I am trying to find the closed form the expression $${_3F_2}\!\left(\begin{array}c\tfrac34,1,1\\\tfrac32,\tfrac74\end{array}\middle|1\right).$$ I was able to convert the expression into the series $${_3F_2}\!\left(\begin{array}c\tfrac34,1,1\\\tfrac32,\tfrac74\end{array}\middle|1\right)=\sum_{n \geq 0} \frac{3 \cdot 4^n}{(4n+3)(2n+1)\binom{2n}{n}}$$ using the series definition of the generalized hypergeometric function and properties of the Pochhammer symbol/gamma function. Since $\frac{1}{(2n+1)\binom{2n}{n}} = \mathrm{B}(n+1,n+1)$ where $\mathrm{B}$ is the beta function, I tried substituting the integral representation of the beta function into the sum and interchanging the order of summation and integration to get \begin{align*} \sum_{n \geq 0} \frac{3 \cdot 4^n}{(4n+3)(2n+1)\binom{2n}{n}} &= \sum_{n \geq 0} \frac{3 \cdot 4^n}{(4n+3)}\mathrm{B}(n+1,n+1)\\ &= \sum_{n \geq 0} \frac{3 \cdot 4^n}{4n+3} \int_0^1 t^n (1-t)^n\,dt\\ &= \int_0^1\sum_{n \geq 0} \frac{3 \cdot 4^n}{4n+3} t^n (1-t)^n\,dt. \end{align*} According the Mathematica, the series can be simplified down to $$\int_0^1 \left(\frac{3 \tanh ^{-1}\left(\sqrt{2} \sqrt[4]{1-t} \sqrt[4]{t}\right)}{4 \sqrt{2} (1-t)^{3/4} t^{3/4}}-\frac{3 \tan ^{-1}\left(\sqrt{2} \sqrt[4]{1-t} \sqrt[4]{t}\right)}{4 \sqrt{2} (1-t)^{3/4} t^{3/4}}\right)\,dt.$$ From here, I do not know how to proceed. Is there a better way of approaching this or is this perhaps already a known hypergeometric function identity, like the one found in this previous post?
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