# Does this integral have a closed form expression?

I have been working with two related integrals. The first one yields a simple expression, but I can't seem to find a simple expression for the second.

Integral 1

\begin{align*} I_x &= \int_z^{1-y+z} \sqrt{\frac{1}{1-x-y+z} + \frac{1}{x-z}} \, dx\\ &= \int_z^{1-y+z} \sqrt{\frac{1-y}{(1-x-y+z)(x-z)}} \, dx\\ &= \sqrt{1-y} \int_z^{1-y+z} \frac{1}{\sqrt{(1-x-y+z)(x-z)}} \, dx\\ \end{align*}

We can substitute $u=\frac{x-z}{1-y}$, so $x=u(1-y)+z$, and $dx=(1-y) \, du$. Then we have:

\begin{align*} I_x &= \sqrt{1-y} \int_0^1 u^{\frac{1}{2}-1}(1-u)^{\frac{1}{2}-1} \, du\\ &= \sqrt{1-y} \, \operatorname{B} \left( \tfrac{1}{2}, \tfrac{1}{2} \right)\\ &= \sqrt{1-y} \, \pi \end{align*}

So this integral has a simple expression because the substitution led to the exact form of the beta function.

Integral 2

$$I_z = \int_{\max(0,x+y-1)}^{\min(x,y)} \sqrt{\frac{1}{1-x-y+z} + \frac{1}{x-z} + \frac{1}{y-z} + \frac{1}{z}} \, dz$$

This seems related to the first integral, but I can't seem to simplify it. Even if we assume one of the four "cases" of integration limits ($\int_0^x$, $\int_0^y$, $\int_{x+y-1}^x$, or $\int_{x+y-1}^y$) and try to proceed, it doesn't seem to help.

Am I missing something simple? Is there an obscure integration technique that would help? Or is there just no way to simplify this?

Update (6-26-14): From Jacquet & Szpankowski, 2003, "Markov Types and Minimax Redundancy for Markov Sources" (link), p.10, we have the following, which may (or may not) help here:

\begin{align*} I_{JS} &= 4 \int_0^1 \frac{1}{\sqrt{(1{-}x)x}} \, dx \int_0^{\min(x,1{-}x)} \frac{1}{(1{-}x{-}y)(x{-}y)} \, dy\\ &= 8 \int_0^{\frac{1}{2}} \frac{\log(1{-}2x) - \log\left(1{-}2\sqrt{(1{-}x)x}\right)}{\sqrt{(1{-}x)x}} \, dx\\ &= 16 \int_0^{\frac{\pi}{4}} \log \left( \frac{\cos(2\theta)}{1{-}\sin(2\theta)} \right) \, d\theta\\ &= 16 \, G\\ \end{align*}

where $G \approx 0.915965594$ is Catalan's constant.

First of all, regarding the limits of integration $a=\max{(0,x+y-1)}$ and $b=\min{(x,y)}$, you do not have to break up your analysis into four separate cases to accommodate the two possible values taken by the min/max functions. These functions can be represented in terms of elementary functions as follows:
$$\max{(a,b)}=\frac{a+b}{2}+\frac12|a-b|=\frac{a+b}{2}+\frac12\sqrt{(a-b)^2},\\ \min{(a,b)}=\frac{a+b}{2}-\frac12|a-b|=\frac{a+b}{2}-\frac12\sqrt{(a-b)^2}.$$
With that in mind, and after making the substitutions $u=2z-x-y$ followed by $w=u+1$, a little algebra reduces the integral to:
\begin{align} \mathcal{I}{\left(x,y\right)} &=\int_{\max{(0,x+y-1)}}^{\min{(x,y)}}\sqrt{\frac{1}{z}+\frac{1}{z-(x+y-1)}+\frac{1}{x-z}+\frac{1}{y-z}}\,\mathrm{d}z\\ &=\int_{\frac{x+y-1+|1-x-y|}{2}}^{\frac{x+y-|x-y|}{2}}\sqrt{\frac{1}{z}+\frac{1}{z-(x+y-1)}+\frac{1}{x-z}+\frac{1}{y-z}}\,\mathrm{d}z\\ &=\frac12\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{1}{\frac{u+x+y}{2}}+\frac{1}{\frac{u+x+y}{2}-(x+y-1)}+\frac{1}{x-\frac{u+x+y}{2}}+\frac{1}{y-\frac{u+x+y}{2}}}\,\mathrm{d}u\\ &=\frac12\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{2}{u+x+y}+\frac{2}{u-x-y+2}+\frac{2}{-u+x-y}+\frac{2}{-u-x+y}}\,\mathrm{d}u\\ &=\frac{\sqrt{2}}{2}\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{1}{u+x+y}+\frac{1}{u-x-y+2}+\frac{1}{-u+x-y}+\frac{1}{-u-x+y}}\,\mathrm{d}u\\ &=\frac{\sqrt{2}}{2}\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{1}{w+x+y-1}+\frac{1}{w-x-y+1}+\frac{1}{1-w+x-y}+\frac{1}{1-w-x+y}}\,\mathrm{d}w\\ &=\frac{\sqrt{2}}{2}\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{2w}{w^2-(x+y-1)^2}+\frac{2(1-w)}{(1-w)^2-(x-y)^2}}\,\mathrm{d}w\\ &=\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{w}{w^2-|x+y-1|^2}+\frac{(1-w)}{(1-w)^2-|x-y|^2}}\,\mathrm{d}w. \end{align}