How to do the series expansion of terms with PolyLog faster?

Question

I have the following expression that I need to series expand, around t=0

(PolyLog[3, 1 + Tanh[J t]] - PolyLog[3, 1 - Tanh[J t]])Tanh[J t]

The amount of time to calculate one extra term is kind of just blowing up exponentially. I need at least 30 terms but for 20 terms only my time exceeded more than 30mins. To further speed up the process, I tried to use Assumptions command, with t>0, but it didn't help either. Infact I further tried to simply and do the series expansion of following expression around t=0

(PolyLog[3, 1 + t] - PolyLog[3, 1 - t])

and found that it is time taking too. I want to know if it is possible to simplify somehow the expression that I have in order to speed up the process? Or if there is some other way to find the Series expansion faster and efficiently?

Edit: I found with a little bit of experimentation that, the expansion of the Polylog is fast. In case instead of PolyLog[3, 1 + Tanh[J t]] if I had PolyLog[3, Tanh[J t]] then the series expansion around t=0 is pretty quick even up to 50 terms.

Answer 1

Not a full answer. Some hints.

Since function is symmetric, treat even and odd SeriesCoefficients differently.

f[t_, J_] = (PolyLog[3, 1 + Tanh[J t]] - 
     PolyLog[3, 1 - Tanh[J t]]) Tanh[J t];

f[x, J] == f[-x, J] // Simplify      (*   True   *)

(tab = Table[{2 n, 
      SeriesCoefficient[f[t, J], {t, 0, 2 n}, 
       Assumptions -> t > 0 && J > 0]}, {n, 1, 7}] // 
    Expand) // TableForm

(tab = Table[{2 n - 1, 
      SeriesCoefficient[f[t, J], {t, 0, 2 n - 1}, 
       Assumptions -> t > 0 && J > 0]}, {n, 1, 7}] // 
    Expand) // TableForm

Leave it to you to find a systematic dependence on 2 n and 2 n-1. This is only valid for t>0 && J>0.