How to calculate $\int \frac{1}{1+\exp(-r\sin(tx))}\,dx$ How would one calculate the indefinite integral: 
$$\int \frac{1}{1+\exp(-r\sin(tx))}\,dx,$$
where $r >0,t>0$ are some constants? Should I use series expansion? Definite integral solution would also be of assistance.
UPDATE:
This problem originates from my application of generalized linear models where my link function is a composition of a sigmoid and a sine function. I'm trying to formulate the Akaike information criterion for sinusoidal binary classifier.
 A: I trust Wolfram alpha in that the indefinite integral is hopeless. However, if you happen to encounter this integrand as a definite integral over a symmetric interval, then things do work out  nicely and you may compute as follows
$$
I=\int_{-a}^a\frac{1}{1+\exp\left(-r\sin(tx)\right)}\mathrm dx=
\int_{-a}^a\frac{1}{1+\exp\left(r\sin(tx)\right)}\mathrm dx
$$
By a standard symmetry argument. Then, 
$$
2I=\int_{-a}^a\frac{1}{1+\exp\left(-r\sin(tx)\right)}+
\frac{1}{1+\exp\left(r\sin(tx)\right)}\mathrm dx\\
=\int_{-a}^a\frac{2+\exp\left(-r\sin(tx)\right)+\exp\left(r\sin(tx)\right)}{2+\exp\left(-r\sin(tx)\right)+\exp\left(r\sin(tx)\right)}\mathrm dx=2a
$$
and $I=a$.
Hope this helped!
A: Note that:
$$\sum _{n=0}^{\infty } (-\exp (-x))^n=\frac{1}{1+e^{-x}}$$
and:
$$\sum _{j=0}^{\infty } \frac{x^j}{j!}=\exp (x)$$
so,
$\color{red}{\int \frac{1}{1+\exp (-r \sin (t x))} \, dx}=\\
=\int \left(\sum _{n=0}^{\infty } (-1)^n e^{-n r \sin (t x)}\right) \, dx\\=
\int \left(\sum _{n=0}^{\infty }
   \left(\sum _{j=0}^{\infty } \frac{(-1)^j n^j r^j \sin ^j(t x)}{j!}\right)\right) \, dx\\
=\sum _{n=0}^{\infty } \left(\sum _{j=0}^{\infty } \int
   \frac{(-1)^j n^j r^j \sin ^j(t x)}{j!} \, dx\right)\\
=\sum _{j=0}^{\infty } \left(\sum _{n=0}^{\infty } \frac{(-1)^{1+j+n} n^j r^j \cos (t x) \,
   _2F_1\left(\frac{1}{2},\frac{1-j}{2};\frac{3}{2};\cos ^2(t x)\right) \sin ^{1+j}(t x) \sin ^2(t x)^{-\frac{1}{2}-\frac{j}{2}}}{t j!}\right)\\
=\color{red}{\sum
   _{j=0}^{\infty } \frac{(-1)^{1+j} r^j \cos (t x) \, _2F_1\left(\frac{1}{2},\frac{1-j}{2};\frac{3}{2};\cos ^2(t x)\right) \sin ^{1+j}(t x) \sin ^2(t
   x)^{\frac{1}{2} (-1-j)} \left(2^j \zeta (-j,0)+\left(-1+2^j\right) \zeta (-j)\right)}{t j!}+C}$
where:
$\, _2F_1\left(\frac{1}{2},\frac{1-j}{2};\frac{3}{2};\cos ^2(t x)\right)$ is Hypergeometric2F1
$ \zeta (-j,0)$,$\zeta (-j)$ is HurwitzZeta and Zeta.
