<s>[INST] Find $\int _0^1\frac{12\arctan ^2 x\ln (\frac{(1-x)^2}{1+x^2})-\ln ^3(\frac{(1-x)^2}{1+x^2})}{x}\:dx$ I want to find and prove that: $$\int _0^1\frac{12\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)-\ln ^3\left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx=\frac{9 \pi ^4}{16}$$
I tried to split the integral:
$$=12\int _0^1\frac{\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx+12\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x^2\right)}{x}\:dx$$
$$-6\int _0^1\frac{\ln \left(1-x\right)\ln ^2\left(1+x^2\right)}{x}\:dx+\int _0^1\frac{\ln ^3\left(1+x^2\right)}{x}\:dx-8\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$
But the first $3$ integrals are very tough, expanding some terms yield very complicated series.
I also attempted to integrate by parts but it was rather messy and I'd prefer not to put it here. I have faith that there must be a trick for finding this one because of its closed forms simplicity.
Note $2$: I found this integral in a certain mathematics group, I've ask for hints but received none.
 [/INST] Integrals of such form (i.e. combination of $\arctan x, \log x, \log(1\pm x), \log(1+x^2$)) have systematic way of calculation. If you randomly write down such an integral, there is likely no slick way.
However, for this particular one, there is something behind. Writing $a=\arctan x, b=\log(\frac{(1-x)^2}{1+x^2})$, I will simultaneously prove OP's question as well as the following generalizations
$$\begin{aligned}I_3 &= \color{red}{\int_0^1 \frac{12a^2b-b^3}{x} dx = \frac{9\pi^4}{16}} \\
I_5 &= \int_0^1 \frac{-80 a^4 b+40 a^2 b^3-b^5}{x} dx = \frac{33 \pi ^6}{8} \\
I_7 &= \int_0^1 \frac{448 a^6 b-560 a^4 b^3+84 a^2 b^5-b^7}{x} dx = \frac{2193 \pi ^8}{32} \end{aligned}$$
Here numerator of $I_n$ is real part of $(2ai-b)^n$. Write
$$I_n=\int_0^1 \frac{\Re[(2ai-b)^n]}{x}dx = -2^n \int_0^1 \frac{\Re[(\log(1-x)-\log(1+ix))^n]}{x}dx $$
then for $n$ odd, $\frac{I_n}{\pi^{n+1}} \in \mathbb{Q}$. More precisely, if $n=2m+1$,

$$\tag{*}I_n = 2^{2n-2}3\pi^{n+1}(-1)^{m+1} \sum_{r=0}^m \binom{n}{2r+1}
\left(\frac{3}{8}\right)^{2r}
\frac{1}{n-2r}B_{n-2r}\left(\frac{3}{8}\right)$$

here $B_n(x)$ is Bernoulli polynomial, from which above cited values of $I_n$ follows.

Proof of $(*)$:
Integrate $\frac{(\log(1-z)-\log(1+iz))^n}{z}$ around quarter circle in the first quadrant, the integral from $0$ to $i$ is
$$\begin{aligned}2^n \int_0^i \frac{(\log(1-x)-\log(1+ix))^n}{x}dx &= 2^n \int_0^1 \frac{(\log(1-ix)-\log(1-x))^n}{x}dx \\
&= -2^n \int_0^1 \frac{(\log(1-x)-\log(1-ix))^n}{x}dx \end{aligned}$$
so the real part of $\int_0^i$ is $-I_n$. Hence, if $C$ is the quarter circle,
$$\begin{aligned} 2I_n &= 2^n\Re \int_C \frac{(\log(1-z)-\log(1+iz))^n}{z}dz \\
&= -2^n\Im \int_0^{\pi/2} (\log(1-e^{ix})-\log(1+ie^{ix}))^n dx 
\end{aligned}$$
Now $$\log(1-e^{ix})-\log(1+ie^{ix}) = \log\left(\frac{\sin(x/2)}{\sin(\pi/4-x/2)}\right)-\frac{3i\pi}{4}$$
Hence if we write $n=2m+1$,
$$I_n = 2^n \sum_{r=0}^m \binom{2m+1}{2r+1} (-1)^r\left(\frac{3\pi}{4}\right)^{2r+1}\int_0^{\pi/4} \log^{2(m-r)}\left(\frac{\sin x}{\sin(\pi/4-x)}\right) dx$$
As a lesser known fact, integrals like those in the summand are always rational multiple of $\pi^{2(m-r)+1}$:
$$\int_0^{\pi/4}\log^{2n}\left(\frac{\sin x}{\sin(\pi/4-x)}\right) dx = \pi^{2n+1}\frac{(-1)^{n+1}2^{2n+1}}{2n+1}B_{2n+1}\left(\frac{3}{8}\right)$$
QED.
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