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 ($m \geq 2$):
$$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$

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 $m=3$.
