\(G\)はカタラン定数です。
\(\ln z\)の偏角は,\(-\pi <\arg z\leq \pi \)とします。
基本的な不定積分
$$
\int \frac {\ln ^3x}{x-a}dx
=\ln \left (1-\frac {x}a\right )\ln ^{3}x+3\operatorname {Li} _2 \left (\frac {x}a\right )\ln^2 x-6\operatorname{Li} _{3}\left (\frac {x}a\right )\ln x+6\operatorname{Li}_4\left (\frac{x}a\right)+Const.
$$
$$
\begin {aligned}
\int \frac {\ln ^{3}x}{x-a}
&=\ln \left (1-\frac {x}a\right )\ln ^{3}x-3\int \frac {\ln \left (1-\frac {x}a\right )\ln ^{2}x}xdx\\
&=\ln \left (1-\frac {x}a\right )\ln ^{3}x+3\operatorname{Li} _{2}\left (\frac {x}a\right )\ln ^{2}x-6\int \frac {\operatorname{Li} _{2}\left (\frac {x}a\right )\ln x}xdx \\
&= \ln \left (1-\frac {x}a\right )\ln ^{3}x+3\operatorname {Li} _2 \left (\frac {x}a\right )\ln ^{2}x-6\operatorname{Li} _{3}\left (\frac {x}a\right )\ln x+\int \frac {\operatorname{Li} _{3}\left (\frac {x}a\right )}xdx\\
&=\textcolor {blue}{\ln \left (1-\frac {x}a\right )\ln ^{3}x+3\operatorname{Li} _{2}\left (\frac {x}a\right )\ln ^{2}x-6\operatorname{Li} _{3} \left (\frac {x}a\right )\ln x+6\operatorname{Li} _{4} \left (\frac {x}a\right )+Const}.
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln ^{3}x}{1+x^{2}}dx&=-6\beta (4)
\end {aligned}
\\
\begin {aligned}
\int _{0}^{1}\frac {\ln ^{3}(1-x)}{1+x^{2}}dx
&=-6\Im \operatorname{Li} _{4}\left (\frac {1+i}2\right )
\end {aligned}
\\
\begin {aligned}
\int _{0}^{1}\frac {\ln ^{3}(1+x)}{1+x^{2}}dx
&=\frac {\pi \ln ^{3}2}2+\Im \left (-3\operatorname{Li} _{2}\left (1+i\right ) \ln ^{2}2+6\operatorname{Li} _{3}\left (1+i\right )\ln 2-6\operatorname{Li} _{4} \left (1+i\right )+6\operatorname{Li} _{4} \left (\frac {1+i}2\right )\right )
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln ^{3}x}{1+x^{2}}dx&=\Im \int _{0}^{1}\frac {\ln ^{3}x}{x-i}dx\\
&=6\Im \operatorname{Li} _{4}(i)\\
&=\textcolor {blue}{-6\beta (4)}.
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln ^{3}(1-x)}{1+x^{2}}dx
&=\Im \int _{0}^{1}\frac {\ln ^{3}(1-x)}{x-i}dx\\
&=\Im \int _{0}^{1}\frac {\ln ^{3}x}{1-i-x}dx\\
&=\textcolor {blue}{-6\Im \operatorname{Li} _{4}\left (\frac {1+i}2\right ) }.
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln ^{3}(1+x)}{1+x^{2}}dx
&=-\Im \int _{0}^{1}\frac {\ln ^{3}(1+x)}{x+i}dx\\
&=-\Im \int _{1}^2\frac {\ln ^{3}x}{x-1+i}dx\\
&=-\Im \left [\ln \left (1-\frac {x}{1-i}\right )\ln ^{3}x+3\operatorname {Li} _2 \left (\frac {x}{1-i}\right )\ln ^{2}x-6\operatorname{Li} _{3} \left (\frac {x}{1-i}\right )\ln x+6\operatorname{Li} _{4} \left (\frac {x}{1-i}\right )\right ]_1^2\\
&=-\Im \left (\ln \left (-i\right )\ln ^{3}2+3\operatorname{Li} _{2}\left (1+i\right ) \ln ^{2}2-6\operatorname{Li} _{3}\left (1+i\right ) \ln 2+6\operatorname{Li} _{4} \left (1+i\right )-6\operatorname{Li} _{4} \left (\frac {1+i}2\right )\right )\\
&=\textcolor {blue}{\frac {\pi \ln ^{3}2}2+\Im \left (-3\operatorname{Li} _{2}\left (1+i\right ) \ln ^{2}2+6\operatorname{Li} _{3}\left (1+i\right )\ln 2-6\operatorname{Li} _{4} \left (1+i\right )+6\operatorname{Li} _{4} \left (\frac {1+i}2\right )\right )}.
\end {aligned}
$$