Integration of $f(x)=\exp(-ax-b\sin^{-1}(cx))$

Question

Is there an analytical solution for the following integral?

$$\int \exp(-ax-b\sin^{-1}(cx)) dx$$

Mathematica was unable to solve it, and I tried myself by integration by parts but without success.

Answer 1

Assume $a,b,c\neq0$ for the key case:

Let $u=\sin^{-1}(cx)$ ,

Then $x=\dfrac{\sin u}{c}$

$\therefore\int e^{-ax-b\sin^{-1}(cx)}~dx$

$=\int e^{-\frac{a\sin u}{c}}e^{-bu}~d\left(\dfrac{\sin u}{c}\right)$

$=-\dfrac{1}{a}\int e^{-bu}~d\left(e^{-\frac{a\sin u}{c}}\right)$

$=-\dfrac{e^{-\frac{a\sin u}{c}-bu}}{a}+\dfrac{1}{a}\int e^{-\frac{a\sin u}{c}}~d\left(e^{-bu}\right)$

$=-\dfrac{e^{-\frac{a\sin u}{c}-bu}}{a}-\dfrac{b}{a}\int e^{-\frac{a\sin u}{c}}e^{-bu}~du$

$=-\dfrac{e^{-\frac{a\sin u}{c}-bu}}{a}-\dfrac{b}{a}\int\left(e^{-bu}+\sum\limits_{n=1}^\infty\dfrac{a^{2n}e^{-bu}\sin^{2n}u}{c^{2n}(2n)!}\right)~du+\dfrac{b}{a}\int\sum\limits_{n=0}^\infty\dfrac{a^{2n+1}e^{-bu}\sin^{2n+1}u}{c^{2n+1}(2n+1)!}~du$

$=-\dfrac{e^{-\frac{a\sin u}{c}-bu}}{a}-\int\sum\limits_{n=0}^\infty\dfrac{C_n^{2n}a^{2n-1}be^{-bu}}{4^nc^{2n}(2n)!}~du-\int\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kC_{n+k}^{2n}a^{2n-1}be^{-bu}\cos2ku}{2^{2n-1}c^{2n}(2n)!}~du+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kC_{n+k+1}^{2n+1}a^{2n}be^{-bu}\sin((2k+1)u)}{4^nc^{2n+1}(2n+1)!}~du$ (according to https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae)

$=-\dfrac{e^{-\frac{a\sin u}{c}-bu}}{a}+\sum\limits_{n=0}^\infty\dfrac{a^{2n-1}e^{-bu}}{4^nc^{2n}(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^ka^{2n-1}be^{-bu}(2k\sin2ku-b\cos2ku)}{2^{2n-1}c^{2n}(n+k)!(n-k)!(b^2+4k^2)}~du-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ka^{2n}be^{-bu}(b\sin((2k+1)u)+(2k+1)\cos((2k+1)u)}{4^nc^{2n+1}(n+k+1)!(n-k)!(b^2+(2k+1)^2)}+C$ (according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_involving_exponential_and_trigonometric_functions)

$=\dfrac{e^{-b\sin^{-1}(cx)}}{a}I_0\left(\dfrac{a}{c}\right)-\dfrac{e^{-ax-b\sin^{-1}(cx)}}{a}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^ka^{2n-1}be^{-bu}(2k\sin(2k\sin^{-1}(cx))-b\cos(2k\sin^{-1}(cx)))}{2^{2n-1}c^{2n}(n+k)!(n-k)!(b^2+4k^2)}~du-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ka^{2n}be^{-bu}(b\sin((2k+1)\sin^{-1}(cx))+(2k+1)\cos((2k+1)\sin^{-1}(cx))}{4^nc^{2n+1}(n+k+1)!(n-k)!(b^2+(2k+1)^2)}+C$