A conjectured identity for tetralogarithms $\operatorname{Li}_4$

I experimentally discovered (using PSLQ) the following conjectured tetralogarithm identity: $$720 \,\text{Li}_4\!\left(\tfrac{1}{2}\right)-2160 \,\text{Li}_4\!\left(\tfrac{1}{3}\right)+2160 \,\text{Li}_4\!\left(\tfrac{2}{3}\right)+270 \,\text{Li}_4\!\left(\tfrac{1}{4}\right)+540 \,\text{Li}_4\!\left(\tfrac{3}{4}\right)+135 \,\text{Li}_4\!\left(\tfrac{1}{9}\right)\\ \stackrel?=19 \pi ^4+30 \left(\pi ^2-\beta ^2\right) \left(10 \alpha ^2-12 \alpha \beta +3 \beta ^2\right)-30 \,\alpha ^2 \left(19 \alpha ^2-24 \alpha \beta +8 \beta ^2\right)\!,\\ \text{where $\alpha=\ln 2, \,\, \beta=\ln 3$.}$$ Numeric computations show that the absolute value of the difference between the left and right sides is smaller than $10^{-10^5}$, so I believe the identity must hold exactly. Moreover, this is the simplest vanishing linear combination with integer coefficients of elementary terms and tetralogarithm values at rational points in the interval $(0,1)$ that I found.

How can we prove it?


How can we prove it?

This might be a viable method:

From "Structural Properties of Polylogarithms, Leonard Lewin, p.30 sec.3, Polylogarithmic Ladders" we have

Lewin

However, due to restrictions on viewing I can't see the conditions required for $x,y,\eta,\xi$, checking in Mathematica it's not convinced its true for most combinations.

If we can understand that identity, we can probably pick a combination of $\{x,y,\eta,\xi\}\in\{\pm 1,\pm 2,\pm 3\}$ such that all of the relevant terms are generated from your example and the identity remains true and then use either $$ \frac{1}{8}\mathrm{Li}_4(z^2)=\mathrm{Li}_4(z)+\mathrm{Li}_4(-z) $$ or $$ \mathrm{Li}_4(z)=-\mathrm{Li}_4\left(\frac{1}{z}\right) - \frac{16\pi^4}{4!}B_4\left(\frac{\log(z)}{2\pi i}\right) $$ where $B_n(x)$ are Bernoulli Polynomials, to reduce the identity to your expression. We should then find that the resulting polynomial factors generated from this expression, when cancelled down, recreate your identity.

It does look like the right hand side of your conjecture will be generated by the Bernoulli polynomials, for example the first and last terms of your L.H.S $$ - 720\frac{16\pi^4}{4!}B_4\left(\frac{\log(\frac{1}{2})}{2\pi i}\right) - 135\frac{16\pi^4}{4!}B_4\left(\frac{\log(\frac{1}{9})}{2\pi i}\right) = 19\pi^4 + 30\pi^2(4\alpha^2+3\beta^2)-60i\pi(2\alpha^3+3\beta^3)-30(\alpha^4+3\beta^4) $$ may well be the origin of the $19\pi^4$!

There are also more identities on page 19 of this link