$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$ Is there a closed form integral for $$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$$ for $-1 < x < 1$?
This integral is related to Legendre polynomials as
$$\frac{-1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)=\sum_{n=1}^{\infty} \frac{P_{n}(x)}{n} z^{n-1}$$
$$I(x) = \sum_{n=1}^{\infty} \frac{P_{n}(x)}{n^2}$$ where $P_{n}(x)$ satisfies $\sum_{n=0}^{\infty} P_n(x) z^2 = (1-2 x z + z^2)^{-
1/2}$ for all $-1< x< 1$ and $-1< z< 1$. Integrating twice with respect to $z$ gives a relation
\begin{align}\sum_{n=0}^{\infty} \frac{P_{n}(x)}{(n+1)(n+2)} &= \int_0^1 dz \int_0^z dz'\frac{1}{\sqrt{1 - 2 z' x +x^2}}\,. 
\\
&=
1 - \sqrt{2(1-x)} +(1-x) \ln\left(1+\sqrt{\frac{2}{1-x}}\right) \\&= 1-2s+2s^2 \ln\frac{1+s}{s} 
\end{align}
where $s=\sin(\theta/2)$ and $x=\cos\theta$. The elements of this sum are asymptotically equal to the sum defining $I$, but I am not sure if this is useful to get a closed form for $I$. Another related relation is
$$I(x) = \int_0^1 \frac{dz}{z} \int_0^z \frac{dz'}{z'} \left[\frac{1}{\sqrt{1 - 2 z' x +x^2}}-1\right]\,. 
$$
 A: Here's an approach using differentiation under the integral sign:
$$\begin{align}I'(x) &= \int_{0}^{1}\frac{\frac{1}{\sqrt{-2xz+z^2+1}}+1}{\sqrt{-2xz+z^2+1}-xz+1}\, dz\\
&=\left.\frac{(x+1)\ln\left(\sqrt{z^2-2xz+1}+x-z\right)-2x\ln\left(\sqrt{z^2-2xz+1}-z+1\right)}{x^2-1}\right]_{z=0}^{z=1}\\
&=\frac{x\ln(2)-x\ln(1-x)-(x+1)\ln(x+1)+(x+1)\ln\left(x+\sqrt{2-2x}-1\right)}{x^2-1}\end{align}$$
Mathematica now gives:
$$\begin{align}I(x) &= \int \frac{x\ln(2)-x\ln(1-x)-(x+1)\ln(x+1)+(x+1)\ln\left(x+\sqrt{2-2x}-1\right)}{x^2-1}\,dx\\
&=\frac{1}{2}\left(4\operatorname{Li}_2\left(\frac{1}{2}\left(2-\sqrt{2-2x}\right)\right)+\operatorname{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right)-\ln^2(1-x)+2\ln\left(2\left(x+\sqrt{2-2x}-1\right)\right)\ln(2-2x)-\ln\left(2-\sqrt{2-2x}\right)\ln(8-8x)+\ln\left(\sqrt{2-2x}+2\right)(\ln(2)-\ln(1-x))-3\ln^2(2)\right)+C\end{align}$$
Now using, as @Steven Clark in the comments said
$$I(0) = \frac{1}{4} \left(2 \operatorname{Li}_2 \left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\left(\operatorname{arsinh} (1)-\ln (2)\right)^2\right)$$
we can solve for $C$.
Upon doing so, one determines the rather horrible expression (which might be possible to simplify)
$$\boxed{I(x) = 2\operatorname{Li}_2\left(1-\frac{1}{2}\sqrt{2-2x}\right)+\frac{1}{2}\text{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right)-2\operatorname{Li}_2\left(1-\frac{1}{\sqrt{2}}\right)+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\ln^2(2-2x)+\ln\left(2x+2\sqrt{2-2x}-2\right)\ln(2-2x)+\frac{1}{2}\ln(2)\ln\left(\sqrt{2-2x}+2\right)-\frac{1}{2}\ln\left(2-\sqrt{2-2x}\right)\ln(8-8x)-\frac{1}{2}\ln\left(\sqrt{2-2x}+2\right)\ln(1-x)-\frac{\pi^2}{24}+\frac{\ln^2(2)}{4}-\frac{1}{2}\ln(2)\ln\left(2+\sqrt{2}\right)-\frac{1}{4}\operatorname{arsinh}^2(1)}$$
EDIT
Through some dilogarithm identities, one can simplify this expression to $$I(x)=\frac{\pi^2}{6}-\operatorname{Li}_{2}{\left(p\right)}+\operatorname{Li}_{2}{\left(-p\right)}-\ln{\left(p\right)}\ln{\left(1-p^{2}\right)}$$
with $p=\sqrt{\frac{1-x}{2}}$ as mentioned in the comments below by @David H
