<s>[INST] Closed-form expression for this definite integral Does this definite integral have a closed-form expression?
\begin{align*}
I &= \int_0^\infty \sqrt{ \frac{1}{2} \frac{1}{x} \left( \frac{1}{(1+x)^2} + \frac{z}{(1+xz)^2} \right) } \, dx \\
&= \frac{1}{\sqrt{2}} \int_0^\infty \frac{1}{x} \sqrt{ \frac{x}{1+x} \left( 1-\frac{x}{1+x} \right) + \frac{xz}{1+xz} \left( 1-\frac{xz}{1+xz} \right) } \, dx
\end{align*}
where $x>=0$, and $z>=0$ is constant.
I have tried the following substitution:
$$
u = \frac{xz}{1+xz}, \quad
x = \frac{1}{z} \frac{u}{(1-u)}, \quad
dx = \frac{1}{z} \frac{1}{(1-u)^2} du
$$
$$ I = \frac{1}{\sqrt{2}} \int_0^1 \frac{1}{\sqrt{u(1-u)}} \sqrt{ 1 + \frac{z}{((1-u)z+u)^2} } \, du $$
...but I can't seem to make any more progress. Using $u=\frac{x}{1+x}$ yields a similar expression which is also difficult to simplify.
Can you find a way to simplify this further?
Update (Nov 4, 2013): The integral can be rearranged to the following form, which might make it easier to match something from a table of known integrals:
$$ I = \frac{1}{\sqrt{2}} \sqrt{\frac{1{+}z}{z}} \int_0^1 u^{-\frac{1}{2}} (1{-}u)^{-\frac{1}{2}} \left( 1+\frac{1{-}z}{z} u \right)^{-1} \left( 1 + 2\frac{1{-}z}{1{+}z}u + \frac{(1{-}z)^2}{z(1{+}z)}u^2 \right)^{\frac{1}{2}} \, du $$
 [/INST] $$
{\rm I}\left(z\right)
\equiv
\int_{0}^{\infty}\sqrt{\vphantom{\LARGE A^{A}}\frac{1}{2x}
\left\lbrack%
\frac{1}{\left(1 + x\right)^{2}} + \frac{z}{\left(1 + xz\right)^2}
\right\rbrack\,}\ {\rm d}x\,,
\qquad
\begin{array}{|rclcl}
\,\,{\rm I}\left(0\right)
& = &
{\rm I}\left(\infty\right)
& = &
{\sqrt{2\,} \over 2}\,\pi
\\[1mm]
\,\,{\rm I}\left(1\right) & = & \pi&& 
\end{array}
$$
\begin{align}
{\rm I}\left(z\right)
&=
{\sqrt{2} \over 2}\int_{0}^{\infty}\!\!
{\sqrt{z\left(z + 1\right)x^{2} + 4zx + 1 + z\,}
 \over
 \left(x + 1\right)\left(1 + xz\right)}\,
{{\rm d}x \over \sqrt{x\,}}
\\[3mm]&=
{\sqrt{2} \over 2}\left(z + 1 \over z\right)^{1/2}\int_{0}^{\infty}
{\sqrt{x^{2} + 4\left(z + 1\right)^{-1}\,x + z^{-1}\,}
 \over
\left(x + 1\right)\left(x + z^{-1}\right)}
{{\rm d}x \over \sqrt{x\,}}&
\end{align}
Equation $x^{2} + 4\left(z + 1\right)^{-1}\,x + z^{-1} = 0$ has the complex roots:
$$
x_{\pm}
=
-\,{2 \over z + 1}
\pm
{\rm i}\,{\left\vert z -1\right\vert \over \left(z + 1\right)\,\sqrt{z\,}}\,,
\qquad\mbox{Notice that}\quad
x_{+}x_{-} = \left\vert x_{\pm}\right\vert^{2} = z^{-1}
$$ 
Then,
${\rm I}\left(z\right)
 =
 \left(\sqrt{2}/2\right)\sqrt{a^{2} + 1\,}\ {\cal I}\left(a\right)$ where
$a = z^{-1/2}$.
\begin{eqnarray*}
{\cal I}\left(a\right)
& \equiv &
\int_{0}^{\infty}
{\sqrt{x^{2} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x + a^{2}\,}
 \over
\left(x + 1\right)\left(x + a^{2}\right)}
{{\rm d}x \over \sqrt{x\,}}
\\
x_{\pm}
& = &
{-2 \pm {\rm i}\,\left\vert\,a^{2} - 1\right\vert
 \over
 a^{2} + 1}\,a^{2}
\end{eqnarray*}
$$
{\rm I}\left(z\right)
=
{\sqrt{2\,} \over 2}\,\left(z + 1 \over z\right)^{1/2}{\cal I}\left(1 \over \sqrt{z\,}\right)\,,
\qquad\qquad
{\cal I}\left(a\right)
=
{\sqrt{2\,} \over \sqrt{a^{2} + 1\,}}\,{\rm I}\left(1 \over a^{2}\right)
$$
$$
{\cal I}\left(a\right)
=
\int_{-\infty}^{\infty}
{\sqrt{x^{4} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x^{2} + a^{2}\,}
 \over
\left(x^{2} + 1\right)\left(x^{2} + a^{2}\right)}
\,{\rm d}x
$$
Integration over the complex plane is possible. You have to take into account that
$$
x^{4} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x^{2} + a^{2}
=
\left(x - x_{\atop -}^{1/2}\right)\left(x + x_{\atop -}^{1/2}\right)
\left(x - x_{\atop +}^{1/2}\right)\left(x + x_{\atop +}^{1/2}\right)
$$
which introduces branch-cuts in the complex plane.
</s>