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The following challenging integrals are coming from the new preprint "The derivation of eighteen special challenging logarithmic integrals" by Cornel Ioan Valean where they are nicely evaluated.

The following equalities hold: $$i) \ \int_0^1 \frac{\log(1-x) \log ^2(x) \log(1+x^2)}{x} \textrm{d}x$$ $$=\frac{1461}{256}\zeta (5)+\frac{7 }{128}\pi ^2 \zeta (3)+\frac{\pi ^5}{96}-\frac{\pi ^3}{8}G-\frac{\pi}{768}\psi ^{(3)}\left(\frac{1}{4}\right);$$ $$ii) \int_0^1 \frac{\log(1+x) \log ^2(x) \log(1+x^2)}{x} \textrm{d}x$$ $$=\frac{\pi}{768}\psi ^{(3)}\left(\frac{1}{4}\right)+\frac{\pi ^3}{8}G-\frac{\pi ^5}{96}-\frac{5}{128}\pi ^2 \zeta (3)-\frac{1515}{256}\zeta (5);$$ $$iii) \ \int_0^1 \frac{\log(1-x) \log ^4(x) \log(1+x^2)}{x} \textrm{d}x$$ $$=\frac{121557}{1024}\zeta (7)+\frac{105}{512}\pi ^2 \zeta (5)+\frac{121}{2560}\pi ^4 \zeta(3)+\frac{9}{320}\pi^7-\frac{5}{32}\pi^5G-\frac{\pi^3}{512}\psi ^{(3)}\left(\frac{1}{4}\right)-\frac{\pi}{20480}\psi ^{(5)}\left(\frac{1}{4}\right);$$ $$iv) \ \int_0^1 \frac{\log(1+x) \log ^4(x) \log(1+x^2)}{x} \textrm{d}x$$ $$=\frac{\pi^3}{512}\psi ^{(3)}\left(\frac{1}{4}\right)+\frac{\pi}{20480}\psi ^{(5)}\left(\frac{1}{4}\right)+\frac{5}{32}\pi^5G-\frac{9}{320}\pi^7-\frac{75}{512}\pi ^2 \zeta (5)-\frac{119}{2560}\pi ^4 \zeta(3)-\frac{122283}{1024}\zeta (7);$$ $$v) \ \int_0^1 \frac{\log(1-x) \log ^6(x) \log(1+x^2)}{x} \textrm{d}x$$ $$=\frac{41184405}{8192} \zeta (9)+\frac{6615 }{4096}\pi ^2 \zeta (7)+\frac{1815}{4096} \pi ^4 \zeta (5)+\frac{2017}{14336} \pi ^6 \zeta (3)+\frac{479 }{3584}\pi ^9-\frac{61}{128} \pi ^7 G$$ $$ -\frac{25 }{4096} \pi ^5\psi ^{(3)}\left(\frac{1}{4}\right)-\frac{3 }{16384} \pi ^3\psi ^{(5)}\left(\frac{1}{4}\right)-\frac{\pi }{458752}\psi ^{(7)}\left(\frac{1}{4}\right); $$ $$ vi) \ \int_0^1 \frac{\log(1+x) \log ^6(x) \log(1+x^2)}{x} \textrm{d}x $$ $$ =\frac{\pi }{458752}\psi ^{(7)}\left(\frac{1}{4}\right)+\frac{3 }{16384} \pi ^3\psi ^{(5)}\left(\frac{1}{4}\right)+\frac{25 }{4096} \pi ^5\psi ^{(3)}\left(\frac{1}{4}\right)+\frac{61}{128} \pi ^7 G-\frac{479 }{3584}\pi ^9$$$$ -\frac{2015}{14336} \pi ^6 \zeta (3)-\frac{1785}{4096} \pi ^4 \zeta (5)-\frac{4725 }{4096}\pi ^2 \zeta (7)-\frac{41229675}{8192} \zeta (9), $$ where $\zeta$ is the Riemann zeta function, $\displaystyle \psi^{(n)}$ denotes the Polygamma function and $G$ represents the Catalan's constant.

Question: For example, it's not hard to see that \begin{equation*} \int_0^1 \frac{\log(1-x) \log ^2(x) \log(1+x^2)}{x} \textrm{d}x \end{equation*} \begin{equation*} =\frac{7}{48} \pi ^2 \zeta (3)-\frac{1}{4} \sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_{2 n}}{n^4}-\frac{1}{2} \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(2)}}{n^3}-\sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_{2 n}^{(3)}}{n^2}. \end{equation*}

How would you calculate individually each of the resulting harmonic series and then get the values of the given integrals? Are these series known in the mathematical literature?

A note that might be useful: one might establish a relation between $i)-ii)$, $iii)-iv)$ and $v)-vi)$ by adding them and observing that the resulting integrals are of the type,

\begin{equation*} \int_0^1 \frac{\log(1-x)\log^{2n}(x)\log(1+x)}{x} \textrm{d}x \end{equation*} \begin{equation*} =\frac{1}{2}(2n)!\left(1-\frac{1}{2^{2n+1}}\right)\sum_{k=1}^{2n} \zeta(k+1)\zeta(2n-k+2)-(2n)!\sum_{k=1}^{n}\left(1-\frac{1}{2^{2k-1}}\right)\zeta(2k)\zeta(2n-2k+3) \end{equation*} \begin{equation*} +\frac{1}{2^{2n+3}} (2n+3-2^{2n+3})(2n)!\zeta(2n+3), \end{equation*} which is a generalization that is established in the book (Almost) Impossible Integrals, Sums, and Series, see page 6.

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