TITLE: Calculating in closed form $\int _0^1\int _0^1\frac{1}{1+x y (x+y)} \ dx \ dy$
QUESTION [22 upvotes]: 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 
$$\int _0^1\int _0^1\frac{1}{1+x y (x+y)} \ dx \ dy$$

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!

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!

REPLY [4 votes]: Let $a$ be real and $|a|\leq1$. Then we set 
$$
I_n(a):=\int^{a}_{0}\int^{a}_{0}x^{n}y^n(x-y)^ndxdy\textrm{, }n=0,1,2,\ldots.
$$
and
$$
f(a,t)=\int^{a}_{0}\int^{a}_{0}\frac{1}{1-txy(x-y)}dxdy
$$
Hence
$$
f(a,t)=\sum^{\infty}_{l=0}t^l\int^{a}_{0}\int^{a}_{0}x^{n}y^n(x-y)^ndxdy=\sum^{\infty}_{n=0}t^nI_n(a).
$$
Under the change of coordinates $x\rightarrow r\sin(\theta)$, $y\rightarrow r\cos(\theta)$, we get 
$$
I_n(a)=2\int^{\pi/4}_{0}\int^{a/\cos(\theta)}_{0}(r\sin(\theta))^{n}(r\cos(\theta))^{n}r^n(\sin(\theta)-\cos(\theta))^nrdrd\theta=
$$
$$
2\int^{\pi/4}_{0}\sin^n(\theta)\cos^n(\theta)(\sin(\theta)-\cos(\theta))^n\left(\int^{a/\cos(\theta)}_{0}r^{3n+1}dr\right)d\theta=
$$
$$
2\frac{a^{3n+2}}{3n+2}\int^{\pi/4}_{0}\sin^n(\theta)\cos^{-2n-2}(\theta)(\sin(\theta)-\cos(\theta))^nd\theta
$$
But (see Wolfram alpha) it holds
$$
\int^{\pi/4}_{0}\sin^n(\theta)\cos^{-2n-2}(\theta)(\sin(\theta)-\cos(\theta))^nd\theta=\frac{2(-1)^n\Gamma(n+1)\Gamma(n+2)}{\Gamma(2n+3)}.
$$
Hence when $n=0,2,4,\ldots$ 
$$
I_n(a)=\frac{4(-1)^na^{3n+2}}{3n+2}\frac{\Gamma(n+1)\Gamma(n+2)}{\Gamma(2n+3)}
$$
and 
$$
I_n(a)=0
$$
when $n=1,3,5,\ldots$.
Consequently
$$
f(a,t)=2\sum^{\infty}_{n=0}\frac{a^{6n+2}}{6n+2}B(2n+1,2n+1)t^{2n}=a^2\cdot { }_4F_{3}\left(\frac{1}{2},\frac{1}{3},1,1;\frac{3}{4},\frac{5}{4},\frac{4}{3};\frac{a^6t^2}{16}\right)
$$
and clearly if $|a|\leq1$,
$|t|\leq 1$, then
$$
\int^{a}_{0}\int^{a}_{0}\frac{1}{1-txy(x-y)}dxdy=a^2\cdot { }_4F_{3}\left(\frac{1}{2},\frac{1}{3},1,1;\frac{3}{4},\frac{5}{4},\frac{4}{3};\frac{a^6t^2}{16}\right)\tag 1
$$
Note that, an interesting formula is
$$
\int^{\arctan(a)}_{0}\int^{b/\cos(\theta)}_{0}(r\sin(\theta))^n(r\cos(\theta))^n r^n (\sin(\theta)-\cos(\theta))^n rdrd\theta=
$$
$$
=\frac{(-1)^n b^{3n+2}}{3n+2}B(a,n+1,n+1)\textrm{, }Re(n)>-1
$$
Also for $|a|\leq 1$ and $n=0,1,2,\ldots$ it holds (see Wolfram alpha):
$$
\int^{a}_{0}\int^{a}_{0}(xy(x+y))^ndxdy=
$$
$$
=\int^{\pi/4}_{0}\int^{a/\cos(\theta)}_{0}(r\sin(\theta))^n(r\cos(\theta))^n r^n (\sin(\theta)+\cos(\theta))^n rdrd\theta=
$$
$$
=2\frac{a^{3n+2}}{3n+2}\cdot { }_2F_1\left(-n,n+1;n+2;-1\right)
$$
Hence
$$
J=\int^{a}_{0}\int^{a}_{0}\frac{1}{1-txy(x+y)}dxdy=2\sum^{\infty}_{n=0}\frac{a^{3n+2}}{3n+2}\cdot \frac{{ }_2F_{1}\left(-n,n+1;n+2;-1\right)}{n+1}t^n=
$$
$$
=2a^2\sum^{\infty}_{n=0}\frac{(-1)^{n+1}}{3n+2}\cdot B\left(-1,n+1,n+1\right)(a^3t)^n\tag 2
$$
where $|a|<1$ and $|t|\leq1$ or if $|a|\leq 1$ and $|t|<1$. 
Also is
$$
B(-1,n+1,n+1)=\int^{-1}_{0}t^{n}(1-t)^ndt=(-1)^{n+1}\int^{1}_{0}t^n(1+t)^ndt
$$
Having in mind the above integral we can write (easy)
$$
\int^{a}_{0}\int^{a}_{0}\frac{1}{1-txy(x+y)}dxdy=a^2\int^{1}_{0}{ }_2F_1\left(1,\frac{2}{3};\frac{5}{3};a^3tx(1+x)\right)dx.
$$
Setting $x(1+x)\rightarrow y$ and then $y\rightarrow 2w$, we arrive to
$$
J=2a^2\int^{1}_{0}\frac{{ }_2F_1\left(1,\frac{2}{3};\frac{5}{3};2a^3tw\right)}{\sqrt{1+8w}}dw.
$$
But 
$$
{ }_2F_1\left(1,\frac{2}{3};\frac{5}{3};x\right)=-\frac{2\log(1-x^{1/3})}{3x^{2/3}}+\frac{2(-1)^{1/3}\log\left(1-e^{-2\pi i/3}x^{1/3}\right)}{3x^{2/3}}-\frac{2(-1)^{2/3}\log\left(1-e^{2\pi i/3}x^{1/3}\right)}{3x^{2/3}}
$$
Hence if we set
$$
F(z)=\int^{1}_{0}\frac{\log(1-zx)}{\sqrt{1+8x^3}}dx,\tag 3
$$
we get
$$
J=-2\cdot 2^{1/3}t^{-2/3}F\left(2^{1/3}at^{1/3}\right)+2\cdot (-2)^{1/3}t^{-2/3}F\left(2^{1/3}e^{-2\pi i/3}at^{1/3}\right)-2\cdot (-1)^{2/3}2^{1/3}t^{-2/3}F\left(2^{1/3}e^{2\pi i/3}at^{1/3}\right)\tag 4
$$