I have been trying to evaluate the series $$\, _4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right) = 1.133928715547935...$$ using integration techniques, and I was wondering if there is any simple way of finding a closed-form evaluation of this hypergeometric series. What is a closed-form expression for the above series?

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    $\begingroup$ In other terms, you are asking for a closed form of $$\sum_{n\geq 0}\frac{27\cdot 16^n}{(2n+3)^3 (2n+1)^2 \binom{2n}{n}^2}. $$ $\endgroup$ – Jack D'Aurizio Feb 1 '17 at 1:57
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    $\begingroup$ It is interesting to point out that if the factor $(2n+3)^3$ was $(2n+3)^2$, it would not be too hard to tackle such series through Parseval's theorem and known Taylor series for $\arcsin(x)$ and $\arcsin(x)^2$. $\endgroup$ – Jack D'Aurizio Feb 1 '17 at 2:04
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    $\begingroup$ Namely $$\sum_{n\geq 0}\frac{16^n}{(2n+3)^2 (2n+1)^2 \binom{2n}{n}^2} = \pi-3.$$ $\endgroup$ – Jack D'Aurizio Feb 1 '17 at 2:41
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    $\begingroup$ Wow, that was unexpected. The question has an affirmative answer, such value of a hypergeometric function has a manageable closed form! $\endgroup$ – Jack D'Aurizio Feb 5 '17 at 20:29
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    $\begingroup$ @TitoPiezasIII I encountered the given hypergeometric series while in the process of researching an integral transform which may be used to evaluate new classes of infinite series involving harmonic numbers. Using this integral transform, I discovered that $$\sum_{n=1}^{\infty} \left( \frac{4^{n}}{n(2n+1)\binom{2n}{n}} \right)^{2}H_{2n}'$$ may be expressed in a natural way in terms of $${}_{4}F_{3}\left(1,1,1,\frac{3}{2};\frac{5}{2};\frac{5}{2};\frac{5}{2};1\right),$$ letting $H_{2n}'=H_{2n}-H_{n}$ for $n \in \mathbb{N}$. $\endgroup$ – John M. Campbell Feb 8 '17 at 20:19

A complete answer now.

If we exploit the identities $$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx \tag{1}$$ $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{1}{2}\sum_{n\geq 1}\frac{4^n x^{2n-1}}{n\binom{2n}{n}},\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2\binom{2n}{n}}\tag{2}$$ we get: $$(\pi-2)=\int_{0}^{\pi/2}\theta^2\sin(\theta)\,d\theta = \frac{1}{2}\sum_{n\geq 1}\frac{16^n}{(2n+1)n^2 \binom{2n}{n}^2}=\frac{1}{2}\sum_{n\geq 0}\frac{16^n}{(2n+3)(2n+1)^2\binom{2n}{n}^2} $$ and in a similar way: $$\begin{eqnarray*}\frac{7\pi}{9}-\frac{40}{27}=\int_{0}^{\pi/2}\theta^2\sin^3(\theta)\,d\theta=\frac{1}{2}\sum_{n\geq 1}\frac{4^n 4^{n+1}}{n^2 (2n+3)\binom{2n}{n}\binom{2n+2}{n+1}}\end{eqnarray*}$$ If we integrate $\arcsin^2(x)$ and exploit $(1)$, we get: $$ \sum_{n\geq 1}\frac{16^n}{(2n+1)^2 n^2 \binom{2n}{n}^2} = 4(\pi-3) $$ and maybe it is enough to integrate $\arcsin^2(x)$ once more to get a closed expression for the series of interest: $$ \sum_{n\geq 0}\frac{16^n}{(2n+3)^3(2n+1)^2\binom{2n}{n}^2}. $$ In such a case it appears a dependence on a dilogarithm, arising from the primitive of $\frac{\arcsin x}{x}\sqrt{1-x^2}$. At the moment I do not know if that is manageable or not, I have to carry out further experiments. Probably a logarithm appears from $\int_{0}^{\pi/2}\theta\cot(\theta)\,d\theta=\frac{\pi}{2}\log(2).$

Now that the path to the answer is a bit more clear, let us put $(1)$ and $(2)$ in a slightly more convenient way: $$ \int_{0}^{\pi/2}\sin(x)^{2n+3}\,dx = \frac{4^{n}(2n+2)}{(2n+3)(2n+1)\binom{2n}{n}}\tag{1bis}$$ $$\arcsin^2(x)=\frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+2}}{(2n+2)(2n+1)\binom{2n}{n}}\tag{2bis}$$ If we integrate both sides of $(2\text{bis})$ we get: $$ -2x+2\sqrt{1-x^2}\arcsin(x)+x\arcsin^2(x) = \frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+3}}{(2n+3)(2n+2)(2n+1)\binom{2n}{n}}\tag{3}$$ We just have to gain an extra $\frac{1}{(2n+3)}$ factor. For such a purpose, we divide both sides of $(3)$ by $x$ and perform termwise integration again, leading to: $$ -4x+2\sqrt{1-x^2}\arcsin(x)+x\arcsin^2(x)+2\int_{0}^{\arcsin(x)}\frac{u\cos^2(u)}{\sin(u)}\,du\\= \frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+3}}{(2n+3)^2(2n+2)(2n+1)\binom{2n}{n}}\tag{4}$$ Now we evaluate both sides of $(4)$ at $x=\sin\theta$ and exploit $(1\text{bis})$ to perform $\int_{0}^{\pi/2}(\ldots)\, d\theta$.
That leads to: $$ \sum_{n\geq 0}\frac{16^n}{(2n+3)^3(2n+1)^2\binom{2n}{n}^2}=(\pi-4)+\int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u\cos^2(u)}{\sin(u)}\,du\,d\theta\tag{5} $$ and we may start buying beers, since the last integral boils down to $\int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u}{\sin u}\,du\,d\theta$, that is well-known. We get: $$\boxed{\begin{eqnarray*}\phantom{}_4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)&=&27\sum_{n\geq 0}\frac{16^n}{(2n+3)^3 (2n+1)^2 \binom{2n}{n}^2}\\&=&\color{red}{\frac{27}{2}\left(7\,\zeta(3)+(3-2K)\,\pi-12\right)}\end{eqnarray*}}\tag{6}$$ where $K$ is Catalan's constant. Please, do not ask me to do the same for other values of $\phantom{}_4 F_3$.
However, this instantly goes in my best of collection.

Addendum (15/08/2017) This result, together with another interesting identity relating $\phantom{}_4 F_3$ and $\text{Li}_2$, is going to appear on Bollettino UMI. You may have a glance at it on Arxiv.

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    $\begingroup$ You knocked-out all math software (Mathematica, Maple, and Mathcad) by computing this closed form!!! BRAVO - Jack - BRAVO. And the method is very instructive!!! Please make sure to publish it. (+1) $\endgroup$ – Hazem Orabi Feb 6 '17 at 8:57
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    $\begingroup$ @HazemOrabi: I am quite fond of this proof, and ready to buy a shirt saying better than Mathematica. I do not know if this proof deserves publishing as-it-is, I actually just used Parseval's theorem in a non conventional form. But if the method leads to an explicit evaluation of other interesting series, that might be the case. $\endgroup$ – Jack D'Aurizio Feb 6 '17 at 9:08
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    $\begingroup$ Hi Jack, I am a professional software developer with 17 years of experience and a member of British Computer Society. Up to my knowledge, all CAD systems had a special library to compute/simplify hyper-geometric functions considering the special values (at least I am sure about this in Wolfram from its technical documentation). With a development of an algorithm, the method you introduced is applicable to be generalized for calculating a closed form for a class of hyper-geometric functions. Generalize-it and Publish-it, I know what I see, it is an achievement. BRAVO again and good luck. Hazem $\endgroup$ – Hazem Orabi Feb 6 '17 at 12:02
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    $\begingroup$ @JackD'Aurizio "Please, do not ask me to do the same for other values of ${_4F_3}$"... I've actually been curious for some time now if somebody somewhere has compiled a list of closed forms for ${_4F_3}$'s at low half-integer parameters, and classified by complexity. At the very least, it would be nice to see the individual results scattered about this site gathered in one place... (BTW, speaking as a person who is impressed by your work on a regular basis, I agree this one belongs on the highlights reel ;) ) $\endgroup$ – David H Feb 9 '17 at 14:34
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    $\begingroup$ @HazemOrabi: I have to thank you again for the encouragement, since I am really going to publish it, together with another interesting identity. $\endgroup$ – Jack D'Aurizio Aug 16 '17 at 0:46

Preliminary. This can be viewed as a generalization of OP's problem, or an extension of Jack's answer's structure, or a brief elaboration on David H's and Hazem Orabi's further requests in comments. But essentially, this is a beautiful garden lying between the territory of generalized hypergeometric series and Multiple Zeta Values.

General Principle. Let $A$ (resp. $M, N, B$) be a vector with all components in $\mathbb Z/2$ (resp. $\mathbb N, \mathbb N, \mathbb C$), $A, M$ and $B, N$ are of same length, $S, T$ vectors that met one of five following conditions ($k,m,n,i,j\in\mathbb Z$):

$$\color{blue}{0.\ S=\{k\},\ T=\emptyset}\ \ \ \ \color{green}{1.\ S=\{k+1/2\},\ T=\emptyset}\ \ \ \ \color{purple}{2.\ S=\{k,m\},\ T=\{n+1/2\}}$$ $$\color{red}{3.\ S=\{k+1/2, m+1/2\},\ T=\{n\}}\ \ \color{orange}{4.\ S=\{k,m,n\},\ T=\{i+1/2,j+1/2\}}$$

Then the hypergeometric series $\, _{q+1}F_q(S,A,B;T,A+M,B-N;1)$, whenever convergent and non-terminating, is expressible via level 4 MZVs. OP's problem belongs to the last class, thus solvable without much effort.

For the statement's proof and examples, see Theorem 1 here.

Equivalent statement. For all rational $R(n)$ satisfying all its poles belong to $\{-\frac12,-1,-\frac32,-2,\cdots \}\cup\{\frac12, \frac32,\frac52,\cdots\}$, all $5$ classes of binomial sums $$\sum _{n=0}^{\infty } R(n) \left(\frac{\binom{2 n}{n}}{4^n}\right)^k\ (k=0, \pm1, -2),\ \pi \sum _{n=0}^{\infty } R(n) \left(\frac{\binom{2 n}{n}}{4^n}\right)^2$$ whevever convergent, are expressible via level 4 MZVs.

$_3F_2$ Phenomenon. Order $2$ is trivial due to Gauss. For order $3$, all $_3F_2(a,b,c;d,e;1)$, whenever all parameters in $\mathbb Z/2$ and the series non-terminating and convergent, is expressible via polylogarithms, Gamma function or level 4 MZVs. This is trivial since we may separate them into 12 classes by parity, where 10 of them have polylog closed-forms and remaining 2 are evaluated by Theorem 1 in paper above. We elaborate 12 examples, one for each case of parity:

$$\small\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;2,2;1\right)=\frac{16}{\pi }-4$$ $$\small \, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},2;1\right)=\pi -2$$ $$\small _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2};1\right)=\frac{\pi ^2}{8} $$ $$\small\, _3F_2(1,1,1;2,2;1)=\frac{\pi ^2}{6}$$ $$\small\, _3F_2\left(1,1,1;\frac{3}{2},2;1\right)=\frac{\pi ^2}{4} $$ $$\small _3F_2\left(1,1,1;\frac{3}{2},\frac{5}{2};1\right)=\frac{3 \pi }{2}-3 $$ $$\small\, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};1\right)=\frac{4 C}{\pi } $$ $$\small\, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};1\right)=2 C$$ $$\small\, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};1\right)=\frac{1}{2} \pi \log (2) $$ $$\small\, _3F_2\left(\frac{1}{2},1,1;2,2;1\right)=4-4 \log (2) $$ $$\small\, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},2;1\right)=2 \log (2) $$ $$\small\, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;1\right)=\frac{\pi }{\Gamma \left(\frac{3}{4}\right)^4}$$

A $_4F_3$ table. One may generate much more $_4F_3$ with half-integer parameters based on the principle above. As an example, the small table below consists of all $_4F_3$ with argument $1$ & all parameters in $\{1/2,1,3/2,2\}$ that has MZV- or Gamma-closed-form that I have known (feel free to add more).

$$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{\pi ^3}{48}+\frac{1}{4} \pi \log ^2(2)$$ $$\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2};1\right)=-16 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3 \pi ^3}{8}+\frac{1}{2} \pi \log ^2(2)$$ $$\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},2;1\right)=-8 C-32 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3 \pi ^3}{4}+4+\pi \log ^2(2)$$ $$\small\, _4F_3(1,1,1,1;2,2,2;1)=\zeta (3)$$ $$\small\, _4F_3\left(1,1,1,1;\frac{3}{2},\frac{3}{2},2;1\right)=2 \pi C-\frac{7 \zeta (3)}{2}$$ $$\small\, _4F_3\left(1,1,1,1;\frac{1}{2},2,2;1\right)=\frac{7 \zeta (3)}{4}+\frac{\pi ^2}{2}-\frac{1}{2} \pi ^2 \log (2)$$ $$\small\, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;1\right)=\frac{1}{2} \pi ^2 \log (2)-\frac{7 \zeta (3)}{4}$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{7 \zeta (3)}{8}$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2},2;1\right)=-\pi +2+\pi \log (2)$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;2,2,2;1\right)=8-\frac{16 \Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{7}{4}\right)}{\pi \Gamma \left(\frac{5}{4}\right)^2}$$ $$\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},2,2;1\right)=16 C-24+4 \pi$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},2;1,\frac{3}{2},\frac{3}{2};1\right)=\frac{\pi }{4}+\frac{1}{4} \pi \log (2)$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},2;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{7 \zeta (3)}{16}+\frac{\pi ^2}{16}$$ $$\small\, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;1\right)=\frac{\pi ^2}{3}-4 \log ^2(2)$$ $$\small\, _4F_3\left(1,1,1,\frac{1}{2};2,2,2;1\right)=-\frac{\pi ^2}{3}+8+4 \log ^2(2)-8 \log (2)$$ $$\small\, _4F_3\left(1,1,1,\frac{1}{2};2,2,\frac{3}{2};1\right)=4 \log (2)-\frac{\pi ^2}{6}$$ $$\small\, _4F_3\left(1,1,1,\frac{1}{2};2,\frac{3}{2},\frac{3}{2};1\right)=4 C-\frac{\pi ^2}{4}$$ $$\small\, _4F_3\left(1,1,1,\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\frac{7 \zeta (3)}{2}-\pi C$$ $$\small\, _4F_3\left(\frac{3}{2},\frac{3}{2},\frac{3}{2},1;2,2,2;1\right)=\frac{8 \pi }{\Gamma \left(\frac{3}{4}\right)^4}-8$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=4 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{\pi ^3}{32}-\frac{1}{8} \pi \log ^2(2)$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},2;1\right)=\frac{\pi ^2}{4}-2 \log (2)$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},2,2;1\right)=2 \pi -8+4 \log (2)$$ $$\small\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;2,2,2;1\right)=-32 C-16 \pi +48+16 \pi \log (2)$$ $$\small\pi \, _4F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;1\right)=16 \pi \log (2)-32 C$$ $$\small\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,2;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=C+2 \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{\pi ^3}{64}-\frac{1}{16} \pi \log ^2(2)$$ $$\small\pi \, _4F_3\left(1,1,\frac{1}{2},\frac{3}{2};2,2,2;1\right)=32 C+8 \pi -16-16 \pi \log (2)$$

$_5F_4,\ _6F_5,\ _7F_6,\ _8F_7$ examples. Observing the power of MZV, clearly $_4F_3$ should not be our limitation. Here are some examples of general principle, case 4, generalizing the original post:

$$\, _5F_4\left(1,1,1,1,\frac{3}{2};2,\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)=108 \pi C-\frac{567 \zeta (3)}{2}-162 \pi +540$$

$$\small \, _6F_5\left(1,1,1,1,1,\frac{3}{2};2,2,\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)=-432 \pi C-432 \pi \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+432 \text{Li}_4\left(\frac{1}{2}\right)+945 \zeta (3)+\frac{123 \pi ^4}{20}+540 \pi -1620+18 \log ^4(2)-\frac{9}{2} \pi ^2 \log ^2(2)$$

$$\scriptsize \, _7F_6\left(1,1,1,1,1,1,\frac{3}{2};2,2,2,\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)=1512 \pi C+2592 \pi \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+3456 \pi \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-2592 \text{Li}_4\left(\frac{1}{2}\right)-1728 \text{Li}_5\left(\frac{1}{2}\right)-3024 \zeta (3)+\frac{5859 \zeta (5)}{4}-\frac{81}{8} \pi \zeta \left(4,\frac{1}{4}\right)+\frac{81}{8} \pi \zeta \left(4,\frac{3}{4}\right)-\frac{369 \pi ^4}{10}\\ \scriptsize-1620 \pi +4536+\frac{72 \log ^5(2)}{5}-108 \log ^4(2)-6 \pi ^2 \log ^3(2)+27 \pi ^2 \log ^2(2)+\frac{123}{5} \pi ^4 \log (2)$$

Examples of other cases:

$$\small \, _7F_6\left(\frac{1}{2},1,\{\frac54\}_5;\frac{3}{2},\{\frac94\}_5;1\right)=-\frac{3125 C}{81}-\frac{96875 \zeta (5)}{96}-\frac{21875 \zeta (3)}{216}+\frac{756250}{243}-\frac{3125 \pi ^2}{648}-\frac{3125 \pi ^4}{864}-\frac{3125 \pi ^3}{864}-\frac{3125 \pi }{972}-\frac{15625 \pi ^5}{4608}-\frac{3125}{486} \log (2)+\frac{3125 }{2304}\left(\zeta \left(4,\frac{3}{4}\right)-\zeta \left(4,\frac{1}{4}\right)\right)$$

$$\small \, _8F_7\left(\{\frac12\}_4,\frac{7}{6},\frac{5}{4},\frac{4}{3},\frac{3}{2};\frac{1}{6},\frac{1}{4},\frac{1}{3},\{\frac52\}_4;1\right)=\frac{2835 \pi \zeta (3)}{32}-\frac{17739 \pi }{128}-\frac{1593 \pi ^3}{512}+\frac{945}{16} \pi \log ^3(2)-\frac{4779}{128} \pi \log ^2(2)+\frac{945}{64} \pi ^3 \log (2)-\frac{3645}{64} \pi \log (2)$$

$$\scriptsize \, _8F_7\left(\{\frac12\}_4,1,1,\frac{4}{3},\frac{5}{3};\frac{1}{3},\frac{2}{3},\{\frac32\}_4,\frac{5}{2};1\right)=-\frac{3}{8} S+\frac{3}{8} T-\frac{105 C}{64}+\frac{105}{16} \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3}{4} \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-3 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3 \zeta \left(4,\frac{3}{4}\right)}{2048}-\frac{3 \zeta \left(4,\frac{1}{4}\right)}{2048}+\frac{35 \pi ^5}{8192}+\frac{105}{128}-\frac{105 \pi ^3}{2048}+\frac{1}{512} \pi \log ^4(2)+\frac{1}{256} \pi \log ^3(2)+\frac{3 \pi ^3 \log ^2(2)}{1024}-\frac{105}{512} \pi \log ^2(2)+\frac{3 \pi ^3 \log (2)}{1024}$$

$$\tiny \pi \, _7F_6\left(\{-\frac12\}_2,\{1\}_5;\{2\}_6;1\right)=-\frac{2560}{9} S+\frac{9728}{27} T-\frac{47104 C}{243}-\frac{14336}{27} \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{32768}{27} \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{16384}{9} \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{256 \pi \zeta (3)}{27}-\frac{64}{9} \pi \zeta (3) \log (2)+\frac{32 \zeta \left(4,\frac{1}{4}\right)}{9}-\frac{32 \zeta \left(4,\frac{3}{4}\right)}{9}+\frac{4}{27} \zeta \left(4,\frac{1}{4}\right) \log (2)-\frac{4}{27} \zeta \left(4,\frac{3}{4}\right) \log (2)+\frac{25 \pi ^5}{9}+\frac{112 \pi ^3}{9}-\frac{46784 \pi }{729}+\frac{117248}{729}-\frac{1}{9} 32 \pi \log ^4(2)+\frac{512}{27} \pi \log ^3(2)+\frac{16}{3} \pi ^3 \log ^2(2)-\frac{448}{9} \pi \log ^2(2)-\frac{128}{9} \pi ^3 \log (2)+\frac{23552}{243} \pi \log (2)$$

Here $S,T$ denotes

$$\Im \sum_{k>j>0} \frac{i^k}{k^4 j},\ \ \Im \sum_{k>j>0} \frac{i^k (-1)^j}{k^4 j}$$

repsectively, which are irreducible level 4 MZVs. See paper linked above for more, where more considerably complicated $_6F_5,\ _7F_6$ and high order $_8F_7,\ _9F_8,\ _{10}F_9$ awaits.

End of story.

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