Computation of $\int_0^1 \frac{\arctan^2 x\ln x}{1+x}dx$ I'm searching for a "simple" proof of:
\begin{align}\int_0^1 \frac{\arctan^2 x\ln x}{1+x}dx=-\frac{233}{5760}\pi^4-\frac{5}{48}\pi^2\ln ^2 2+\text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{16}\zeta(3)\ln 2+\frac{1}{24}\ln^4 2+\pi \Im\left(\text{Li}_3\left(1+i\right)\right)-\frac{1}{4}\text{G}\pi\ln 2\end{align}
The context:
I have written a script for Pari GP to search heuristically for certain close-forms.
Testing my script with the above integral i was lucky enough to find something.
The script searchs for integer linear relation between the integral and some constants:
\begin{align}\pi^4,\pi\ln^3 2,\pi^2\ln^2 2,\pi^3\ln 2,\text{Li}_4\left(\frac{1}{2}\right),\zeta(3)\ln 2,\zeta(3)\pi,\ln^4 2,\pi \Im\left(\text{Li}_3\left(1+i\right)\right), \Im\left(\text{Li}_3\left(1+i\right)\right)\ln 2,\text{G}^2,\text{G}\ln^2 2,\text{G}\pi^2,\text{G}\pi\ln 2\end{align}
NB:
Using the algorithm with $\displaystyle\int_0^1 \dfrac{\ln^2(1+x^2)\ln x}{1+x}dx$ gives someting too.
Addendum:
PARI GP script:
beta(n)={intnum(x=0,1,(-log(x))^(n-1)/(1+x^2))};
lindep4(x)={
NAME=["x","Pi^4","Pilog(2)^3","Pi^2log(2)^2","Pi^3log(2)","polylog(4,1/2)","zeta(3)log(2)","zeta(3)Pi","log(2)^4","Piimag(polylog(3,1+I))","log(2)imag(polylog(3,1+I))","Catalan^2","Catalanlog(2)^2","CatalanPi^2","Catalanlog(2)Pi","beta(4)","imag(polylog(4,1+I))"];
VAL=[x,Pi^4,Pilog(2)^3,Pi^2log(2)^2,Pi^3log(2),polylog(4,1/2),zeta(3)log(2),zeta(3)Pi,log(2)^4,Piimag(polylog(3,1+I)),log(2)imag(polylog(3,1+I)),Catalan^2,Catalanlog(2)^2,CatalanPi^2,Catalan*log(2)*Pi,beta(4),imag(polylog(4,1+I))];
L=lindep(VAL);
for(i=2,length(L),if(-L[i]/L[1]>0,print1("+",-L[i]/L[1],NAME[i]));if(-L[i]/L[1]<0,print1(-L[i]/L[1],NAME[i])));
}
for example:
\p 100
lindep4(intnum(x=0,1,atan(x)^2*log(x)/(1+x)))

NB: To improve the script add $\beta(4)$ value and some polygamma values.
PS: the PARI GP script has been updated to take into account more integrals.
Try this one: $\displaystyle \int_0^1\frac{\ln^2(1+x^2)\ln x}{1+x^2}dx$
 A: A framework proposed by Cornel (answer to the main integral)
The skeleton of the solution may be immediately obtained by using the strategy given for the generalization in Sect. $1.24$, page $14$, (Almost) Impossible Integrals, Sums, and Series (more precisely, see pages $142-145$).
So, following the suggested strategy, we have that
$$\int_0^1 \frac{\arctan^2(x)\log(x)}{1+x}\textrm{d}x$$
$$=\frac{1}{2}\int_0^1 \frac{\arctan^2(x)\log(x)}{x}\textrm{d}x-\frac{1}{2}\int_0^{\infty} \frac{\arctan^2(x)\log(x)}{x(1+x)}\textrm{d}x+\frac{\pi}{2}\int_0^1 \frac{\arctan(x)\log(x)}{1+x}\textrm{d}x$$
$$-\frac{\pi^2}{8}\int_0^1\frac{\log(x)}{1+x}\textrm{d}x.$$
The resulting integrals are either known or manageable.

*

*Behind the first resulting integral lies a very difficult harmonic series which is calculated by real methods here  https://math.stackexchange.com/q/3803762.


*The integral $\displaystyle \int_0^{\infty} \frac{\arctan^2(x)\log(x)}{x(1+x)}\textrm{d}x$ is reducible to the following integrals, $\displaystyle \int_0^1  \frac{x\log(1-x)}{1+x^2}\textrm{d}x$, $\displaystyle \int_0^1  \frac{x\log(1+x)}{1+x^2}\textrm{d}x$, $\displaystyle \int_0^1  \frac{x\log^2(x)\log(1-x)}{1+x^2}\textrm{d}x$, $\displaystyle \int_0^1  \frac{x\log^2(x)\log(1+x)}{1+x^2}\textrm{d}x$,
$\displaystyle \int_0^1  \frac{\log(x)\log(1-x)}{1+x^2}\textrm{d}x$,$\displaystyle \int_0^1  \frac{\log(x)\log(1+x)}{1+x^2}\textrm{d}x$, where most of them are known or easily reducible to known harmonic series like the ones given at the previous link (see the third resulting integral). Here one also needs to calculate $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}^{(3)}}{n}$ which is derived in a generalized form in the second theorem of the paper https://www.researchgate.net/publication/333339284_A_simple_strategy_of_calculating_two_alternating_harmonic_series_generalizations.
End of story
A: Another framework proposed by Cornel (answer to the second integral,  $\displaystyle \int_0^1\frac{\ln^2(1+x^2)\ln x}{1+x^2}\textrm{d}x$)
Observe that $$\int_0^1 \frac{1}{1+x^2}\log^3\left(\frac{2x}{1+x^2}\right)\textrm{d}x$$
$$=\log^3(2)\int_0^1\frac{1}{1+x^2}\textrm{d}x+3\log^2(2)\int_0^1\frac{\log(x)}{1+x^2}\textrm{d}x+3\log(2)\int_0^1\frac{\log^2(x)}{1+x^2}\textrm{d}x$$
$$+\int_0^1\frac{\log^3(x)}{1+x^2}\textrm{d}x-3\log^2(2)\int_0^1\frac{\log(1+x^2)}{1+x^2}\textrm{d}x-6\log(2)\int_0^1\frac{\log(x)\log(1+x^2)}{1+x^2}\textrm{d}x$$
$$-3\int_0^1\frac{\log^2(x)\log(1+x^2)}{1+x^2}\textrm{d}x+3\log(2)\int_0^1\frac{\log^2(1+x^2)}{1+x^2}\textrm{d}x$$$$-\int_0^1\frac{\log^3(1+x^2)}{1+x^2}\textrm{d}x+3\color{blue}{\int_0^1\frac{\log(x)\log^2(1+x^2)}{1+x^2}\textrm{d}x}.$$

*

*Note that the integral in the left-hand side may be beautifully reduced by the variable change $\displaystyle x\mapsto \frac{2x}{1+x^2}$ to $\displaystyle \int_0^1 \frac{1}{1+x^2}\log^3\left(\frac{2x}{1+x^2}\right)\textrm{d}x=\frac{1}{2}\int_0^1 \frac{\log^3(x)}{\sqrt{1-x^2}}\textrm{d}x$, where the last integral is a form involving the derivative of the Beta function.


*Note that all the other resulting integrals in the right-hand side are already known.


*To easily make the connection with the known integrals, for the integrals $\displaystyle \int_0^1\frac{\log^2(1+x^2)}{1+x^2}\textrm{d}x$ and $\displaystyle \int_0^1\frac{\log^3(1+x^2)}{1+x^2}\textrm{d}x$ make the variable change $x \mapsto \tan(x)$ to have a view in terms of trigonometric functions. A relevant link: About the integral $\int_{0}^{\pi/4}\log^4(\cos\theta)\,d\theta$


*Also, you might like to know the following generalized integral, $\displaystyle \int_0^1\frac{\log^{2n}(x)\log(1+x^2)}{1+x^2}\textrm{d}x$, is nicely presented and calculated by Ali Shadhar in his book, An Introduction To The Harmonic Series And Logarithmic Integrals: For High School Students Up To Researchers (see page $149$). The integral easily and naturally reduces to forms involving derivatives of the Beta function.
End of story
