Integrate $\int_0^\infty \frac x{ \sec x\cosh x \>+\>1}dx$ I am interested in whether it is possible to evaluate the integral
$$\int_0^\infty \frac x{ \sec x\cosh x +1}dx$$
For reference, the analogous integral below is manageable
$$\int_0^\infty \frac 1{ \sec x\cosh x +1}dx= -\pi \sum_{k=1}^\infty (-1)^k \text{csch}\>k\pi
$$
which can be evaluated with the residues in the upper-half plane given the symmetry. But a similar approach for the integral in question is not applicable due to the odd integrand. I would like to know of any other  possibilities.
 A: This is not an answer but, may be, it could give you some ideas.
Suppose that we write
$$\frac{x}{\sec (x) \cosh (x)+1}=\sum_{n=0}^\infty (-1)^n\, x \Big[\cos (x)\, \text{sech}(x)\Big]^{n+1}$$
The antiderivative
$$J_n=\int x \Big[\cos (x)\, \text{sech}(x)\Big]^{n+1}\,dx$$ express as nasty combinations of hypergeometric functions but the definite integrals
$$K_n=\int_0^\infty x \Big[\cos (x)\, \text{sech}(x)\Big]^{n+1}\,dx$$ do not seem to be too bad in terms of polygamma functions with complex arguments.
$$K_0=\frac{1}{16} \left(\psi ^{(1)}\left(\frac{1-i}{4}\right)+\psi
   ^{(1)}\left(\frac{1+i}{4}\right)-\psi
   ^{(1)}\left(\frac{3-i}{4}\right)-\psi
   ^{(1)}\left(\frac{3+i}{4}\right)\right)$$
$$K_1=\frac{\log (2)}{2}+\frac {\Delta_1} {16}$$
$$\Delta_1=-2 \psi ^{(0)}\left(\frac{1-i}{2}\right)-2 \psi
   ^{(0)}\left(\frac{1+i}{2}\right)+2 \psi
   ^{(0)}\left(\frac{2-i}{2}\right)+2 \psi ^{(0)}\left(\frac{2+i}{2}\right)+i
   \left(\psi ^{(1)}\left(\frac{1-i}{2}\right)-\psi
   ^{(1)}\left(\frac{1+i}{2}\right)-\psi
   ^{(1)}\left(\frac{2-i}{2}\right)+\psi
   ^{(1)}\left(\frac{2+i}{2}\right)\right)$$ The next ones become relly too long to be typed but they do not present specific problems.
What I am afraid is that the convergence will be extremely slow.
Edit
Using what @KStarGamer wrote in comment
$$I=\int_0^\infty \frac x{ \sec (x)\cosh (x) +1}\,dx=-2\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} (-1)^{n+k} \frac{n k}{(n^2+k^2)^2}$$
$$I=\frac i 8 \sum_{n=1}^\infty (-1)^{n+1}\left(\psi ^{(1)}\left(\frac{1-i n}{2}\right)-\psi
   ^{(1)}\left(\frac{1+i n}{2}\right)-\psi ^{(1)}\left(-\frac{i
   n}{2}\right)+\psi ^{(1)}\left(\frac{i n}{2}\right)\right)$$
It could be worth to notice that, if
$$a_n=\frac i 8 (-1)^{n+1}\left(\psi ^{(1)}\left(\frac{1-i n}{2}\right)-\psi
   ^{(1)}\left(\frac{1+i n}{2}\right)-\psi ^{(1)}\left(-\frac{i
   n}{2}\right)+\psi ^{(1)}\left(\frac{i n}{2}\right)\right)$$
$$a_n=(-1)^n\sum_{k=0}^\infty \left|\left(1-2^{2 k+2}\right) B_{2 k+2}\right|\, n^{-(2k+3)}$$
