Closed form for this integral $I=\int_0^{1}\frac{{\arcsin}({x^2})}{\sqrt{1-x^2}}dx$ I’m trying to find a closed form for this integral.Any help is appreciated.Thanks
 $$I=\int_0^{1}\frac{{ \arcsin}({x^2})}{\sqrt{1-x^2}}dx$$
 A: As already said in comments, it does not seem that the antiderivative of the integrand exists.
Concerning the integral, a CAS found something you will not like very much $$I=\frac{\pi }{4}  \,
   _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{5}{4};1,\frac{3}{2},\frac{3}{
   2};1\right)$$ where appears the generalized hypergeometric function.
Numerically, $$I\approx 0.9552018064811796875605004$$
A: To make it more clear why this integral is so problematic, we can represent it as a double integral, using the definition of $\arcsin$:
$$\arcsin z=z \int_0^1 \frac{dy}{\sqrt{1-z^2y^2}}$$
Then the integral in the OP will have the form:
$$I=\int_0^{1}\frac{{ \arcsin}({x^2})}{\sqrt{1-x^2}}dx=\int_0^1 \int_0^1 \frac{x^2 dx dy}{\sqrt{1-x^2}\sqrt{1-y^2x^4}}=$$
$$=\int_0^1 \int_0^1 \frac{x^2 dx dy}{\sqrt{1-x^2}\sqrt{1-yx^2}\sqrt{1+yx^2}}=$$
$$=\int_0^1 \int_0^1 \frac{dx dy}{\sqrt{1-x^2}\sqrt{1-y^2x^4}}-\int_0^1 \int_0^1 \frac{\sqrt{1-x^2}dx dy}{\sqrt{1-y^2x^4}}$$
If we had $\arcsin x$ instead of $\arcsin x^2$, then we would at least be able to use complete elliptic integrals of the first and second kind.
But with $\arcsin x^2$ we don't get any nice special functions, thus, apparently, the generalized Hypergeometric function from Claude Leibovici's answer is the only way to go.
