Indefinite integral whose solution involves W Lambert, $\int e^{bcos(ct)-at} dt $

Question

How do I solve the following indefinite integral?

$$\int e^{b\cos(ct)-at} dt $$

where $a,b,c$ are constants. The hint is to consider the Lambert W-function.

Answer 1

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

$\int e^{b\cos ct-at}~dt$

$=\int\left(e^{-at}+\sum\limits_{n=1}^\infty\dfrac{b^{2n}e^{-at}\cos^{2n}ct}{(2n)!}\right)~dt+\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^{-at}\cos^{2n+1}ct}{(2n+1)!}~dt$

$=\int\sum\limits_{n=0}^\infty\dfrac{C_n^{2n}b^{2n}e^{-at}}{4^n(2n)!}~dt+\int\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{C_{n+k}^{2n}b^{2n}e^{-at}\cos2kct}{2^{2n-1}(2n)!}~dt+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_{n+k+1}^{2n+1}b^{2n+1}e^{-at}\cos((2k+1)ct)}{4^n(2n+1)!}~dt$

(according to https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae)

$=-\dfrac{I_0(b)e^{-at}}{a}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{b^{2n}e^{-at}(2kc\sin2kct-a\cos2kct)}{2^{2n-1}(n+k)!(n-k)!(4c^2k^2+a^2)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{b^{2n+1}e^{-at}((2k+1)c\sin((2k+1)ct)-a\cos((2k+1)ct))}{4^n(n+k+1)!(n-k)!(c^2(2k+1)^2+a^2)}+C$

(according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_involving_exponential_and_trigonometric_functions)