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Does the following iterated integral have a simple closed-form expression in terms of $z$?

$$ I = \int_0^\infty \int_0^\infty \sqrt{\frac{1 + x^2 y^2 + x^2 z^2 + y^2 z^2}{(x^2 + y^2 + z^2 + x^2 y^2 z^2)^3}} \, dy \, dx $$

where $x \in \mathbb{R^+}$ and $y \in \mathbb{R^+}$. Assume $z \in \mathbb{R^+}$ is a constant.

I tried substituting:

$$ u = \frac{1}{2} \left[ 1 + \frac{y^2(1 + x^2 z^2)-(x^2+z^2)}{y^2(1 + x^2 z^2)+(x^2+z^2)} \right] $$

which simplifies the integral to this:

\begin{align} I &= \int_0^\infty \frac{\pi}{2} \frac{1}{x^2+z^2} \operatorname{_2F_1} \left( -\tfrac{1}{2}, \tfrac{1}{2}; 1; 1{-}\left[ \frac{x^2+z^2}{1+x^2 z^2} \right]^2 \right) \, dx \\ &= \int_0^\infty \frac{1}{x^2+z^2} \operatorname{E} \left( \sqrt{1{-}\left[ \frac{x^2+z^2}{1+x^2 z^2} \right]^2} \right) \, dx \end{align}

$\operatorname{E}(k)$ here is the complete elliptic integral of the second kind (using elliptic modulus $k$, not the parameter $m=k^2$).

I can't seem to simplify this any further. Is there a way to proceed with this integral over $\operatorname{E}(k)$? (I have looked through tables of integrals of elliptic integrals, but nothing seems to match.) Or is there another way to simplify the iterated integral, e.g. via a substitution for both $x$ and $y$?

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Phew! May I ask where this integral comes from? It might very well not have a simple closed form. –  Bruno Joyal Oct 23 '13 at 2:53
@Marie It comes from computing the normalizing constant of a Jeffreys prior for a probabilistic model over a bivariate Bernoulli distribution using odds-ratio parameters. –  Tyler Streeter Oct 23 '13 at 3:39
I have tried to expand the elliptic integral into a power series, and looks like each term integrates into a polynomial of $(z-1/z)^2$. No further results yet. –  Chen Wang Feb 5 '14 at 23:12

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