$\int e^{\sin x+2x}dx$ During solving  this differential equation: $$\frac{dy}{dx}+y\cos x=e^{2x}$$
I get this answer: $$y=e^{-\sin x}\left(\int e^{\sin x+2x}dx+c\right)$$
I wonder what is the answer of:$$\int e^{\sin x+2x}dx$$
also I checked answer of the book. It also didn't calculated that integral 

$\int e^{\sin x+2x}~dx$
$=\int\left(e^{2x}+\sum\limits_{n=1}^\infty\dfrac{e^{2x}\sin^{2n}x}{(2n)!}\right)~dx+\sum\limits_{n=0}^\infty\dfrac{e^{2x}\sin^{2n+1}x}{(2n+1)!}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{C_n^{2n}e^{2x}}{4^n(2n)!}~dx+\int\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kC_{n+k}^{2n}e^{2x}\cos2kx}{2^{2n-1}(2n)!}~dx+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_{n+k+1}^{2n+1}e^{2x}\sin((2k+1)x)}{4^n(2n+1)!}~dx$
(according to https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae)
$=\dfrac{I_0(1)e^{2x}}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^ke^{2x}(\cos2kx+k\sin2kx)}{4^n(n+k)!(n-k)!(k^2+1)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{e^{2x}(2\sin((2k+1)x)-(2k+1)\cos((2k+1)x))}{4^n(n+k+1)!(n-k)!((2k+1)^2+4)}+C$
(according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_involving_exponential_and_trigonometric_functions)
