An anti-derivative of a product of three poly-logarithms and a simple fraction.

Question

It is fairly well known that integrating a function that involves products of polylogarithms is not always possible. The motivation for this question is to reduce a large class of such integrals to known functions and to "irreducible" quantities, i.e. such which cannot be expressed through the known functions.

Let $0 \le p\le q \le r$ be integers and let $t$ be real. Consider a following anti-derivative. \begin{equation} {\mathcal I}^{(p,q,r)}(t):= \int \frac{Li_p(t) Li_q(t) Li_r(t)}{t} dt \end{equation} By using integration by parts we have found the quantity above for some particular values of parameters in question. We have: \begin{eqnarray} &&{\mathcal I}^{(0,2 q+1,r)}(t) = \\ &&\sum _{j=0}^q \text{Li}_{r-2 j}(t) \left(\sum _{l=2 j+1}^{j+q} (-1)^{l-1} \binom{l-1}{2 j} \text{Li}_l(t) \text{Li}_{2 j-l+2 q+2}(t)+\frac{1}{2} (-1)^{j+q} \binom{j+q}{2 j} \text{Li}_{j+q+1}(t){}^2\right)+\\ &&\sum _{j=0}^q \text{Li}_{-2 j+r-1}(t) \sum _{l=2 j+2}^{j+q+1} (-1)^{l-1} \binom{l-1}{2 j+1} \text{Li}_l(t) \text{Li}_{2 j-l+2 q+3}(t)+\\ &&\sum _{j=0}^q (-1)^{j+q+1} \left(\binom{j+q}{2 j+1}+\frac{1}{2} \binom{j+q}{2 j}\right) \int \frac{\text{Li}_{j+q+1}(t){}^2 \text{Li}_{-2 j+r-1}(t)}{t} \, dt \end{eqnarray} Here $q\ge 1$ and $r\ge 2q+3$.

When running the Mathematica code below

Simplify[D[
   Table[
       Sum[(Sum[
          PolyLog[l, t] PolyLog[2 q + 2 + 2 j - l, t] (-1)^(l - 1)
            Binomial[l - 1, 0 + 2 j], {l, 1 + 2 j, 
           q + 0 + j}] + (-1)^(q + j)/
          2 Binomial[q + j, 2 j] PolyLog[q + 1 + j, t]^2) PolyLog[
        r - 2 j, t], {j, 0, q}] +
     Sum[(Sum[
         PolyLog[l, t] PolyLog[2 q + 3 + 2 j - l, t] (-1)^(l - 1)
            Binomial[l - 1, 1 + 2 j], {l, 2 + 2 j, 
          q + 1 + j}]) PolyLog[r - 1 - 2 j, t], {j, 0, q}] +
     Sum[(-1)^(
       q + 1 + j) (1/2 Binomial[q + j, 2 j] + 
         Binomial[q + j, 2 j + 1]) Integrate[
        PolyLog[q + 1 + j, t]^2  PolyLog[r - 1 - 2 j, t]/t, t], {j, 0,
        q}], {q, 1, 10}, {r, 2 q + 3, 20}]
   , t] - 
  Table[(PolyLog[0, t] PolyLog[2 q + 1, t] PolyLog[r, t])/
   t, {q, 1, 10}, {r, 2 q + 3, 20}]]

we get

{{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 
  0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 
  0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 
  0, 0}, {0, 0, 0, 0}, {0, 0}, {}, {}}

as it should be.

Now, apart from the question as to how the result looks like for all possible values of $p,q,r$ another question would be whether the residual integrals in the right hand side (meaning those that involve squares of poly-logarithms) can be reduced to poly-logarithms themselves or whether instead they are some "new" functions.