I am trying to compute an integral $$\int_{-\infty}^{\infty} \frac{1}{ce^{kx} + \frac{1}{c}e^{-kx}}N(x; \mu, \sigma^2) \, dx,$$ where $N(x; \mu, \sigma^2)$ is a normal pdf w.r.t. $x$ with mean $\mu$ and variance $\sigma^2$.
Just by the looks of the function $\frac{1}{ce^{kx} + \frac{1}{c}e^{-kx}}$ it seems that this integral should converge -- for $x$ going to plus and minus infinity this fraction is 0, so "the area under the curve" is finite. Wolfram Alpha ran out of time and did not give an answer.
Dropping some of the prefactors, I get that $$\int_{-\infty}^{\infty} \frac{1}{ce^{kx} + \frac{1}{c}e^{-kx}}N(x; \mu, \sigma^2) \, dx \sim \int _{-\infty}^{\infty} \frac{e^{\frac{-x^2+x(2\mu - 2k\sigma^2)}{2\sigma^2}}}{1 + \frac{1}{c^2}e^{-2kx}} \, dx,$$ but I do not see where to go from here. I know that for specific values of $x$, $k$, $\mu$, and $\sigma$ it is possible to find the answer numerically, but I would like to have an expression for it.
I have also tried to look for approximations (e.g. I know that for logistic-normal integrals there exist approximations), but I am not sure whether there is a special name for functions of the form $\frac{1}{ce^{kx} + \frac{1}{c}e^{-kx}}$.
Is there a way to solve this integral or write an approximation in terms of $x$, $k$, $\mu$, and $\sigma$?
An approximation for your integral can be obtained by applying the saddle point method to $$\begin{align} \left\langle f(x) \right\rangle& := \frac{1}{\sqrt{2\pi}\, \sigma} \int\limits_{-\infty}^\infty\! dx \, e^{-\frac{(x-\mu)^2}{2 \sigma^2}}f(x)\\ &= \left\langle\sum\limits_{n=0}^\infty \frac{f^{(n)}(\mu)}{n!}(x-\mu)^n \right\rangle\\ &=\sum\limits_{k=0}^\infty \frac{f^{(2k)}(\mu)}{(2k)!} \left\langle (x-\mu)^{2k} \right\rangle \\ &= f(\mu) + \sum\limits_{k=1}^\infty \frac{f^{(2k)}(\mu)}{(2k)!} (2k-1)!! \, \sigma^{2k} \\ &= f(\mu) +f^{\prime \prime}(\mu) \sigma^2 /2 \, +f^{IV}(\mu) \sigma^4/8 + \ldots,\end{align}$$ giving $$\frac{1}{\sqrt{2\pi} \, \sigma} \int\limits_{-\infty}^\infty \! dx \, e^{-\frac{(x-\mu)^2}{2 \sigma^2}} \frac{1}{c e^{kx}+e^{-kx}/c} \simeq \frac{1}{ce^{k \mu} + e^{-k \mu} /c}+ \ldots$$ as the leading term in your case.