Prove $\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} ( \text{agm} (\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} ))^{-1}$ The following definite integral turns out to be expressible as the Arithmetic-Geometric Mean: $$I_4(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} \left( \text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)\right)^{-1}$$
$$I_4(1,1)=\frac{\pi}{2 \sqrt2}$$
I would like to remind you that:
$$I_2(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2} \left( \text{agm} \left(a,b \right)\right)^{-1}$$
$$I_2(1,1)=\frac{\pi}{2}$$
Which is why I have two questions:

How do we prove the identity for $I_4(a,b)$?
Is it possible to also express other integrals of this type using agm? Such as $I_8(a,b)$?


Because of the relation of the agm to elliptic integrals, we can also write:
$$I_4(a,b)=\frac{1}{a b \sqrt{a^2+b^2}} K \left( \frac{a-b}{\sqrt{2(a^2+b^2)}} \right)$$
Here the parameter convention is $$K(k)=\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}}$$
This seems to be the best way to prove the identity, but I don't know which substitution to use.
Another way to express this integral would be through the hypergeometric function:
$$I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right)$$
And for every integral of this type we have:
$$I_m(a,b)=\frac{I_m(1,1)}{a^{m-1}} {_2F_1} \left(\frac{1}{2},\frac{m-1}{m};1;1-\frac{b^m}{a^m} \right)$$
The outline for the proof can be found in this question for $m=3$ and is easily adapted to the general case.
Here we assume $a \geq b$.
As for the arithmetic geometric mean, I tried to get it into a simpler form, but the only transformation I was able to achieve is this:
$$\text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)=\text{agm} \left(\frac{a+b}{2}+i \frac{a-b}{2},\frac{a+b}{2}-i \frac{a-b}{2} \right)$$
Since we have complex conjugates, it's quite obvious, that they will give real numbers at first iteration, an it would give the left hand side.

Since $K(k)=\frac{\pi}{2} ~{}_2F_1(1/2,1/2;1;k^2)$ and it was shown in the question that
$$
I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right),
$$
the first question is equivalent to proving the equality

$$
\frac{\sqrt{2} }{b \sqrt{b^2+1}}{_2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{(1-b)^2}{2 \left(b^2+1\right)}\right)}={_2F_1\left(\frac{1}{2},\frac{3}{4};1;1-b^4\right)}\tag{1}.
$$

$\it{Proof}$. The proof can be accomplished by combining 3 transformations for hypergeometric function. 
By Pfaff's transformation
$$
{_2F_1\left(\frac{1}{2},\frac{3}{4};1;1-b^4\right)}=\frac{1}{b^2}{~_2F_1\left(\frac{1}{2},\frac{1}{4};1;\frac{b^4-1}{b^4}\right)}.
$$
By quadratic transformation 2.11(5) from Erdelyi, Higher transcendental functions (put $z=\frac{b^2-1}{b^2+1}$)
$$
{~_2F_1\left(\frac{1}{2},\frac{1}{4};1;\frac{b^4-1}{b^4}\right)}=\sqrt{\frac{2b^2}{1+b^2}}{~_2F_1\left(\frac{1}{4},\frac{1}{4};1;\left(\frac{b^2-1}{b^2+1}\right)^2\right)}.
$$
By transformation 2.11(2) (put $z=\frac{(1-b)^2}{2 \left(b^2+1\right)}$)
$$
{~_2F_1\left(\frac{1}{4},\frac{1}{4};1;\left(\frac{b^2-1}{b^2+1}\right)^2\right)}={~_2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{(1-b)^2}{2 \left(b^2+1\right)}\right)}.
$$
