Integral with residue method $$  I= \int_{-\infty}^{\infty}{e^{ax}\over(1+e^x)\cosh x}dx $$
with $0<a<1$. 
The poles (simple I think) are in : $z=i(\pi+2k\pi)$ and $z=i(\pi/2+k\pi)$. I tried  to integrate on $\Gamma =\gamma_1+\gamma_2$, where the integral on $\gamma_1 $ is $  \int_{-R}^{R}{e^{ax}\over(1+e^x)\cosh x}dx $, and $\gamma_2$ is the arc with radius $R$ on the positive half plane. The second integral goes to zero when $R$ goes to infinity and the first integral is then equal to the sum of all residues in the positive imaginary half plane. But I'm not able to calculate those residues, so I guess this is not the right method. 
 A: Tl;dr:
Via contour integration we can show
$$\int_{\mathbb R} \frac{e^{ax}}{(1+e^x)\cosh x} = \frac{\pi  \sec ^2\left(\frac{\pi  a}{4}\right)}{2
   \tan \left(\frac{\pi  a}{4}\right)+2}$$
(Most of this answer is just showing integrals along certain arcs vanish for the OP's sake)

First consider the integrand
$$f(x) = \frac{e^{ax}}{(1+e^x)\cosh(x)}$$
and note $f$ is almost periodic; indeed, we have
$$f(x+2\pi i) = \frac{e^{ax+2a\pi i}}{(1+e^{x+2 \pi i})\cosh(x+2\pi i)} = e^{2a\pi i}f(x)$$
So we should be considering a rectangle contour $\Gamma = \gamma_1 + \gamma_2 + \gamma_3 + \gamma_4$ where:


*

*$\gamma_1$ is a line segment from $-\rho$ to $\rho$

*$\gamma_2$ is a line segment from $\rho$ to $\rho+2\pi i$

*$\gamma_3$ is a line segment from $\rho+2\pi i$ to $-\rho+2\pi i$

*$\gamma_4$ is a line segment from $-\rho + 2\pi i$ to $-\rho$
Now note that we have
$$
\begin{align}
\left |\int_{\gamma_2} \frac{e^{az}}{(1+e^z)\cosh(z)} dz\right| &= \left |\int_{0}^{2\pi} \frac{e^{a(\rho + it)}}{(1+e^{\rho + it})\cosh(\rho + it)} i\cdot dt\right|
\\& \le \int_{0}^{2\pi} \left |\frac{e^{a\rho}}{(1+e^{\rho + it})\cosh(\rho + it)} \right|dt \tag{1}
\end{align}$$
Now note
$$\begin{align}|(1+e^{\rho + it})\cosh(\rho + it)| &\ge (|e^{\rho + it}|-1)|\cosh(\rho + it)|
\\&= (e^\rho-1)\sqrt{\sin^2(t)\sinh^2(\rho)+\cos^2(t)\cosh^2(\rho)}
\\&= (e^\rho-1)\sqrt{\tfrac{1}{2}(\cos(2t) + \cosh(2\rho))}
\\&\le (e^\rho-1)\sqrt{\tfrac{1}{2}(\cosh(2\rho))}
\end{align}$$
So, continuing from $(1)$, we have that
$$
\begin{align}
\int_{0}^{2\pi} \left |\frac{e^{a\rho}}{(1+e^{\rho + it})\cosh(\rho + it)} \right|dt &\le 2\pi \frac{e^{a\rho}}{(e^\rho-1)\sqrt{\frac{1}{2}(\cosh(2\rho))}} 
\end{align}$$
Which is quickly seen to vanish as $\rho \to \infty$ for all $a \in (0,1)$.
We thus conclude the integral along $\gamma_2$ vanishes; by a nearly identical argument we can show the integral along $\gamma_4$ vanishes. Since the integral along $\gamma_1$ is what we want, all the remains is $\gamma_3$; however, our "periodicity" observation earlier yield
$$\int_{\gamma_3}f(z) = -e^{2 a \pi i } \int _{\gamma_1} f(z)$$
(the negative sign comes from us changing the direction of integration).
Putting this all together, as $\rho \to \infty$ we have that
$$\int_{\Gamma} f(z) = (1-e^{2 a \pi i })\int_{\gamma_1}f(z)$$
Or, rearranging and noting $\gamma_1 = \mathbb R$, that
$$\int_{\mathbb R} f(x) = \tfrac{1}{1-e^{2 a \pi i }}\int_{\Gamma} f(z)$$
All that remains is to calculate $\int_{\Gamma} f(z)$ via the residue theorem; it is quick to see the only pole from the exponential in our contour is at $z = \pi i$ with residue $e^{i a \pi}$ while the hyperbolic cosine yields poles at $\tfrac{i \pi}{2}$ and at $\tfrac{3 i \pi}{2}$ with residues $\frac{-1-i}{2}e^{\frac{i
   \pi  a}{2}}$ and $\frac{-1+i}{2}e^{\frac{3i
   \pi  a}{2}}$ respectively so our final answer is 
$$\int_{\mathbb R} f(x) = \frac{2 \pi i (e^{i a \pi} +\frac{-1-i}{2}e^{\frac{i
   \pi  a}{2}} + \frac{-1+i}{2}e^{\frac{3i
   \pi  a}{2}})}{1-e^{2 a \pi i }} = \frac{\pi  \sec ^2\left(\frac{\pi  a}{4}\right)}{2
   \tan \left(\frac{\pi  a}{4}\right)+2}$$
