Deriving Maclaurin series for $\frac{\arcsin x}{\sqrt{1-x^2}}$. Intrigued by this brilliant answer from Ron Gordon, I was attempting to find the Maclaurin series for 
$$f(x)=\frac{\arcsin x}{\sqrt{1-x^2}}=g(x)G(x)$$
with $g(x)=\frac{1}{\sqrt{1-x^2}}$ and $G(x)$ its primitive. So I attempted to multipy series, which yielded this:
$$f(x)=\sum_{n=0}^{\infty}x^{2n+1} (-1)^n\sum_{k=1}^{n}\frac{1}{k+1} { -\frac{1}{2}\choose n-k}{ -\frac{1}{2}\choose k},$$
which I'm unable to simplify further. How to proceed? Or is this approach doomed?

Note that
$$\int_0^1\frac{dt}{1-x^2+x^2t^2}=\frac{1}{x\sqrt{1-x^2}}\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)=\frac{\arcsin(x)}{x\sqrt{1-x^2}}$$
so we can write
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\left(\int_0^1(1-t^2)^n\,dt\right)x^{2n+1}.$$
But
$$\int_0^1(1-t^2)^n\,dt=\int_0^1\sum_{k=0}^n(-1)^k\binom{n}{k}t^{2k}\,dt=\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}.$$
Hence,
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!}x^{2n+1}.$$
Also see here for proof of $\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}$.
