My question is simple, but maybe hard to answer. I would like to have a series expansion for $\text{Li}_3 (1-x)$ at $x \sim 0$ in the following form:
$$\text{Li}_3 (1-x) = \sum_{n=0} c_n x^n + \log x \sum_{m=1} c_m x^m. \tag{1}$$
The first few terms are:
$$\text{Li}_3(1-x) = \zeta (3)-\frac{\pi ^2 x}{6}+\left(\frac{3}{4}-\frac{\pi ^2}{12}\right) x^2+\left(\frac{7}{12}-\frac{\pi ^2}{18}\right) x^3+\left(\frac{131}{288}-\frac{\pi ^2}{24}\right) x^4+\left(\frac{53}{144}-\frac{\pi ^2}{30}\right) x^5+ \left(-\frac{x^2}{2}-\frac{x^3}{2}-\frac{11 x^4}{24}-\frac{5 x^5}{12}\right) \log x+O\left(x^6\right), \tag{2}$$
however Mathematica fails to go beyond $\mathcal{O}(x^{15})$. The coefficients for the $\log$ term are easy to guess:
$$c_m = \frac{H_{m-1}}{m}. \tag{3}$$
I also managed to obtain the coefficient with the $\pi^2$:
$$- \frac{\pi^2}{6n}, \tag{4}$$
but I am struggling with the remaining coefficient, especially with so few coefficients. Any idea what this coefficient could be?