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I am trying to find the closed form expression for $${_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac12\right).$$ I encountered this expression when evaluating the integral $$\int_0^1 \frac{\ln^2(1-x)}{\sqrt{1-x^2}}\,dx.$$ I tried utilizing the series representation of the generalized hypergeometric function Pochhammer symbol and applying Legendre duplication, which left me with the infinite series $$\sum_{n=0}^\infty \frac{\binom{2n}{n}}{8^n(2n+1)^3}.$$ Using Mathematica, it can be approximated to be $${_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac12\right) \approx 1.01015469252627.$$

I have reason to believe that the closed form may contain the inverse tangent integral function, as the hypergeometric function is similar to @Cleo's answer. Are there any neat properties of the $_4F_3$ hypergeometric function that can be exploited in order to obtain the closed form result?

FundamentalTheorem
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    check out @Vladimir Reshetnikov's comment on my answer here: https://math.stackexchange.com/a/1438525/219995 – Noam Shalev - nospoon Sep 03 '23 at 07:22
  • @Noam Shalev - nospoon Thank you so much! I have reduced it down to $\frac{\pi ^3}{192 \sqrt{2}}+\frac{i \text{Li}_3(1-i)-i \text{Li}_3(1+i)}{2 \sqrt{2}}$. However, I am wondering if there is a way to remove the imaginary numbers the way Cleo did in her answer – FundamentalTheorem Sep 03 '23 at 07:48
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    Simpler is: $\frac{\pi ^3}{192 \sqrt{2}}+\frac{\Im(\text{Li}_3(1+i))}{\sqrt{2}}$ – Mariusz Iwaniuk Sep 03 '23 at 09:45

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