Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$ This integral looks a lot like an elliptic integral, but with cubes instead of squares:
$$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$
Let's consider $a,b>0$ for now.
$$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{3} a^2}$$
I obtained the general series solution the following way. Choose $a,b$ such that $a \geq b$, then:
$$I(a,b)=\frac{1}{a^2} \int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+b^3/a^3)}}=\frac{1}{a^2} I \left(1, \frac{b}{a} \right)$$
$$\frac{b^3}{a^3}=p, \qquad I \left(1, \frac{b}{a} \right)=I_1(p)$$
$$I_1(p)=\int_0^\infty\frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=2 \frac{d}{dp} J(p)$$
$$J(p)=\int_0^\infty\sqrt{\frac{t^3+p}{t^3+1}}dt=\int_0^\infty\sqrt{1+\frac{p-1}{t^3+1}}dt=$$
$$|p-1| \leq 1$$
$$=\sum_{k=0}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) (p-1)^k \int_0^\infty \frac{dt}{(t^3+1)^k}$$
Now this is the most problematic part. The first integral of this series diverges. However, it's a constant in $p$, so if we differentiate, it formally disappears:
$$I_1(p)=2 \sum_{k=1}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) k (p-1)^{k-1} \int_0^\infty \frac{dt}{(t^3+1)^k}$$
Now, every integral in this series converges. The integtals can be computed using the Beta function, if we substitute: $$t^3=\frac{1}{u}-1$$
Finally, we rewrite:
$$I_1(p)=\frac{\Gamma (1/3)}{3 \sqrt{\pi}} \sum_{k=1}^\infty \frac{k^2}{k!^2} \Gamma \left(k- \frac{1}{2}\right) \Gamma \left(k- \frac{1}{3}\right) (1-p)^{k-1}$$
Or, using the Pochhammer symbol:

$$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{(k+1)^2}{(k+1)!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k$$

My questions are:

Is the method I used valid (see the 'problematic part')? How to get this series into a Hypergeometric function form?
Is there any 'arithmetic-geometric mean'-like transformation (Landen's transformation) for this integral? How to go about finding it?

If the method I used is correct, it can be used for any integral of the form ($m \geq 2$):
$$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$
 A: It was already shown that
$$
I_1(p)=\int_0^\infty \frac{dx}{\sqrt{(x^3+1)(x^3+p)}}=\frac{2 \pi}{3 \sqrt{3}} {_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right).
$$
By transformation 2.11(5) from Erdelyi, Higher transcendental functions (put $z=\frac{1-\sqrt{p}}{1+\sqrt{p}}$)
$$
{_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right)=\left(\frac{2}{1+\sqrt{p}}\right)^{4/3}{_2F_1} \left(\frac{2}{3},\frac{2}{3};1;\left(\frac{1-\sqrt{p}}{1+\sqrt{p}}\right)^{2} \right).
$$
By Pfaff's transformation
$$
{_2F_1} \left(\frac{2}{3},\frac{2}{3};1;\left(\frac{1-\sqrt{p}}{1+\sqrt{p}}\right)^{2} \right)=\left(\frac{(1+\sqrt{p})^2}{4\sqrt{p}}\right)^{2/3}{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;\frac{(1-\sqrt{p})^2}{-4\sqrt{p}} \right).
$$
As a result
$$
I_1(p)=\frac{2 \pi}{3 \sqrt{3}p^{1/3}}{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;\frac{(1-\sqrt{p})^2}{-4\sqrt{p}} \right).
$$
Now we will use a generalization of AGM found by Borwein and Borwein, A Cubic Counterpart of Jacobi's Identity and the AGM, Transactions of the American Mathematical Society, Vol. 323, No. 2, (1991), pp.691-701 (after correcting for some typos):
$$
a_{n+1}=\frac{a_n+2b_n}{3} ,\quad b_{n+1}=\sqrt[3]{b_n\frac{a_n^2+a_nb_n+b_n^2}{3}},\quad a_0=1,\quad b_0=s,
$$
$$
\quad AG_3(1,s)=\lim_{n\to\infty} a_n=\frac{1}{{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;1-s^3 \right)}.
$$
Using this we get

\begin{align}
I_1(p)=\frac{2 \pi}{3 \sqrt{3}~p^{1/3}\cdot AG_3\left(1,\left(\frac{1+\sqrt{p}}{2~\sqrt[4]{p}}\right)^{2/3}\right)}.
\end{align}

A: More generally, with $|p-1|<1$, some experimentation shows that,
$$\int_0^\infty \frac{dt}{\sqrt{(t^m+1)(t^m+p)}} = \pi\,\frac{\,_2F_1\big(\tfrac12,\tfrac{m-1}{m};1;1-p\big)}{m\sin\big(\tfrac{\pi}{m}\big)}$$
where the question was just the case $m=3$.
A: Using the advice from @tired in the comments, we can write:
$$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k=$$
$$=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{1}{(1)_k} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k \frac{(1-p)^k}{k!}=\frac{2 \pi}{3 \sqrt{3}} {_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right)$$
So this is just the usual Gauss hypergeometric function.
This answers my first question, but I'm hoping to get an answer to my second question as well.

If we speak of this integral as a mean, it's very close to both Arithmetic Geometric Mean and the Logarithmic Mean:
$$M(a,b)=\frac{a}{\sqrt{{_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-\frac{b^3}{a^3} \right)}}$$
$$a \geq b$$

I remind that the AGM can be written as:
$$\text{agm}(a,b)=\frac{a}{{_2F_1} \left(\frac{1}{2},\frac{1}{2};1;1-\frac{b^2}{a^2} \right)}$$
$$a \geq b$$
And numerically we have:
$$M(a,b) \leq \text{agm}(a,b)$$
