ds{ {\rm J}\pars{a} = \sum_{n = -\infty}^{\infty}{1 \over n^{4} + a^{4}} = -\,{1 \over a^{4}} + 2\sum_{n = 0}^{\infty}{1 \over n^{4} + a^{4}} }$
\begin{align} &{\rm J}\pars{a} + {1 \over a^{4}} = 2\sum_{n = 0}^{\infty}{1 \over 2\ic a^{2}} \pars{{1 \over n^{2} - \ic a^{2}} - {1 \over n^{2} + \ic a^{2}}} = {2 \over a^{2}}\Im\sum_{n = 0}^{\infty}{1 \over n^{2} - \ic a^{2}} \\[3mm]&= {2 \over a^{2}}\Im\sum_{n = 0}^{\infty} {1 \over \pars{n + \tilde{n}}\pars{n - \tilde{n}}} \qquad\mbox{where}\qquad \tilde{n} = \expo{\ic\pi/4}\verts{a} \end{align}
\begin{align} &{\rm J}\pars{a} + {1 \over a^{4}} = {2 \over a^{2}}\Im\bracks{ \Psi\pars{\tilde{n}} - \Psi\pars{-\tilde{n}} \over \tilde{n} - \pars{-\tilde{n}}} = {1 \over \verts{a}^{3}}\Im\braces{\expo{-\ic\pi/4}\bracks{ \Psi\pars{\tilde{n}} - \Psi\pars{-\tilde{n}}}}\tag{1} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function $\bf\mbox{6.3.1}$.
Also, $$ \Psi\pars{\tilde{n}} - \Psi\pars{-\tilde{n}} =\overbrace{\Psi\pars{\tilde{n} + 1} - {1 \over \tilde{n}}} ^{\ds{=\ \Psi\pars{\tilde{n}}}}\ -\ \Psi\pars{-\tilde{n}} =-\,{1 \over \tilde{n}} +\ \overbrace{\bracks{\Psi\pars{\tilde{n} + 1} - \Psi\pars{-\tilde{n}}}} ^{\ds{=\ \pi\cot\pars{\pi\bracks{-\tilde{n}}}}} $$ where we used the Recurrence Formula $\ds{\bf\mbox{6.3.5}}$ and the Euler Reflection Formula $\ds{\bf\mbox{6.3.7}}$. We replace this result in expression $\pars{1}$: \begin{align} {\rm J}\pars{a} + {1 \over a^{4}}& ={1 \over \verts{a}^{3}}\,\Im\braces{\expo{-\pi\ic/4}\bracks{-\,{1 \over \expo{\pi\ic/4}\verts{a}} - \pi\cot\pars{\pi\tilde{n}}}} ={1 \over \verts{a}^{4}} - {\pi \over \verts{a}^{3}} \,\Im\bracks{\expo{-\pi\ic/4}\cot\pars{\pi\tilde{n}}} \end{align}
\begin{align}{\rm J}\pars{a}&=-\,{\pi \over \verts{a}^{3}} \,\Im\bracks{\expo{-\pi\ic/4}\cot\pars{\pi\tilde{n}}} =-\,{\pi \over a^{3}} \,\Im\bracks{\expo{-\pi\ic/4} \cot\pars{{\pi\root{2} \over 2}\, a + {\pi\root{2} \over 2}\, a\ic}} \\[3mm]&=-\,{\root{2}\pi \over 2a^{3}}\,\Im\bracks{\pars{1 - \ic}\, {1 - \mu\nu\ic \over \mu + \nu\ic}} \qquad\mbox{where}\qquad\left\lbrace\begin{array}{rcl} \mu & \equiv & \tan\pars{{\root{2}\pi \over 2}\,a} \\[2mm] \nu & \equiv & \tanh\pars{{\root{2}\pi \over 2}\,a} \end{array}\right. \\[5mm] {\rm J}\pars{a}&={\root{2}\pi \over 2}\, {\mu\pars{1 - \nu^{2}} + \nu\pars{1 + \mu^{2}} \over a^{3}\pars{\mu^{2} + \nu^{2}}} \end{align}
$$\color{#00f}{\begin{array}{|l|}\hline\\ \quad{\rm J}\pars{a} =\sum_{n = -\infty}^{\infty}{1 \over n^{4} + a^{4}} \\[3mm]\quad ={\root{2}\pi \over 2}\, {\tan\pars{\root{2}\pi a/2}\sech^{2}\pars{\root{2}\pi a/2} +\tanh\pars{\root{2}\pi a/2}\sec^{2}\pars{\root{2}\pi a/2} \over a^{3}\bracks{\tan^{2}\pars{\root{2}\pi a/2} + \tanh^{2}\pars{\root{2}\pi a/2}}} \quad \\ \\ \hline \end{array}} $$ We can check, after a "painful manipulation", that the above expression satisfies $\ds{\lim_{a \to 0} \underbrace{{1 \over 2}\bracks{{\rm J}\pars{a} - {1 \over a^{4}}}} _{\ds{=\ \sum_{n = 1}^{\infty}{1 \over n^{4}}}} = {\pi^{4} \over 90}}$.