For all integer $m\ge 0$ one would use a partial fraction decomposition of the rational polynomial
to simplify the kernel.
$$
r_m \equiv \frac{-i(x-i)^{2m}[1+5(i-x)^2]+i(x+i)^{2m}[1+5(i+x)^2]}{ 2(1+x^2)^{2m}[(1-a)^2+2a(1+a)x^2+a^2x^4]}
$$
and (for short) $p_a(x)\equiv 1-2a+a^2+2a(1+a)x^2+a^2x^4$
like
$$
r_0=-\frac{10x}{p_a(x)},
$$
$$
r_1=-\frac{38x}{(1+x^2)^2(4a-1)}+\frac{4x(a-5)}{(1+x^2)(4a-1)^2}-\frac{2xa(2a^2x^2-10ax^2+13a-20-74a^2)}{p_a(x)(4a-1)^2};
$$
$$
r_2=\frac{2x(-156a^2+36a+15+304a^3)}{(1+x^2)^2(4a-1)^3}
-\frac{4xa(16a^3-80a^3+3a-15)}{(1+x^2)(4a-1)^4}
-\frac{4x(80a-39)}{(1+x^2)^3(4a-1)^2}
+\frac{152x}{(1+x^2)^4(4a-1)}
+\frac{2xa^2(-30ax^2-160x^2a^3+\cdots)}{p_a(x)(4a-1)^4}.
$$
This seems to reduce the problem to sums of integrals of the
form
$$
I_j\equiv \int_0^\infty \frac{x}{(1+x^2)^j}[\coth(\pi x)-1]
= 2\int_0^\infty dx \frac{x}{(1+x^2)^j(e^{2\pi x}-1)}
$$
and
$$
J_s(a)\equiv \int_0^\infty \frac{x^s}{p_a(x)}[\coth(\pi x)-1],\quad 0\le s\le 3
.
$$
$$
\int_0^\infty dx\frac{x}{e^{\mu x}-1} = \frac{\pi^2}{6\mu^2}.
$$
By repeated differentiation of the Gradsteyn-Rzyhik formula 3.415.1 w.r.t. $\beta$
all $I_j$ have closed forms in terms of derivatives of the digamma-function:
$$
\int_0^\infty dx\frac{x}{(x^2+\beta^2)(e^{\mu x}-1)} = \frac12[\ln \frac{\beta \mu}{2\pi}
-\frac{\pi}{\beta \mu}-\psi(\frac{\beta \mu}{2\pi})]
.
$$
$$
\int_0^\infty dx \frac{x}{(x^2+\beta^2)^2 (e^{\mu x}-1)}
=
-\frac{1}{4\beta^2}
-\frac{\pi}{4\beta^3\mu}
+\frac{\mu}{8\beta\pi}\psi'\left(\frac{\beta\mu}{2\pi}\right)
.
$$
$$
\int_0^\infty dx \frac{x}{(x^2+\beta^2)^3 (e^{\mu x}-1)}
=
-\frac{1}{8\beta^4}
-\frac{3\pi}{16\beta^5\mu}
+\frac{\mu}{32\beta^3\pi}\psi'\left(\frac{\beta\mu}{2\pi}\right)
-\frac{\mu^2}{64\beta^2\pi^2}\psi''\left(\frac{\beta\mu}{2\pi}\right)
.
$$
$$
\int_0^\infty dx \frac{x}{(x^2+\beta^2)^4 (e^{\mu x}-1)}
=
-\frac{1}{12\beta^6}
-\frac{5\pi}{32\beta^7\mu}
+\frac{\mu}{64\beta^5\pi}\psi'\left(\frac{\beta\mu}{2\pi}\right)
-\frac{\mu^2}{128\beta^4\pi^2}\psi''\left(\frac{\beta\mu}{2\pi}\right)
+\frac{\mu^3}{768\beta^3\pi^3}\psi'''\left(\frac{\beta\mu}{2\pi}\right)
.
$$
See https://arxiv.org/abs/2308.14154 for recurrences of these terms for higher exponents than 4 in the denominator of the integral.
Here we are only using $\mu=2\pi$, $\beta=1$, such that
$\psi(1)=-\gamma$, $\psi'(1)=\pi^2/5$, $\psi''(1) = -2\zeta(3)$, $\psi'''(1)=\pi^4/15$ etc.
To use an equivalent scheme for the $J_s(a)$ one would need to split the biquadratic $p_a(x)$ into
factors. I have not looked further into that more complicated part of the problem.
