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.