In this article Arithmetic-Geometric Mean it is claimed that
let $\displaystyle
I_p(a,b)=\int_0^{+\infty}\frac{x^{p-2}\,\mathrm dx}{(x^p+a^p)^{1/p}(x^p+b^p)^{(p-1)/p}},\
M_1(a,b)=\frac{a+2b}3,\ M_2(a,b)=\sqrt[3]{\frac{a^2b+ab^2+b^3}3}$
then $I_3(a,b)=I_3(M_1(a,b),M_2(a,b))$.
In the source Hypergeometric Analogues of the Arithmetic-Geometric Mean Iteration the original proof involves differential equation. But I wonder if this identity can be proved directly by substitution as it is similar to Landen's transformation.
So my question is: Is it possible?
By the way, there is another identity that
let $I_3^\prime(a,b)=I_3(a,\sqrt[3]{a^3-b^3})$ then $I^\prime_3(M_1(a,b),M_2(a,b))=3I^\prime_3(a,b)$.
Update: A possible approach
The post evaluate the integral... suggests a relation between $I_3(a,b)$ and $\displaystyle
\int_0^{\infty}\frac{\mathrm dx}{\sqrt{x^6+p x^3+1}}
$. Using substitution $x=z\pm\sqrt{z^2-1}$ from reciprocal square root of sixth degree polynomial, the latter integral can be transformed to an elliptic one, namely $\displaystyle
\int_1^{\infty}\frac{\mathrm dx}{\sqrt{(x-1)(4x^3-3x+p/2)}}
$, which is easier to deal with.
Related:
If the integral were $\displaystyle
\int\frac{\mathrm dx}{(1-x^3)^{2/3}}
$, then the substitution $\displaystyle
x=\frac{-6z}{3\sqrt{4z^3-1/27}-1}$ from Dixon elliptic functions would transform it into $\displaystyle
\int\frac{\mathrm dz}{\sqrt{4z^3-1/27}}
$. Maybe we can do similar things to $I_3(a,b)$.
