Evaluating $\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$ 
How to evaluate this integral?
$$\int_{ - \infty }^\infty  {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$$

Maybe we can start by $$\int_0^\infty  {\frac{{dx}}{{({x^2} + 1)\cosh ax}}}  = \frac{1}{2}\left[ {{\psi _0}\left( {\frac{a}{{2\pi }} + \frac{3}{4}} \right) - {\psi _0}\left( {\frac{a}{{2\pi }} + \frac{1}{4}} \right)} \right]$$ in this. Then take the derivative with respect to $a$, but I'm failed to solve it!

The given integral equals
$$ I=\frac{1}{8}\int_{0}^{+\infty}\frac{2\cosh(7\pi x)}{(1+x^2)\cosh^3(3\pi x)}\,dx = \frac{1}{8}\int_{\mathbb{R}}\frac{\cosh(7\pi x)}{(1+x^2)\cosh^3(3\pi x)}\,dx $$
and by computing the residues of the integrand function at $x=i$ and at $(2k+1)\frac{i}{2}$ we get:
$$ I = \frac{\pi}{8}+\frac{2\pi i}{8}\sum_{k\geq 0}\lim_{z\to\frac{2k+1}{2}i}\frac{d}{dz}\frac{\cosh(7\pi z)(z-(2k+1)i/2)}{(1+z^2)\cosh^3(3\pi z)}+R$$
that simplifies to:
$$ I = \frac{\pi}{8}-\frac{28}{27\pi}\sum_{k\geq 0}\frac{(2k+1)}{ (4k^2+4k-3)^2 }+R = \color{blue}{\frac{\pi}{8}-\frac{7}{27\pi}+R}$$
where $R$ is the contribute given by the poles at $\frac{i}{6},\frac{5i}{6},\frac{7i}{6},\frac{11i}{6}$ and so on.
