I am looking for a closed form for
$$I=\int_0^1\frac{(x^3-3x^2+x)\log(x-x^2)}{(x^2-x+1)^3}\mathrm dx\approx 0.851035604949$$
Wolfram does not evaluate $I$.
I suspect $I$ has a closed form, because if $I$ has a closed form then the answer to my question
In Pascal's triangle without the $1$s, what is the sum of squares of reciprocals? has a closed form. In that question, I give reasons for why I suspect a closed form answer. But of course I could be wrong.
My attempt
Here is the graph of $y=\frac{(x^3-3x^2+x)\log(x-x^2)}{(x^2-x+1)^3}$.
I thought about translating the function to an odd function in order to take advantage of rotational symmetry, but the curve clearly does not have rotational symmetry.
I tried substitution and Maclaurin series, but got nowhere.
Best Answer
$$I=\int_0^1\frac{x(1-3x+x^2)\ln(x-x^2)}{(1-x+x^2)^3}dx=\frac{2\pi}{9\sqrt 3}-\frac13+\frac{4}{3\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$ Where $\operatorname{Cl}_2(x)$ is the Clausen function.
$$I=\int_0^1\frac{x(1-3x+x^2)\ln(x-x^2)}{(1-x+x^2)^3}dx$$ $$\overset{x\to 1-x}=\int_0^1 \frac{(1-x)(-1+x+x^2)\ln(x-x^2)}{(1-x-x^2)^3}dx$$ $$\Rightarrow 2I=\int_0^1 \frac{(3x-3x^2-1)\ln(x-x^2)}{(1-x-x^2)^3}dx$$
$$\int_0^1 \frac{(3x-3x^2-1)\ln x}{(1-x+x^2)^3}dx\overset{x\to 1-x}=\int_0^1 \frac{(3x-3x^2-1)\ln (1-x)}{(1-x+x^2)^3}dx$$ $$\Rightarrow I=\int_0^1 \frac{(3x-3x^2-1)\ln x}{(1-x+x^2)^3}dx$$
$$=2\int_0^1 \frac{\ln x}{(1-x+x^2)^3}dx-3\int_0^1 \frac{\ln x}{(1-x+x^2)^2}dx$$
$$\left(\frac{x(1-x)(2x-1)}{3(1-x+x^2)^2}\right)'=\frac{2}{(1-x+x^2)^3}-\frac{3}{(1-x+x^2)^2}+\frac{\frac23}{1-x+x^2}$$ $$\Rightarrow I=\int_0^1 \ln x\left(\frac{x(1-x)(2x-1)}{3(1-x+x^2)^2}\right)'dx-\frac23\int_0^1 \frac{\ln x}{1-x+x^2}dx$$
$$\int_0^1 \ln x\left(\frac{x(1-x)(2x-1)}{3(1-x+x^2)^2}\right)'dx\overset{IBP}=\frac13\int_0^1 \frac{(1-x)(1-2x)}{(1-x+x^2)^2}dx=\frac{2\pi}{9\sqrt 3}-\frac13$$
$$\int_0^1 \frac{\ln x}{1-x+x^2}dx=\int_0^1 \frac{\ln x}{1-2\cos\left(\frac{\pi}{3}\right)x+x^2}dx=$$ $$=\frac{1}{\sin\left(\frac{\pi}{3}\right)}\sum_{n=1}^\infty \sin\left(\frac{n\pi}{3}\right)\int_0^1 x^{n-1} \ln x\, dx=-\frac{2}{\sqrt 3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)}{n^2}=-\frac{2}{\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$