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In the course of certain research (along the lines of my preprint https://arxiv.org/abs/1905.09228), I have obtained--as the result of a three-dimensional integration--a ("bound entanglement") probability formula \begin{equation} \label{bound} P= \frac{9 \sqrt{243-64 \sqrt{3}}-4 \Big(16 \coth ^{-1}\left(\frac{9}{\sqrt{81-\frac{64}{\sqrt{3}}}}\right)+A+B+C\Big)}{81 \sqrt{3}} \approx 0.08655423366978987, \end{equation} where \begin{equation} A=2 \log \left(\frac{1024}{243} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right) \log \left(27-\sqrt{729-192 \sqrt{3}}\right)-3 \log (48) \log (108) \end{equation} and \begin{equation} B=2 \log ^2\left(27+\sqrt{729-192 \sqrt{3}}\right)+3 \log \left(\frac{2187}{256}\right) \log \left(27+\sqrt{729-192 \sqrt{3}}\right) \end{equation} and, with the polylogarithmic function being employed, \begin{equation} C=8 \text{Li}_2\left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-8 \text{Li}_2\left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right). \end{equation} The Mathematica code for this formula is

1/(81 Sqrt[3]) (9 Sqrt[243 - 64 Sqrt[3]] - 4 (16 ArcCoth[9/Sqrt[81 - 64/Sqrt[3]]] - 3 Log[48] Log[108] + 2 Log[1024/243 (9 + Sqrt[81 - 64/Sqrt[3]])] Log[27 - Sqrt[729 - 192 Sqrt[3]]] + 3 Log[2187/256] Log[27 + Sqrt[729 - 192 Sqrt[3]]] + 2 Log[27 + Sqrt[729 - 192 Sqrt[3]]]^2 + 8 PolyLog[2, 1/18 (9 - Sqrt[81 - 64/Sqrt[3]])] - 8 PolyLog[2, 1/18 (9 + Sqrt[81 - 64/Sqrt[3]])]))

Interestingly, all the integers occurring above have prime decompositions involving only 2 and/or 3. I have done some work trying to utilize such decompositions in transformations of the logarithms, but have not obtained any striking simplifications (assuming such is possible).

Also, let us note that \begin{equation} \sqrt{729-192 \sqrt{3}}=3 \sqrt{81-\frac{64}{\sqrt{3}}}. \end{equation}

It should certainly be noted, as well that \begin{equation} \log \left(27+\sqrt{729-192 \sqrt{3}}\right)-\log \left(27-\sqrt{729-192 \sqrt{3}}\right)=2 \coth ^{-1}\left(\frac{9}{\sqrt{81-\frac{64}{\sqrt{3}}}}\right). \end{equation}

Also, \begin{equation} \log \left(27+\sqrt{729-192 \sqrt{3}}\right)-\coth ^{-1}\left(\frac{9}{\sqrt{81-\frac{64}{\sqrt{3}}}}\right)=3 \log (2)+\frac{3 \log (3)}{4}. \end{equation}

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Making use of the two identities added at the end of the question, we arrive at the considerably simpler formula, \begin{equation} P=\frac{16 (-4-9 \log (3)+\log (256)) \coth ^{-1}\left(\frac{9}{\sqrt{81-\frac{64}{\sqrt{3}}}}\right)}{81 \sqrt{3}} + \end{equation} \begin{equation} \frac{32 \left(\text{Li}_2\left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\text{Li}_2\left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)\right)+9 \sqrt{3} \sqrt{81-\frac{64}{\sqrt{3}}}}{81 \sqrt{3}}. \end{equation} The polylogs remain as in the original formula.

A discussion in https://mathoverflow.net/questions/347747/simplify-the-difference-of-two-dilogarithms-as-in-the-logarithmic-counterpart indicates that the polylogarithmic expression can not be further meaningfully simplified.

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