Integral $\int_0^1 \frac{\sqrt{1-x^4}}{1+x^2}dx$=? As title, how could one calculate $$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^2}dx$$?
I thought of using $x=\sqrt{\sin u}$ substitution but to no avail.
 A: By substituting $x=\tan\theta$ we are left with
$$ I=\int_{0}^{\pi/4}\sqrt{(1-\tan^2\theta)(1+\tan^2\theta)}\,d\theta =\int_{0}^{\pi/4}\frac{\sqrt{\cos(2\theta)}}{\cos^2\theta}\,d\theta=\int_{0}^{\pi/2}\frac{\sqrt{\cos\theta}}{1+\cos\theta}\,d\theta\tag{1}$$
that can be written in the following form:
$$ I=\int_{0}^{1}\frac{\sqrt{x}}{(1+x)\sqrt{1-x^2}}\,dx =  \int_{0}^{1}x^{1/2}(1-x^2)^{-3/2}\,dx-\int_{0}^{1}x^{5/2}(1-x^2)^{-3/2}\,dx\tag{2}$$
or in the following form:
$$ \frac{1}{2}\left[\int_{0}^{1}x^{-1/4}(1-x)^{-3/2}\,dx-\int_{0}^{1}x^{3/4}(1-x)^{-3/2}\,dx\right]\tag{3}$$
that by Euler's Beta function equals:

$$ I = \color{red}{\frac{\sqrt{\pi}}{2}\left[\frac{\Gamma\left(\frac{1}{4}\right)^2}{\pi\sqrt{8}}-\frac{\pi\sqrt{8}}{\Gamma\left(\frac{1}{4}\right)^2}\right]}\tag{4}$$

that can be numerically evaluated very efficiently through the AGM :

$$ I = \color{red}{\frac{\sqrt{\pi}}{2}\left[\frac{\sqrt{\pi}}{\text{AGM}(1,\sqrt{2})}-\frac{\text{AGM}(1,\sqrt{2})}{\sqrt{\pi}}\right]}=\frac{\pi}{2\,\text{AGM}(1,\sqrt{2})}-\frac{\text{AGM}(1,\sqrt{2})}{2}. \tag{5}$$

A: Let $z=x^{2}$
\begin{align}
\int\limits_{0}^{1} \frac{\sqrt{1-x^{4}}}{1+x^{2}} dx &= 
\int\limits_{0}^{1} (1-x^{2})^{1/2}(1+x^{2})^{1/2} dx \\
&= \frac{1}{2} \int\limits_{0}^{1} (1-z)^{1/2} (1+z)^{-1/2} z^{-1/2} dz \\
&= \frac{\Gamma(1/2) \Gamma(3/2)}{2 \Gamma(2)} {}_{2}\mathrm{F}_{1}(1/2,1/2;2;-1) \\
&\approx 0.7119586
\end{align}
We used Gauss's hypergeometric function.
