I'm interested in evaluating integrals of positive integer powers of the dilogarithm function. I'd like to see the general case tackled if possible, or barring that then as many particular cases as possible.
Problem. For each $n\in\mathbb{N}$, evaluate the definite integral $\mathcal{I}_n$: $$\mathcal{I}_n:=\int_{0}^{1}\left[\operatorname{Li}_2{(x)}\right]^n\,\mathrm{d}x=\,???$$
The first two cases aren't that difficult to evaluate, but for reference I'll just state their values without proof:
$$\begin{align} \int_{0}^{1}\operatorname{Li}_2{(x)}\,\mathrm{d}x &=\zeta{(2)}-1\\ &=0.6449340668482264\dots; \end{align}$$
and
$$\begin{align} \int_{0}^{1}\operatorname{Li}_2{(x)}^2\,\mathrm{d}x &=\frac52\zeta{(4)}-4\zeta{(3)}-2\zeta{(2)}+6\\ &=0.6077123379430154\dots. \end{align}$$
The next case, $n=3$, is much harder than the previous two. However, by combining my answer to this question with Omran Kouba's answer to the same question, and assuming both answers are correct, I was able to infer that
$$\zeta{(5)}+6\zeta{(3)}+\frac{\pi^4}{15}-15=\frac{1}{3!}\int_{0}^{1}\left[\operatorname{Li}_2{(x)}-\zeta{(2)}\right]^3\,\mathrm{d}x.$$
Then by using the Binomial Theorem to expand the integrand on the RHS and by using the previous values of $\mathcal{I}_n$ for $n<3$, I was able to solve for the integral $\mathcal{I}_3$ and find the following value:
$$\begin{align} \mathcal{I}_3 &=\int_{0}^{1}\operatorname{Li}_2{(x)}^3\,\mathrm{d}x\\ &=6\zeta{(5)}+36\zeta{(3)}+\frac{\pi^6}{216}+\frac{19\pi^4}{60}+3\pi^2-2\pi^2\zeta{(3)}-90\\ &=0.6738641012555397\dots. \end{align}$$
But surely there is a less circuitous way to prove this. Can anybody offer a more direct proof that the integral $\mathcal{I}_3$ has the conjectured value indicated above?
Prove: $$\int_{0}^{1}\operatorname{Li}_2{(x)}^3\,\mathrm{d}x\stackrel{?}{=}6\zeta{(5)}+36\zeta{(3)}+\frac{\pi^6}{216}+\frac{19\pi^4}{60}+3\pi^2-2\pi^2\zeta{(3)}-90.$$
And finally, what of the $n>3$ cases? What is the largest $n$ for which $\mathcal{I}_n$ can be evaluated? Can $\mathcal{I}_4$ be evaluated? At the bottom of Omran Kouba's response to the question I referred to above, he says that, to his knowledge, the values of the integral of integer powers of the dilogarithm for powers larger than $2$ are not known. Clearly, for that statement to be true then, at the very least, it would have to be amended to reflect the fact that the value of the integral of the third power of the dilogarithm is indeed known (assuming I'm not somehow the first person to know it!). But are the values for $n>3$ really unknown?
Update: As SuperAbound was able to show, finding $\int_{0}^{1}\operatorname{Li}_2{(x)}^3\,\mathrm{d}x$ via repreated integration by parts actually isn't that difficult at all! For comparison, let's see how much harder finding $\int_{0}^{1}\operatorname{Li}_2{(x)}^4\,\mathrm{d}x$ is if we try to recycle the same method.
To evaluate $\int_0^1\operatorname{Li}_2{(x)}^4\mathrm{d}x$, integrate by parts using
$$\frac{d}{dx}\operatorname{Li}_2{(x)}^{4}=-\frac{4\ln{(1-x)}\operatorname{Li}_2{(x)}^3}{x}.$$
Then,
$$\begin{align} \mathcal{I}_4 &=\int_{0}^{1}\operatorname{Li}_2{(x)}^4\,\mathrm{d}x\\ &=\left[x\operatorname{Li}_2{(x)}^4\right]_{0}^{1}-\int_{0}^{1}x\left(-\frac{4\ln{(1-x)}\operatorname{Li}_2{(x)}^3}{x}\right)\,\mathrm{d}x\\ &=\operatorname{Li}_2{(1)}^4+4\int_{0}^{1}\ln{(1-x)}\operatorname{Li}_2{(x)}^3\,\mathrm{d}x\\ &=\zeta{(2)}^4+4\int_{0}^{1}\ln{(1-x)}\operatorname{Li}_2{(x)}^3\,\mathrm{d}x. \end{align}$$
Next, evaluate $\int_{0}^{1}\ln{(1-x)}\operatorname{Li}_2{(x)}^3\,\mathrm{d}x$ by integrating by parts again using
$$\int\ln{(1-x)}\,\mathrm{d}x=(x-1)\ln{(1-x)}-x+constant,$$
$$\frac{d}{dx}\operatorname{Li}_2{(x)}^3=-\frac{3\ln{(1-x)}\operatorname{Li}_2{(x)}^2}{x}.$$
Then,
$$\begin{align} \int_{0}^{1}\ln{(1-x)}\operatorname{Li}_2{(x)}^3\,\mathrm{d}x &=\left[\left((x-1)\ln{(1-x)}-x\right)\operatorname{Li}_2{(x)}^3\right]_{0}^{1}\\ &~~~~~-\int_{0}^{1}\left((x-1)\ln{(1-x)}-x\right)\left(-\frac{3\ln{(1-x)}\operatorname{Li}_2{(x)}^2}{x}\right)\,\mathrm{d}x\\ &=-\operatorname{Li}_2{(1)}^3\\ &~~~~~-3\int_{0}^{1}\left[\frac{\ln^2{(1-x)}}{x}-\ln^2{(1-x)}+\ln{(1-x)}\right]\operatorname{Li}_2{(x)}^2\,\mathrm{d}x\\ &=-\zeta{(2)}^3-3\int_{0}^{1}\frac{\ln^2{(1-x)}\operatorname{Li}_2{(x)}^2}{x}\,\mathrm{d}x\\ &~~~~~+3\int_{0}^{1}\ln^2{(1-x)}\operatorname{Li}_2{(x)}^2\,\mathrm{d}x-3\int_{0}^{1}\ln{(1-x)}\operatorname{Li}_2{(x)}^2\,\mathrm{d}x. \end{align}$$
The third integral in the last line above has the value,
$$\int_{0}^{1}\ln{(1-x)}\operatorname{Li}_2{(x)}^2\,\mathrm{d}x=2\zeta{(5)}+\frac{19}{2}\zeta{(4)}+12\zeta{(3)}+6\zeta{(2)}-4\zeta{(3)}\zeta{(2)}-30.$$
(to be continued...)