To evaluate $\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$

There is a known AGM-like transformation for your integral $$ g(a,b,c) = \int_0^\infty \frac{dx} { \root 3 \of {x^3+a^3} \root 3 \of {x^3+b^3} \root 3 \of {x^3+c^3} }. $$ It was obtained only a few years ago, and is considerably more complicated and harder to prove than Landen's transformation; it's inspired by the familiar AGM, but does not reduce to it, and must be developed on its own terms.

Define $$ I(a,b,c) = \int_0^\infty \frac{dt}{\root 3 \of {t(t+a^3)(t+b^3)(t+c^3)}}. $$ The change of variables $x^3 = 1/t$, $dx = -t^{-4/3} dt/3$ gives $$ g(a,b,c) = \frac13\int_0^\infty \frac{dt} {\root 3 \of {t(1+a^3t)(1+b^3t)(1+c^3t)}} = \frac1{3abc} I(a^{-1},b^{-1},c^{-1}). $$

Now the AGM-like identity is $$ I(a,b,c) = I(a',b',c') $$ where the transformation $\psi: (a,b,c) \mapsto (a',b',c')$ is given by $$ a' = \frac13(a+b+c), \quad b', c' = \left[\frac12 \left( \frac{a^2 (b+c) + b^2(c+a) + c^2(a+b)}{3} \pm \frac{(a-b)(b-c)(c-a)}{\sqrt{-27}} \right)\right]^{1/3}. $$ Note the $\sqrt{-27}$ in the denominator: if $a,b,c$ are distinct positive reals then $b',c'$ are complex conjugates (chosen to be the principal cube roots), but then another application of $\psi$ returns positive $a'',b'',c''$ that are much closer than $a,b,c$. Repeated application of $\psi$ yields a sequence that converges cubically to $(M,M,M)$ for some $M=M(a,b,c)$ which is the analogue for these "Picard periods" of the AGM; and then $$ I(a,b,c) = I(M,M,M) = \frac{\pi \sqrt{4/3}}{M}. $$

For example, if we want to evaluate $g(3.1,4.1,5.9)$ then we write $$ g(3.1,4.1,5.9) = \frac1{3 \cdot 3.1 \cdot 4.1 \cdot 5.9} I\left(\frac1{3.1},\frac1{4.1},\frac1{5.9}\right). $$ Applying $\psi^3$ to $(1/3.1, 1/4.1, 1/5.9)$ yields $M = 0.2453101037492669116299747362\ldots$, so $I(1/3.1, 1/4.1, 1/5.9) = \pi \sqrt{4/3} / M = 14.78780805610937446473708974\ldots$, and $g(3.1,4.1,5.9) = 0.0657332322345471756512603615-$ which agrees with the numerical computation of $g(a,b,c)$ (try telling

intnum(x=0,[+1],((x^3+3.1^3)*(x^3+4.1^3)*(x^3+5.9^3))^(-1/3))

to gp).

The reference where I found formulas equivalent to the above recipe for $I(a,b,c)$ is

Keiji Matsumoto and Hironori Shiga: A variant of Jacobi type formula for Picard curves, J. Math. Soc. Japan 62 #1 (2010), 305$-$319. (doi: 10.2969/jmsj/06210305)

This paper in turn attributes the formulas for $\psi$ and $M$ to Theorem 2.2 on page 134 of

Kenji Koike and Hironori Shiga: Isogeny formulas for the Picard modular form and a three terms arithmetic geometric mean, J. Number Theory 124 (2007), 123$-$141.