Calculating in closed form $\int _0^1\int _0^1\frac{1}{1+x y (x+y)} \ dx \ dy$

Question

Integrating with respect to a variable and then to the other one, things look pretty
complicated, but I'm sure you have ideas that might simplify the job to do here.
This time we're talking about

$$\int _0^1\int _0^1\frac{1}{1+x y (x+y)} \ dx \ dy$$

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The bounty moment: after 2 years and 8 months from the releasing moment of the question, it's time for a 300 points bounty for finding the simplest closed-form of the integral!

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Supplementary question: Here is an extension of the question for those with a need for more challenging questions.

Calculate

$$\int _0^1\int _0^1\cdots\int _0^1\frac{1}{1+x_1 x_2\cdots x_n (x_1+x_2+\cdots +x_n)} \ \textrm{d}x_1 \ \textrm{d}x_2\cdots \textrm{d}x_n, \ n\ge 2.$$

Last but not least, special greetings to Cleo!

Answer 1

Why do you think that there is a closed form? Most integrals do not have one, and Maple cannot find one. The numerical value is 0.79896482380785081628946784922318984550713669761340, which is not recognised by the Inverse Symbolic Calculator (https://isc.carma.newcastle.edu.au/advancedCalc). Given these facts, you should assume that there is no closed form unless you have a very good reason to think otherwise.

Answer 2

By symmetry (i.e. by exploiting the fact that our integral is twice the integral over the sub-region $0\leq y\leq x\leq 1$) we just have to compute:

$$ I=2 \int_{0}^{1}\int_{0}^{1}\frac{x}{1+x^3 y(1+y)}\,dy\,dx = 2 \int_{0}^{1}\int_{0}^{2}\frac{x}{(1+x^3y)\sqrt{1+4y}}\,dy\,dx$$ Integrating with respect to $x$ first,

$$\begin{eqnarray*} I &=& 2\int_{0}^{2}\frac{2 \sqrt{3}\arctan\left(\frac{\sqrt{3}}{1-2y^{1/3}}\right)-2\log\left(1+y^{1/3}\right)+\log\left(1-y^{1/3}+y^{2/3}\right)}{6 y^{2/3}\sqrt{1+4y}}\,dy\\&=&\int_{0}^{2^{1/3}}\frac{2\sqrt{3}\arctan\left(\frac{\sqrt{3}}{1-2y}\right)-3\log(1+y)+\log(1+y^3)}{\sqrt{1+4y^3}}\,dy\end{eqnarray*}$$ but the resulting integrals in just one variable do not look so appealing.

Am I missing some crucial simplification that follows from replacing $y$ with a Jacobi elliptic function (maybe $\text{dn}$) or with the Weierstrass elliptic function $\wp(z)$ (corresponding to $g_2=0,g_3=-1$) then exploiting some weird/mystical product formulas?