Evaluate $\sum\limits_{n=1}^\infty \left(\frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2}\right)$

I am trying to find the pattern for the coefficients of the closed forms for this series:$$\sum_{n=1}^\infty \left( \frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2} \right)$$The series seems to converge for every natural $k$, and it is easy to evaluate each of the two series using this, ending up with a bunch of digamma functions, as shown here. However, when plugging the series in wolfram for some values of $k$ this is what I got:\begin{align}k=1: &&& \frac{-5+2\pi\tanh\left(\frac{\pi}{2}\right)-\pi\coth(\pi)}{4} \\k=2: &&& \frac{-42-10\pi\tanh\left(\frac{3\pi}{2}\right)+15\pi\coth(\pi)}{60} \\k=3: &&& \frac{-341+120\pi\tanh\left(\frac{3\pi}{2}\right)-90\pi\coth(2\pi)}{720} \\k=4: &&& \frac{-3189-884\pi\tanh\left(\frac{5\pi}{2}\right)+1105\pi\coth(2\pi)}{8840} \\k=5: &&& \frac{-58076+19890\pi\tanh\left(\frac{5\pi} {2}\right)-16575\pi\coth(3\pi)}{198900} \\k=6: &&& \frac{-21576583-6277050\pi\tanh\left(\frac{7\pi}{2}\right)+7323225\pi\coth(3\pi)}{87878700}\end{align}The arguments of $\tanh$ and $\coth$ can be guessed, respectively, to $\frac{k+(k+1\mod2)}{2}\pi$ and $\frac{k+(k\mod2)}{2}\pi$, but the other coefficients are less easy to confront. The linear sequence in the numerators ($5,42,341,\dots$) does not appear even in the OEIS, and same goes for the denominators.

However, it is possible to guess the coefficients of $\tanh$ and $\coth$. Hence the question:

How to show that\begin{align}\sum_{n=1}^\infty \left( \frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2} \right)= \\=-\frac{a_k}{b_k}+\frac{\pi}{2k}\tanh\left(\frac{k}{2}\pi\right)-\frac{\pi}{2k+2}\coth\left(\frac{k+1}{2}\pi\right) \text{,} &&& k\ \text{odd}\\=-\frac{a_k}{b_k}-\frac{\pi}{2k-2}\tanh\left(\frac{k+1}{2}\pi\right)+\frac{\pi}{2k}\coth\left(\frac{k}{2}\pi\right) \text{,} &&& k\ \text{even}\\\end{align}and what are $a_k$ and $b_k$?

Appearently the result in terms of digamma functions can be simplified a lot to

$$\sum_{n=1}^\infty \left( \frac{1}{(k+n)^2+n^2}-\frac{1}{(k+1-n)^2+n^2} \right)=$$$$=\frac{i}{2k}\left(\psi\left(1+\bar{z}k\right)-\psi\left(1+zk\right)\right)+\frac{i}{2(k+1)}\left(\psi(z-\bar{z}k)-\psi(\bar{z}-zk)\right) \tag{*}$$where $z=\frac{1+i}{2}$

I feel like it is almost done, but I can't perform the last steps to get rid of the digamma functions.However, I strongly believe it has to do with the following two identities for $u\in \mathbb{C}$:$$\psi(1-u)-\psi(u)=\pi\cot(\pi u) $$$$\psi\left(\frac12+u\right)-\psi\left(\frac12-u\right)=\pi\tan(\pi u)$$

Any ideas on how to finish?