I need to solve an integral
$$\int_a^b \frac{f(s + x)}{f(s+x) + f(s)} f(s) d s, $$
where $f$ is the pdf of a normal distribution with mean $\mu$ and std. deviation $\sigma$. I am aware that the value will be proportional to the conditional expectation
$$ E(\frac{f(s + x)}{f(s+x) + f(s)} | a < s < b) $$
I tried to simplify the fraction term, which (I believe) simplifies to
$$ \frac{f(s + x)}{f(s+x) + f(s)} = \frac{1}{1 + f(s) / f(s+x)} = \frac{1}{1 + exp\{-\frac{1}{2\sigma^2}(2x(s - \mu) - x^2)\}}.$$
I have found that this last term is equal to the cdf of a logistic distribution with mean $\mu' = \mu +x/2$ and std. deviation $\sigma' = \frac{\sigma^2}{x}\frac{\pi}{\sqrt{3}}$. Denote this cdf by $G(s)$.
Therefore, the integral becomes
$$\int_a^b G(s) f(s) d s. $$
I tried to apply integration by parts which brings me to
$$\int_a^b G(s) f(s) d s = [G(s)F(s)]_a^b - \int_{a}^b g(s)f(s) d s. $$
At this point I'm a bit stuck. I would like to get rid of the remaining integral.
One approximate solution might be to approximate the logistic pdf $g$ with an (appropriately chosen) normal pdf $g'$. Then, it should be possible to solve the integral because the product of two normal pdfs is proportional to a normal pdf, which means that one can approximately write the integral to be equal to the difference of two normal cdfs evaluated at b and a, scaled by a factor. But doing this approximation seems very messy, and it's not clear how good it will be.
I would be interested if there are any more straightforward ways to solve the integral above.