# If possible, meaningfully simplify an expression involving logs, polylogs and hyperbolic trigonometric functions

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


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

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.