How to find the indefinite integral of sin(sin(x))dx? (I got this function by mistake, when I miswrote other function. Now I'm curious how to find the antiderivative of what I miswrote)

I have no a clue how to calculate it and neither does Wolfram Alpha or any other site that I tried. Trig formulas from school course don't seem to be useful too.
From this answer https://math.stackexchange.com/a/877417/65203 we know the Fourier series development

$$\sin(\sin x)=\sum_{k=0}^\infty J_{2k+1}(1)\sin((2k+1)x).$$

Then by term-wise integration

$$\int\sin(\sin x)\,dx=\sum_{k=0}^\infty\frac{J_{2k+1}(1)}{2k+1}\cos((2k+1)x)+C.$$
The coefficients are quickly decaying

$$0.440051,\\0.00652112,\\0.0000499515,\\2.14618×10^{-7},\\5.8325×10^{-10}, \\1.0891×10^{-12},\\\cdots$$
$\int\sin\sin x~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$

$=-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}x}{(2n+1)!}d(\cos x)$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2x)^n}{(2n+1)!}d(\cos x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}x}{(2n+1)!}d(\cos x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}x}{(2n+1)!k!(n-k)!}d(\cos x)$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$