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Answering this question, @user64494 came with a very nice answer.

My problem is that the front factor is

   HypergeometricPFQ[{1/2, 1/2, 1, 1}, {3/4, 5/4, 3/2}, -1]

With an absolute error of $2.6\times 10^{-8}$, the $ISC$ proposes $$\sim \frac 1 {10}\,\frac{\Gamma \left(\frac{1}{8}\right) \,\,\Gamma \left(\frac{13}{20}\right)}{\Gamma \left(\frac{31}{40}\right)}$$

Is anyone able to identify it ? This is just for the sake of the art sake.

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3 Answers 3

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Clearly Yes exist closed-form. Using AskConstants

$$\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{4},\frac{5}{4},\frac{3}{2};-1\right)=\cosh ^{-1}\left(1+\sqrt{2}\right) \sec ^{-1}\left(\sqrt[4]{2}\right)$$

Block[{$MaxExtraPrecision = 700}, 
N[HypergeometricPFQ[{1/2, 1/2, 1, 1}, {3/4, 5/4, 3/2}, -1] - 
ArcCosh[1 + Sqrt[2]]*ArcSec[2^(1/4)], 200]]

(* 0.*10^-887 *)

Update:

Using Maple 2024:

simplify(evalc(Re(convert(hypergeom([1/2, 1/2, 1, 1], [3/4, 5/4, 3/2], -1), elementary))))

gives: $$2 \arcsin \! \left(\frac{\sqrt{2}\, \sqrt{2-\sqrt{2}}}{2}\right) \left(-\ln \! \left(2\right)+\ln \! \left(\left(2+\sqrt{2}\right) \sqrt{2-\sqrt{2}}+2^{\frac{3}{4}}\right)\right)$$

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  • $\begingroup$ How long did it take to return the output? $\endgroup$
    – azerbajdzan
    Commented Apr 17 at 15:45
  • $\begingroup$ @azerbajdzan On my low end computer 170 seconds. $\endgroup$
    – Mariusz Iwaniuk
    Commented Apr 17 at 15:51
  • $\begingroup$ Can you do the same with HypergeometricPFQ[{1/2,1,1},{5/4,7/4},-1] and HypergeometricPFQ[{1/2,2/3,1,7/6},{11/12,17/12,3/2},-1]? I am not sure I can manage to install the tool. It is also quite big 600MB zipped. $\endgroup$
    – azerbajdzan
    Commented Apr 17 at 16:30
  • 1
    $\begingroup$ @azerbajdzan, no need to use AskConstants! Using FunctionExpand: FullSimplify@FunctionExpand@HypergeometricPFQ[{1/2, 1, 1}, {5/4, 7/4}, -1] gives a nice expression :) $\endgroup$
    – Domen
    Commented Apr 17 at 17:00
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    $\begingroup$ @azerbajdzan $$\, _3F_2\left(\frac{1}{2},1,1;\frac{5}{4},\frac{7}{4};-1\right)=\frac{3 \left(\sqrt{2+\sqrt{2}} \tan ^{-1}\left(\sqrt{2 \left(1+\sqrt{2}\right)}\right)-\sqrt{2-\sqrt{2}} \log \left(1+\sqrt{2}+\sqrt{2 \left(1+\sqrt{2}\right)}\right)\right)}{2\ 2^{3/4}}$$,for: HypergeometricPFQ[{1/2,2/3,1,7/6},{11/12,17/12,3/2},-1] AskConstants and Maple Fails. $\endgroup$
    – Mariusz Iwaniuk
    Commented Apr 18 at 14:40
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Clearly no.

   N[HypergeometricPFQ[{1/2, 1/2, 1, 1}, {3/4, 5/4, 3/2}, -1], 64]
   0.8741268390368287270525776890493684248126217635582897428794914747

   N[1/10 (Gamma[1/8] Gamma[13/20])/Gamma[31/40], 64]
   0.8741268129594580557535792112246672815907840597914766996278009398
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Piggybacking on Mariusz's answer, I used AskConstants to find closed forms of $\, _4F_3\left(\ldots;\ldots;-x\right)$ for $x = 1/4, 1/2, 1$. After massaging the results at noticing patterns, it seems that

$ \, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{4},\frac{5}{4},\frac{3}{2};-x\right)=\frac{\tan ^{-1}\left(\sqrt{2 x \left(1+\sqrt{1+\frac{1}{x}}\right)}\right) \tanh ^{-1}\left(\sqrt{2 x \left(-1+\sqrt{1+\frac{1}{x}}\right)}\right)}{2 \sqrt{x}}$

for all complex $x$ away from the negative real line.

Functions to compare:

pFq[x_] := HypergeometricPFQ[{1/2, 1/2, 1, 1}, {3/4, 5/4, 3/2}, -x]
formula[x_] := 
  ArcTan[Sqrt[2x(1+Sqrt[1+1/x])]]ArcTanh[Sqrt[2x(-1+Sqrt[1+1/x])]]/(2Sqrt[x])

Some series terms:

Series[pFq[x], {x, 0, 3}]


PowerExpand[Normal[Series[formula[x], {x, 0, 3}]], Assumptions -> x > 0]

Some numerical verification:

Plot[{pFq[x], formula[x]}, {x, 0, 10}, PlotStyle -> {AbsoluteThickness[3], Directive[AbsoluteThickness[3], Dashed]}]

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    $\begingroup$ A better version of the formula (ArcTanh[Sqrt[2] Sqrt[x - Sqrt[x^2 - x]]] ArcTanh[Sqrt[2] Sqrt[x + Sqrt[x^2 - x]]])/(2 Sqrt[x]) which works for x<=1, i.e. in the whole range where the original hypergeometric function is real. $\endgroup$
    – azerbajdzan
    Commented Apr 17 at 19:49

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