<s>[INST] 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: $\displaystyle \int_0^1\frac{\ln^2(1+x^2)\ln x}{1+x^2}dx$
 [/INST] 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
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