The general formula for rational $x$ can be derived as
follows. Consider, first, a convergent sum of the form $\sum_k
\frac{p(k)}{q(k)}$, where $p$ and $q$ are polynomials, $\deg q\geq
\deg p+2$, and $q$ only has simple roots. If $\rho$ is a variable that
runs over the roots of $q$, then
$$ \frac1{q(x)} = \sum_{q(\rho)=0}\frac{1}{q'(\rho)}\frac1{x-\rho}, $$
so the sum can be written as
$$ \sum_k \frac{p(k)}{q(k)} =
\sum_{q(\rho)=0} \frac1{q'(\rho)} \sum_k \frac{p(k)\bmod k-\rho}{k-\rho} + \lfloor p(k)/(k-\rho)\rfloor. $$
Now, $p(k)\bmod k-\rho = p(\rho)$ is a constant,
$$ \sum_{k\geq0} \frac{z^k}{k-\rho} = \Phi(z,1,-\rho), $$
by definition of Lerch's transcendent function, and the fact that the original sum is convergent assures us we can ignore the terms $\sum_\rho\frac{\lfloor p(k)/(k-\rho)\rfloor}{q'(\rho)}$, as they must necessarily sum to zero.

Therefore, since $\Phi$ has an asymptotic expansion
$$ \Phi(z,1,-\rho) = -\log(1-z) - \gamma - \psi(-\rho), $$
($\gamma$ is Euler's gamma, and $\psi$ is the digamma function), and
since the fact that the original sum converges assures us that we can
ignore all terms of $\Phi(z,1,-\rho)$ independent of $\rho$ as we take
the limit $z\to1-$ to recover the sum, we therefore have
$$ \sum_{k\geq0} \frac{p(k)}{q(k)} = -\sum_{q(\rho)=0} \frac{p(\rho)}{q'(\rho)}\psi(-\rho). $$

The function $f$ has the form
$$ f(x) = -\frac{\coth\frac\pi2}{\pi} + \frac2\pi\sum_{n\geq0} S(n, \sin(2\pi nx), \cos(2\pi nx)), $$
where $S(n,\gamma,\delta) = \frac{p(n,\gamma)/q(n,\delta)}{\cosh\pi-\delta}$ is given as in sos440's answer by
$$ p(n,\gamma) = (2n^2x^2-1)\sinh\pi+2nx\gamma, \qquad q(n)=(4n^2-1)(4n^4x^4+1). $$

When $x=r/s$ is rational, this can be simplified. Rewrite the sum as
$$ \sum_{n\geq0}S(n,\sin2\pi nx, \cos2\pi nx) = \sum_{0\leq\alpha<s} \frac{1}{\cosh\pi-\cos2\pi\alpha x} \sum_{m\geq0}
\frac{p(ms+\alpha, \sin2\pi\alpha x)}{q(ms+\alpha)}, $$
and now the inner sum over $m$ is a convergent sum of terms rational in $m$, so can be written as
$$ \sum_{q(\rho)=0}\frac{-1}{q'(\rho)} \frac1{s} \sum_{0\leq\alpha<s} \frac{p(\rho, \sin2\pi\alpha
x)}{\cosh\pi-\cos2\pi\alpha x} \psi\left(\frac{\alpha-\rho}{s}\right),
$$
where $\rho$ runs over the roots of $q(n)$: $\{\pm\frac12,
\frac{e^{\pi i j/4}}{x\sqrt{2}}\}$, $j$ odd.

This is pretty much a closed form, it is a finite sum of digamma terms, but
you might not be able to simplify it much, especially by hand. For
example, Mathematica now gives:
$$
f({\textstyle\frac52}) =
\left(
68-68 \sqrt{5}-500 \sqrt{10-2 \sqrt{5}} \cosh\frac{\pi }{5}+272 \cosh\frac{2 \pi }{5}-\left(125 \sqrt{10-2 \sqrt{5}}+125 \sqrt{5 \left(10-2 \sqrt{5}\right)}\right) \cosh\pi-\left(160+160 \sqrt{5}\right) \sinh\frac{\pi }{5}+160 \sinh\frac{3 \pi }{5}-160 \sinh\frac{7 \pi }{5}
\right)/\left(
689(-1+\sqrt{5}-4 \cosh\frac{2 \pi }{5})\sinh\pi\right)
$$