Help needed finding a closed form or approximation for an integral(hypergeometric function)

$$ B\left( T+\tfrac{1}{2},\tfrac{3}{2}\right)\cdot {}_2 F_1\left(\tfrac{1}{2},T+\tfrac{1}{2};T+2,z\right) = \int_{0}^{1}\frac{x^T}{\sqrt{x(1-x)(1-xz)}}\,dx$$ 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 $$\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n(n+1)}\left[H_{n-1/2}-H_n\right] =\color{red}{2\pi-4}. $$