I think that there are nice possibilities to get rid of the ##\operatorname{arcsin}.##
I used
\begin{align*}
I&=2\int_{0}^a \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\,d\alpha=2\int_{0}^a\int_0^1 \sqrt{\dfrac{f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}\,dt\,d\alpha \\
&=2\int_{0}^a\int_0^1 \dfrac{1}{\sqrt{\left(f(\alpha)f(-\alpha)\right)^{-1}-t^2}}\,dt\,d\alpha\\
&=2\int_{0}^a\int_0^1\dfrac{1}{\sqrt{\dfrac{32(y+1)}{(y^2+2y-7)(y^2+7)}-t^2}}\,dt\,d\alpha
\end{align*}
where ##a=\operatorname{arcsin}(1/\sqrt{8})\, , \,y^2=9-16 x\, , \,x=\sin^2(\alpha)## and
$$
f(\alpha)f(-\alpha)=\dfrac{1}{32}\dfrac{(y^2+2y-7)(y^2+7)}{y+1}
$$
However, Fubini or not, it leads me directly into the same complicated - now polynomial - integrations as dealt with on MSE.
I have found another funny formula on Wikipedia that at least provide the additional ##\pi## and the logarithm which is apparently part of the deal:
$$
\operatorname{arcsin}R(x) = \int_0^1 \dfrac{1}{\pi \,t}\log\left(\dfrac{t^2+2R(x)t+1}{t^2-2R(x)t+1}\right) \;dt
$$
Looking at the math on MSE shows that there is no easy way to deal with the integrations, even without the trig function.
