This indefinite integral has no closed form $$I(x)=\int_0^xe^{\cos(t)}dt$$ The usual expansions of $ I (x) $ are generally obtained by expanding $ e ^ x $ and integrating $ \int \cos ^ \alpha (x) dx $, thus obtaining a series in terms of the hypergeometric function. Since I am not very familiar with the hypergeometric function, and its properties are not very well known, I was wondering if there was an expansion for $ I (x) $ in terms of "more known" functions.
This is my attempt:
Applying the Faà di Bruno's formula for the composite function $f(x)=\exp(\cos(x))$ we can show that $f^{(2n+1)}(0)=0$ and $f^{(2n)}(0)$ are the number of partitions of a $2n$-set into even blocks A005046 therefore we can express $f (x)$ with the following taylor series:
$$\frac{e^{\cos(x)}}{e}=1+\sum_{n=1}^{\infty} (-1)^n\Bigg(\sum_{k=1}^{2n}\sum_{h=0}^{k-1}\frac{(-1)^h(h-k)^{2n}}{2^{k-1}k!}\binom{2k}{h} \Bigg)\frac{x^{2n}}{(2n)!}$$
$$\frac{1}{e}\int_0^xe^{\cos(t)}dt=x+\sum_{n=1}^{\infty} (-1)^n\Bigg(\sum_{k=1}^{2n}\sum_{h=0}^{k-1}\frac{(-1)^h(h-k)^{2n}}{2^{k-1}k!(2n+1)!}\binom{2k}{h} \Bigg)\frac{x^{2n+1}}{(2n)!}$$
