$$f(a,b)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt{x^2+a^2}\sqrt{x^2+b^2}}$$
To use Landen's transformation
$$f(a,b)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt{x^2+(\frac{a+b}{2})^2}\sqrt{x^2+ab}}$$
$$f(a,b)=f\left(\frac{a+b}{2},\sqrt{ab}\right)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt{x^2+c^2}\sqrt{x^2+c^2}}=\int_0^{+\infty} \frac{\;\mathrm dx}{x^2+c^2}=\frac{\pi}{2c}$$
$$c=\operatorname{AGM}(a,b)$$
$\operatorname{AGM}$ is Arithmetic–geometric mean
I would like to find similar transform for:
$a,b,c>0$
$$g(a,b,c)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$$
Do you know similar kind of transform method for that integral?
Or which other methods can be used to evaluate that improper integral?
Thanks a lot for answers and helps.
UPDATE:
I would like to share my results. Maybe someone can give some ideas to go forward.
$$g(a,b,c)=\frac{1}{F(a,b,c)}=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$$ $$\frac{1}{F(a.k,b.k,c.k)}=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3k^3}\sqrt[3]{x^3+b^3k^3}\sqrt[3]{x^3+c^3k^3}}$$
$$\frac{1}{F(a.k,b.k,c.k)}=\int_0^{+\infty} \frac{\;\mathrm dx}{k^3\sqrt[3]{\frac{x^3}{k^3}+a^3}\sqrt[3]{\frac{x^3}{k^3}+b^3}\sqrt[3]{\frac{x^3}{k^3}+c^3}}$$ $ku=x$ $$\frac{1}{F(a.k,b.k,c.k)}=\int_0^{+\infty} \frac{\;k du}{k^3\sqrt[3]{u^3+a^3}\sqrt[3]{u^3+b^3}\sqrt[3]{u^3+c^3}}$$
$$F(a.k,b.k,c.k)=k^2F(a,b,c)$$
$k=\frac{1}{a}$
$$F\left(1,\frac{b}{a},\frac{c}{a}\right)=\frac{1}{a^2}F(a,b,c)$$
$$F\left(a,b,c\right)=a^2F\left(1,\frac{b}{a},\frac{c}{a}\right)$$
$\frac{b}{a}=t$
$\frac{c}{a}=z$
$$F(a,at,az)=a^2F(1,t,z)=a^2H(t,z)$$
$$\frac{1}{F(a,at,az)}=\frac{1}{a^2F(1,t,z)}=\frac{1}{a^2H(t,z)}=\frac{1}{a^2}\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+1}\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$
Now the problem is for 2 variables. $$\frac{1}{H(t,z)}=V(t,z)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+1}\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$
We can get a partial differential equation from here
$$\frac{\partial V(t,z)}{\partial t}=-\int_0^{+\infty} \frac{t^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+t^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$
$$\frac{\partial V(t,z)}{\partial z}=-\int_0^{+\infty} \frac{z^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+z^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$
$$\frac{\partial^2 V(t,z)}{\partial t \partial z }=\int_0^{+\infty} \frac{t^2z^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+t^3)(x^3+z^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}=$$
$$=\frac{1 }{z^3-t^3}\int_0^{+\infty} \frac{t^2z^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+t^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}+\frac{1 }{t^3-z^3}\int_0^{+\infty} \frac{t^2z^2\; \mathrm dx}{\sqrt[3]{x^3+1}(x^3+z^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$
$$(t^3 -z^3 )\frac{\partial^2 V(t,z)}{\partial t \partial z }=z^2\frac{\partial V(t,z)}{\partial t }- t^2\frac{\partial V(t,z)}{\partial z }$$
Now I am looking for a transform that
$t=k(s,y)$ $z=l(s,y)$ To get same equation
$$(s^3 -y^3 )\frac{\partial^2 V(s,y)}{\partial s \partial y }=y^2\frac{\partial V(s,y)}{\partial s }- s^2\frac{\partial V(s,y)}{\partial y }$$
UPDATE 2: I tried to use a variable change .
$$\frac{1}{H(t,z)}=V(t,z)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+1}\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$
$$u=\frac{x}{\sqrt[3]{x^3+1}}$$
$$\frac{du}{1-u^3}=\frac{dx}{\sqrt[3]{x^3+1}}$$
$$\frac{1}{H(t,z)}=V(t,z)=\int_0^{1} \frac{\;du}{(1-u^3)\sqrt[3]{\frac{u^3}{1-u^3}+t^3}\sqrt[3]{\frac{u^3}{1-u^3}+z^3}}=$$
$$=\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{t^3+(1-t^3)u^3}\sqrt[3]{z^3+(1-z^3)u^3}}$$
$$=\frac{1}{\sqrt[3]{(1-t^3)}\sqrt[3]{(1-z^3)}}\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{(\cfrac{t}{\sqrt[3]{(1-t^3)}})^3+u^3}\sqrt[3]{(\cfrac{z}{\sqrt[3]{(1-z^3)}})^3+u^3}}$$
If we combine the results above, we can get
$$g(a,b,c)=\frac{1}{F(a,b,c)}=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$$
$$u=\frac{x}{\sqrt[3]{x^3+a^3}}$$
$$\frac{du}{1-u^3}=\frac{dx}{\sqrt[3]{x^3+a^3}}$$
$$g(a,b,c)=\frac{1}{F(a,b,c)}=\int_0^{1} \frac{\;du}{(1-u^3)\sqrt[3]{\frac{a^3u^3}{1-u^3}+b^3}\sqrt[3]{\frac{a^3u^3}{1-u^3}+c^3}}=\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{b^3-(b^3-a^3)u^3}\sqrt[3]{c^3-(c^3-a^3)u^3}}$$
$$g(a,b,c)=\frac{1}{F(a,b,c)}=\frac{1}{bc}\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{1-(1-(a/b)^3)u^3}\sqrt[3]{1-(1-(a/c)^3)u^3}}$$
$$\frac{1}{T(x,y)}=\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{1-x^3u^3}\sqrt[3]{1-y^3u^3}}$$
$$F(a,b,c)=bc .T(\sqrt[3]{1-(a/b)^3},\sqrt[3]{1-(a/c)^3})=ac.T(\sqrt[3]{1-(b/a)^3},\sqrt[3]{1-(b/c)^3})=ab.T(\sqrt[3]{1-(c/a)^3},\sqrt[3]{1-(c/b)^3})$$
It seems If we solve $\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{1-x^3u^3}\sqrt[3]{1-y^3u^3}}$, It will be enough.
Maybe It can be other group of integrals or can be found a relation with elliptic integrals such as $\int_0^{1} \frac{\;du}{\sqrt{1-u^2}\sqrt{1-x^2u^2}}$. I do not know yet.
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
to gp).
The reference where I found formulas equivalent to the above recipe for $I(a,b,c)$ is
This paper in turn attributes the formulas for $\psi$ and $M$ to Theorem 2.2 on page 134 of