A suprising conjectural closed-form of $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ and integral variations

I am going to establish two relations involving the following two integrals

$$I=\int_0^1\frac{x\text{Li}_2(x)\ln(1-x)}{1+x^2}\ dx$$

$$K=\int_0^1\frac{x\ln(x)\ln^2(1-x)}{1+x^2}\ dx$$

and solve for $I$ and $K$ by elimination.

The first relation

Start with using $\text{Li}_2(x)=-\int_0^1\frac{x\ln(y)}{1-xy}\ dy$

$$I=\int_0^1\frac{x\text{Li}_2(x)\ln(1-x)}{1+x^2}\ dx=-\int_0^1\ln(y)\left(\int_0^1\frac{x^2\ln(1-x)}{(1+x^2)(1-xy)}dx\right)dy$$

$$=-\int_0^1\ln(y)\left(\frac{1}{1+y^2}\left(G-\frac{\pi\ln(2)}{8}\right)+\frac{y}{1+y^2}\left(\frac{5\pi^2}{96}-\frac{\ln^2(2)}{8}\right)+\frac{\text{Li}_2\left(\frac{y}{y-1}\right)}{y(1+y^2)}\right)\ dy$$

$$=G^2-\frac{\pi\ln(2)}{8}G+\frac{5\pi^4}{4608}-\frac{\pi^2\ln^2(2)}{384}-\underbrace{\int_0^1\frac{\ln(y)\text{Li}_2\left(\frac{y}{y-1}\right)}{y(1+y^2)}\ dy}_{A}$$

$$A=\underbrace{\int_0^1\frac{\ln(y)\text{Li}_2\left(\frac{y}{y-1}\right)}{y}\ dy}_{A_1}-\underbrace{\int_0^1\frac{y\ln(y)\text{Li}_2\left(\frac{y}{y-1}\right)}{1+y^2}\ dy}_{A_2}$$

By integration by parts we have

$$A_1=-\frac12\int_0^1\frac{\ln^2(y)\ln(1-y)}{y(1-y)}\ dy=\frac12\sum_{n=1}^\infty H_n\int_0^1y^{n-1}\ln^2(y)\ dy=\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$$

For $A_2$, use Landen's identity

$$A_2=-\underbrace{\int_0^1\frac{y\ln(y)\text{Li}_2(y)}{1+y^2}\ dy}_{J}-\frac12\underbrace{\int_0^1\frac{y\ln(y)\ln^2(1-y)}{1+y^2}\ dy}_{K}$$

For $J$, use $\text{Li}_2(y)=-\int_0^1\frac{y\ln(t)}{1-ty}\ dt$ again

$$J=-\int_0^1\ln(t)\left(\int_0^1\frac{y^2\ln(y)}{(1+y^2)(1-yt)}dy\right)dt$$

$$=-\int_0^1\ln(t)\left(\frac{G}{1+t^2}+\frac{\pi^2 t}{48(1+t^2)}-\frac{\text{Li}_2(t)}{t}+\frac{t\text{Li}_2(t)}{1+t^2}\right)dt$$

$$=G^2+\frac{\pi^4}{2304}-\frac{\pi^4}{90}-J\Longrightarrow J=\frac12G^2-\frac{41\pi^4}{7680}$$

Collect all the pieces, we get

$$2I+K=G^2-\frac{\pi\ln(2)}{4}G-\frac{\pi^2\ln^2(2)}{192}-\frac{43\pi^4}{2880}\tag1$$

The second relation

We have the reflection identity $$\text{Li}_2(x)+\ln(x)\ln(1-x)=\zeta(2)-\text{Li}_2(1-x)$$

multiply both sides by $\frac{x\ln(1-x)}{1+x^2}$ then $\int_0^1$ we get

$$I+K=\zeta(2)\underbrace{\int_0^1\frac{x\ln(1-x)}{1+x^2}\ dx}_{B}-\underbrace{\int_0^1\frac{x\ln(1-x)\text{Li}_2(1-x)}{1+x^2}\ dx}_{C}$$

For both of $B$ and $C$, use $\Im \frac{1}{1-ix}=\frac{x}{1+x^2}$

$$B=\Im \int_0^1\frac{\ln(1-x)}{1-ix}\ dx\overset{1-x=t}{=}\Im\int_0^1\frac{\ln(t)}{1-i+it}\ dt$$

$$=-\Re\int_0^1\frac{i\ln(t)}{1-i+it}\ dt=-\Re\text{Li}_2\left(\frac{i}{i-1}\right)=\frac{\ln^2(2)}{8}-\frac{5\pi^2}{96}$$

where the last result follows from using the generalzaition

$$\int_0^1\frac{y\ln^{n}(x)}{1-y+yx}dx=(-1)^{n-1}n!\operatorname{Li}_{n+1}\left(\frac{y}{y-1}\right)$$ which can be found in the book Almost Impossible Integrals, sums and series in page 5.

$$C=\Im\int_0^1\frac{\ln(1-x)\text{Li}_2(1-x)}{1-ix}\ dx\overset{1-x=t}{=}\Im\frac{1}{1-i}\int_0^1\frac{\ln(t)\text{Li}_2(t)}{1-at}\ dt,\quad a=\frac{i}{i-1}$$

By using Cornel's identity

$$\int_0^1\frac{\ln(t)\text{Li}_2(t)}{1-at}\ dt=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$

we have

$$C=-\frac12G^2+\frac{\pi\ln(2)}{8}G+\frac{\pi^2\ln^2(2)}{128}-\frac{223\pi^4}{46080}+\frac{5}{128}\ln^4(2)+\frac{15}{16}\text{Li}_4\left(\frac12\right)$$

Combine the results of $B$ and $C$ we get

$$I+K=\frac12G^2-\frac{\pi\ln(2)}{8}G+\frac{5\pi^2\ln^2(2)}{384}-\frac{59\pi^4}{15360}-\frac{5}{128}\ln^4(2)-\frac{15}{16}\text{Li}_4\left(\frac12\right)\tag2$$

Solving $(1)$ and $(2)$ we get

$$I=\frac12G^2-\frac{\pi\ln(2)}{8}G-\frac{7\pi^2\ln^2(2)}{384}-\frac{511\pi^4}{46080}+\frac{5}{128}\ln^4(2)+\frac{15}{16}\text{Li}_4\left(\frac12\right)$$

and

$$K=\frac{\pi^2\ln^2(2)}{32}-\frac{167\pi^4}{23040}-\frac{5}{64}\ln^4(2)-\frac{15}{16}\text{Li}_4\left(\frac12\right)$$

The interesting thing about this solution is that I didn't use any harmonic series. Big thanks to Cornel for his magical identities I used in my solution.