another product of log integral

Assuming one exists, and I think it does, find a closed form for:

$$\displaystyle \int_{0}^{1}\log(1+x)\log(1-x^{3})dx$$

From it, I did manage to derive:

$$\displaystyle -\gamma+2\gamma\log(2)-\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi\left(\frac{n+4}{3}\right)}{n(n+1)}$$

But, now I am stuck on the sum.

By merely switching the signs, an integral I found was:

$$\displaystyle\int_{0}^{1}\log(1-x)\log(1+x^{3})dx=-1/2\psi_{1}(1/3)+\log^{2}(2)-2\log(2)+\frac{5\pi^{2}}{36}-\frac{\pi}{\sqrt{3}}+6$$

But, the other way around appears to be more difficult for some reason.

This is probably not new to some, but while playing around with this I did manage to find various fun identities, such as:

$$\int_{0}^{1}x^{2n}\log(1+x)dx=\frac{1}{2(3n-1)}\left(H_{6n-2}-H_{3n-1}\right)$$ when $n$ is odd and

$$\int_{0}^{1}x^{2n}\log(1+x)dx=\frac{2\log(2)}{6n+1}-\frac{1}{6n+1}\left(H_{6n+1}-H_{3n}\right)$$ when $n$ is even.

Making the substitution $x \mapsto \frac{1-x}{1+x}$ gives $$I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}.$$

Now, we separate the integral into 7 pieces, of which only one is not elementary (containing dilogarithms): $$ \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2).$$

$1$st integral

$$\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12.$$

$2$nd integral

$$\int_0^1 \frac{\ln x}{(1+x)^2}dx=\sum_{n\geq1} \frac{(-1)^n}{n}=-\ln2.$$

$3$rd integral

$$\int_0^1 \frac{\ln(1+x)}{(1+x)^2}dx=-\frac{\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac{dx}{(1+x)^2}dx=\frac12-\frac{\ln2}{2}.$$

$4$th integral

$$\int_0^1 \frac{\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2x}{(1+x)(3+x^2)}dx \\=-\ln2+\ln3+2\operatorname{Re} \int_0^1 \frac{dx}{(1+x)(x+i\sqrt{3})}=-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}.$$

$5$th integral

$$\int_0^1 \frac{\ln x \ln(1+x)}{(1+x)^2}dx=-\frac{\ln x\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac1{1+x}\left(\frac{\ln(1+x)}{x}+\frac{\ln x}{1+x}\right)dx \\=-\ln2+\int_0^1 \frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{\pi^2}{12}-\frac12\ln^22-\ln2.$$

$6$th integral

$$\int_0^1 \frac{\ln^2(1+x)}{(1+x)^2}dx=-\frac{\ln^2(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2\ln(1+x)}{(1+x)^2}dx \\=1-\ln2-\frac12\ln^22.$$

$7$th integral

$$\int_0^1 \frac{\ln(1+x)\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(1+x)\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac1{1+x}\left(\frac{2x\ln(1+x)}{3+x^2}+\frac{\ln(3+x^2)}{1+x}\right)dx \\=-\ln^22-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}+2J$$

where $$ J=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx$$

Now, it is not hard to prove that $$\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right).$$

Putting $a=i\sqrt{3}$ then gives, after taking the real part, and assuming the principle value of the logarithm, $$\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})$$

Now we can "simplify" a bit: we have $\displaystyle \,\, \operatorname{Re}\text{Li}_2(e^{i\pi/3})=\frac{\pi^2}{36}$

and $\displaystyle \,\, \operatorname{Im}\text{Li}_2(e^{i\pi/3})=\frac1{2\sqrt{3}}\psi_1\left(\frac13\right)-\frac{\pi^2}{3\sqrt{3}}.$

Furthermore, using known dilogarithm identities we have $\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right),$

and since $\displaystyle \,\, \operatorname{Re}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)^2\, 3^n}=\frac14\text{Li}_2\left(-\frac13\right),$

we have $$\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)$$

and $$\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi}{12}\ln\left(\frac34\right)-\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).$$

Finally, putting everything together gives the expression I posted as a comment: $$I=\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )$$

I really hope that the dilogarithms can be simplified some more, but I can't see how for now.

I used your suggestion, tired, and it led to some fun results involving Euler- trig sums and polylogs. I am confident that the use of these will eventually lead to the solution to the integral listed above as well as other integral(s).

I list several I derived using the famous identities for $H_{n}$. I was mainly interested in the real parts due to the cos in the sums because they pertain to the integral in question. If anyone knows of any other closed forms, please feel free. i.e. the trilog in the 6th identity from the top or the dilog in the last one.

$$Li_{2}\left(\sqrt{3}e^{\pm \pi i/6}\right)=\frac{\pi^{2}}{9}\pm \left(2/3Cl_{2}(\frac{\pi}{3})+\frac{\pi}{3}\log(3)\right)i$$

$$\int_{0}^{1}x^{n}\log(1+x)dx=\frac{1}{2n}\left[H_{2n}-H_{n}\right], \;\ \text{n odd}$$

$$\int_{0}^{1}x^{n}\log(1+x)dx=\frac{1}{2n+1}\left(2\log(2)+H_{2n+1}-H_{n}\right), \;\ \text{n even}$$

$$\sum_{n=0}^{\infty}\frac{\cos(2\pi n/3)}{(2n+1)^{2}}=\frac{5\sqrt{3}}{12}Cl_{2}(\frac{\pi}{3})+\frac{\pi^{2}}{48}$$

$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n}x^{2n}=-\log(1+x)\log(1-x)+\sum_{n=1}^{\infty}\frac{H_{n}}{n}x^{2n}-\frac{1}{2}Li_{2}(x^{2})$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}\cos(\frac{2\pi n}{3})=\frac{\pi^{2}}{18}\log(3)-\frac{\pi}{9}Cl_{2}(\frac{\pi}{3})+\frac{5}{9}\zeta(3)-\Re\left[Li_{3}\left(\sqrt{3}e^{-\pi i/6}\right)\right]$$

$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n}\cos(\frac{2\pi n}{3})=\frac{1}{8}\log^{2}(3)-\frac{7}{72}\pi^{2}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n}\cos(\frac{2\pi n}{3})=\frac{1}{8}\log^{2}(3)-\frac{5}{72}\pi^{2}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{2n+1}\cos(\frac{2\pi n}{3})=\frac{-1}{4}\log^{2}(2)+\frac{5}{288}\pi^{2}-\frac{1}{32}\log^{2}(3)-\frac{\pi\sqrt{3}}{6}\log(2)+\frac{\pi\sqrt{3}}{48}\log(3)+\Re\left[e^{2\pi i/3}Li_{2}\left(\frac{\sqrt{3}}{2}e^{\pi i/6}\right)\right]$$

I have not gotten around to deriving $$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^{2}}\cos(\frac{2\pi n}{3})$$.

EDIT:

$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^{2}}\cos\left(\frac{2\pi n}{3}\right)=\frac{17}{18}\zeta(3)+\frac{7\pi}{9}Cl_{2}(\frac{\pi}{3})+\frac{5\pi^{2}}{18}\log(3)+2 \Re\left[Li_{3}\left(\sqrt{3}e^{\pi i/6}\right)\right]+\pi \Im\left[Li_{2}(-\sqrt{3}i)-Li_{2}(\sqrt{3}i)\right]-\Re\left[Li_{3}\left(\sqrt{3}e^{-\pi i/6}\right)\right]$$

Perhaps there is none, but if anyone does know of a closed form for any of those polylogs, please contribute.