Help needed finding a closed form or approximation for an integral(hypergeometric function) I am seeking (preferably) a closed form for the integral
\begin{equation}
\int_{0}^{1} \frac{\ln\left[x(1-x)\right]}{\sqrt{x(1-x)(1-zx)}} dx, \;\;\; z<1
\end{equation}
I know from previous work that 
\begin{equation}
B(b,c-b)\,_2F_1(a,b;c;z) = \int_0^1 x^{b-1} (1-x)^{c-b-1}(1-zx)^{-a} \, dx
\end{equation}
so I am hoping that my integral will be some other form of a hypergeometric function.  I have tried expressing the logarithm as a hypergeometric function using the relationship $\ln(1-z)=-zF(1,1;2;z)$ and using properties in the book "Higher Transcendental Functions."  I have also scoured nist.gov and "Table of integrals, series, and products." for some kind of useful identity but I cant find anything.  The closest thing I found is equation 7.512.9 from Table of Integrals Series and Products, which is a result for
\begin{equation}
\int_0^{1}x^{\gamma-1}(1-x)^{\rho-1}(1-zx)^{-\sigma}F(\alpha,\beta;\gamma;x)dx
\end{equation}
which I wont type because my integral doesn't conform.  
At this point, an approximation would be fine.  But if someone can see a way to solve this thing analytically, that would be amazing.  Either way, I appreciate the help!
 A: $$ B\left( T+\tfrac{1}{2},\tfrac{3}{2}\right)\cdot {}_2 F_1\left(\tfrac{1}{2},T+\tfrac{1}{2};T+2,z\right)  =  \int_{0}^{1}\frac{x^T}{\sqrt{x(1-x)(1-xz)}}\,dx$$
hence by differentiating both sides with respect to $T$, then evaluating at $T=0$ we get that
$$ -\tfrac{\pi}{2}\left(1+2\log 2\right)\cdot {}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};2,z\right)+\tfrac{1}{2}\sum_{n\geq 0}\frac{\Gamma\left(n+\tfrac{1}{2}\right)^2\left[1+2\log 2+H_{n-1/2}-H_n\right]}{n!\Gamma(n+2)}\, z^n $$
exactly equals $\int_{0}^{1}\frac{\log x}{\sqrt{x(1-x)(1-xz)}}\,dx$, and $\int_{0}^{1}\frac{\log(1-x)}{\sqrt{x(1-x)(1-xz)}}\,dx$ has an analogous closed form. The previous line can be simplified as follows:
$$\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n(n+1)}\left[H_{n-1/2}-H_n\right] z^n $$
and this is not, strictly speaking, a hypergeometric function, but it is pretty simple to approximate such object with hypergeometric functions by considering the asymptotic expansion of $H_{n-1/2}-H_n$. Campbell has shown that similar series has closed forms for many specific values of $z$, and together with Sondow we proved that Fourier-Legendre series expansions provide a very effective technique for evaluating $\int K(x)g(x)\,dx$. The above series is an instance: at $z=1$ we have
$$\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n(n+1)}\left[H_{n-1/2}-H_n\right] =\color{red}{2\pi-4}. $$
A: Not an approximation but an upper bound for the integral
$$
\int_{0}^{1} \frac{\ln\left[x(1-x)\right]}{\sqrt{x(1-x)(1-zx)}} dx\leq \int_{0}^{1} \frac{\ln\left[1/4\right]}{\sqrt{x(1-x)(1-zx)}}$$
$$=\ln 8\dfrac{F\left[\sqrt{1-z},\sqrt{\dfrac{1}{1-z}}\right]-F\left[1,\sqrt{\dfrac{1}{1-z}}\right]}{\sqrt{z-1}}
$$
In which the following elliptic function was used:
$$F(z,k)=\int_{0}^{z}\dfrac{du}{\sqrt{1-u^2}\sqrt{1-k^2u^2}}.$$
