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$$ \int_{0}^{1} \dfrac{\log x \log (1+x) \log (1+x+x^{2})}{(1-x)(1+x^{2})}\,\mathrm{d}x \approx -0.223434.$$

I am not sure if there exists a closed form but it seems worth trying. I am completely clueless on how to start with this beast. It is worth saying that$$1-x^3= (1-x)(1+x+x^2).$$ That seems to go no where. I think the integral can be represented as the derivative of the integral representation of the Hypergeometric function but I am not comfortable with that. Any ideas ?

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    $\begingroup$ That should depend on third derivatives of beta functions, by writing $(1+x)$ as $\frac{1-x^2}{1-x}$ and $1+x+x^2$ as $\frac{1-x^3}{1-x}$. $\endgroup$ – Jack D'Aurizio Apr 19 '17 at 16:58
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    $\begingroup$ We need @Cleo to provide a closed form :( . $\endgroup$ – Zaid Alyafeai Apr 19 '17 at 19:43
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    $\begingroup$ This ISC has no answer for -0.2234343508699 isc.carma.newcastle.edu.au $\endgroup$ – GEdgar Aug 22 '17 at 17:30
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    $\begingroup$ @phdmba7of12 I think if the OP provides a numerical approximation and explicitly asks for a closed form, then the goal of the question should be fairly clear. $\endgroup$ – Simply Beautiful Art Aug 22 '17 at 18:38
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    $\begingroup$ I don't understand the downvotes to be honest. $\endgroup$ – Zaid Alyafeai Aug 22 '17 at 20:21

I couldn’t find a closed form with traditional functions and constants. Therefore, as a note :

Be $\,m\in\mathbb{N}\,$, $\,a\in\mathbb{R}\,$ with $\,a>-3$ . $\mod{}$ means "Modulo" (see: division with rest).

And be $\,\delta_{i,j}\,$ the Kronecker delta which means: $\,\delta_{i,j}=1\,$ for $\,i=j\,$ and $\,\delta_{i,j}=0\,$ otherwise.

Using the Taylor series for $\,\ln(1+x) \,$, $\,\ln(1-x^m)\,$, $\,\displaystyle\frac {1}{1-x}\,$, $\,\displaystyle \frac {1}{1+x^2}$ we get:

$\displaystyle \int\limits_0^1 \frac{\ln(1+x) \ln(1-x^m)}{(1-x)(1+x^2)}x^a dx =$

$\displaystyle =\frac{m}{2}\sum\limits_{n=0}^\infty \frac{1}{n+3+a} \sum\limits_{k=0}^n (1+\sqrt{2}\sin(\frac{\pi}{4}(2n-2k+1)))\sum\limits_{v=0}^k \frac{(-1)^{k-v+1}\delta_{0,(\text{$v+1$ mod $m$)}}}{(k-v+1)(v+1)}$

With the derivation with respect to $\,a\,$ and then $\,a:=0\,$ follows:

$\displaystyle \int\limits_0^1 \frac{\ln(x) \ln(1+x) \ln(1+x+x^2)}{(1-x)(1+x^2)} dx = $

$\displaystyle =\int\limits_0^1 \frac{\ln(1+x) \ln(1-x^3)}{(1-x)(1+x^2)}\ln(x) dx - \int\limits_0^1 \frac{\ln(1+x) \ln(1-x)}{(1-x)(1+x^2)}\ln(x) dx $

$\displaystyle =\frac{1}{2}\sum\limits_{n=0}^\infty \frac{1}{(n+3)^2} \sum\limits_{k=0}^n (1+\sqrt{2}\sin(\frac{\pi}{4}(2n-2k+1)))\sum\limits_{v=0}^k \frac{(-1)^{k-v+1}(1-3\cdot\delta_{0,(\text{$v+1$ mod $3$) }})}{(k-v+1)(v+1)}$


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