Evaluate $\int_{0}^{1}K\left ( \sqrt{1-x^2} \right ) K\left ( \sqrt{1-x^4} \right )\text{d}x$ How can we get a closed expression for this integral,
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2}  \right )
K\left ( \sqrt{1-x^4}  \right )\text{d}x$$
where an complete elliptic integral $K(x)$ defined by $\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2}  }\text{d}t$ for $|x|<1$ appears and $x$ is the elliptic modulus?

*

*Observation 1: We can make a use of
$$
\int_{0}^{1}x^n K\left ( \sqrt{1-x^2}  \right )\text{d}x
=\frac{\pi}{4} \frac{\Gamma\left ( \frac{n+1}{2}  \right )^2 }{
\Gamma\left ( \frac{n+2}{2}  \right )^2}.
$$
However, I don't know exactly the behaviour of function $K\left ( \sqrt{1-x^4}  \right)$ around $x=0$. Expanding him at other points seems to also be a messy calculation.

*Observation 2: We may notice,
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2}  \right )
K\left ( \sqrt{1-x^4}  \right )\text{d}x=2\int_{0}^{1} \frac{K\left ( \frac{x}{\sqrt{1+x^2} }  \right )
K\left ( \sqrt{1-x^4}  \right )  }{
\sqrt{1+x^2} }\text{d}x.
$$
I am grateful for all your help.

We have
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2}  \right )
K\left ( \sqrt{1-x^4}  \right )\text{d}x
=\frac{\Gamma\left ( \frac14 \right )^4}{16}
 {}_4F_3\left ( \begin{array}{c|}
\frac14,\frac14,\frac14,\frac14\\
\frac12,\frac12,1
\end{array}\text{ }1 \right )-\frac{\Gamma\left ( \frac34 \right )^4}{4}
 {}_4F_3\left ( \begin{array}{c|}
\frac34,\frac34,\frac34,\frac34\\
1,\frac32,\frac32
\end{array}\text{ }1 \right ).
$$
