Closed form of an integral using Mathematica or otherwise

Define $$I=-\int_0^1\int_0^1 \frac{x^2(1-x)y^2(1-y)(2(1-xy)+(1+xy)\log(xy))^3}{(1-xy)^7}\ dxdy$$ Now using Wolfram Alpha $$I\approx 0.00133186$$.

Using Mathematica or otherwise, I need to find a closed form of $$I$$, possibly in terms of some special functions such as the Riemann zeta function.

Now we use $$\frac{1}{(1-xy)^7}=\sum_{k=0}^{\infty} (-1)^k \binom{k+7}{k} (xy)^k$$ So we obtain, $$I=-\frac{d}{d\varepsilon}\int_0^1\int_0^1 x^2(1-x)y^2(1-y)(2(1-xy)+(1+xy)(xy)^\varepsilon)^3\sum_{k=0}^{\infty} (-1)^k \binom{k+7}{k} (xy)^k \ dxdy\Biggr|_ {\varepsilon=0}$$ Any help will be highly appreciated. Thank you!

We can substitute $$\left\{~ u = 1 - xy,~ v = y ~\right\}$$ using IntegrateChangeVariables to get a nice expression for the integral:

integrand = (
x^2 (1 - x) y^2 (1 - y) (2 (1 - x*y) + (1 + x*y) Log[x*y])^3)/(1 -
x*y)^7;
int = Inactive[Integrate][integrand, {x, 0, 1}, {y, 0, 1}];

icv = IntegrateChangeVariables[int, {u, v}, {u == 1 - x*y, v == y}]


And now Activate (and multiply by the -1 out front):

result = -1*Activate[icv]

(*62/45 + (58 \[Pi]^4)/225 - (4 Zeta[3])/3 - 24 Zeta[5]*)



Giving $$\frac{4 \zeta (3)}{3}+24 \zeta (5)-\frac{62}{45}-\frac{58 \pi ^4}{225}$$

where $$\zeta (s)$$ is the Riemann zeta function.

result also agrees with your numerical approximation:

N @ result

(*0.00133186*)