Who's up to the challenge? 62

Changing the variables of integration gives $\begin{array}{rcl} \displaystyle I \; = \; \int_0^1\int_0^1 \frac{\ln^4\,xy}{(1+xy)^2}\,dx\,dy & = & \displaystyle \int_0^1\,dx \int_0^x \frac{ \ln^4\,z}{(1+z)^2}\,\frac{dz}{x} \\ & = & \displaystyle \int_0^1 \frac{\ln^4 z}{(1+z)^2} \left(\int_z^1 \frac{dx}{x}\right)\,dz \; = \; -\int_0^1 \frac{\ln^5\,z}{(1+z)^2}\,dz \\ & = & \displaystyle -F^{(5)}(1) \end{array}$ where $F(a) \; = \; \int_0^1 \frac{x^{a-1}}{(1+x)^2}\,dx \; = \; \tfrac12 + \tfrac12(a-1)\psi\big(\tfrac12a\big) - \tfrac12(a-1)\psi\big(\tfrac12(a+1)\big)$ for $a > 0$ . But then $F^{(5)}(a) \; = \; \tfrac1{64}(a-1)\psi^{(5)}\big(\tfrac12a\big) + \tfrac{5}{32}\psi^{(4)}\big(\tfrac12a\big) - \tfrac1{64}(a-1)\psi^{(5)}\big(\tfrac12(a+1)\big) - \tfrac{5}{32}\psi^{(4)}\big(\tfrac12(a+1)\big)$ so that $I \; = \; \tfrac{5}{32}\psi^{(4)}(1) - \tfrac{5}{32}\psi^{(4)}(\tfrac12) \; = \; \tfrac{225}{2}\zeta(5)$ making the answer $225 + 2 + 5 = \boxed{232}$ .

We know that $\large \frac{1}{(1+x)^{2}} = \sum_{r=1}^{\infty} x^{r-1}(-1)^{r-1}$ for $-1 < x<1$

So rewriting the double integral in a series format we have:-

$\large \int_{0}^{1}\int_{0}^{1}\sum_{r=1}^{\infty}r(-1)^{r-1}x^{r-1}y^{r-1}(\ln(x)+\ln(y))^{4}dxdy$

Using Binomial Theorem on $\large (\ln(x)+\ln(y))^{4}$

we have:-

$\large \int_{0}^{1}\int_{0}^{1}\sum_{r=1}^{\infty}r(-1)^{r-1}x^{r-1}y^{r-1}(\sum_{m=0}^{4} \binom{4}{m}(\ln(x))^{4-m}(\ln(y))^{m})dxdy$

Now let us generalize a result:-

$\large \int_{0}^{1} x^{k-1}(\ln(x))^{n} dx$

Substituting $x=e^{-t}$ we have

$\large \int_{0}^{\infty} (-1)^{n}e^{-tk}t^{n} dt$

Again substituting $kt=z$ we have

$\large \int_{0}^{\infty} \frac{(-1)^{n}e^{-z}z^{n}}{k^{n+1}} dz$

$\large =\frac{(-1)^{n}\Gamma(n+1)}{k^{n+1}}$

$\large =\frac{(-1)^{n}n!}{k^{n+1}}$

So solving the double integrals as iterated integrals by using the above result we have

$\Large \sum_{r=1}^{\infty}\sum_{m=0}^{4} (-1)^{r-1} \cdot r \cdot \binom{4}{m}\frac{(-1)^{4-m}(-1)^{m}(4-m)!m!}{r^{4-m+1}r^{m+1}}$

$\Large =\sum_{r=1}^{\infty}\sum_{m=0}^{4} (-1)^{r-1} \binom{4}{m}\frac{(4-m)!m!}{r^{5}}$

$\Large =\sum_{r=1}^{\infty}\sum_{m=0}^{4} (-1)^{r-1}\frac{4!}{r^{5}}$

$\Large =\sum_{r=1}^{\infty}\frac{(-1)^{r-1}(5 \cdot 4!)}{r^{5}}$

$\Large =\sum_{r=1}^{\infty}\frac{120(-1)^{r-1}}{r^{5}}$

$\large = 120\eta(5)$

Here $\eta(s)$ denotes the Dirichlet eta Function

Using the relation

$\large \eta(s) = (1-\frac{1}{2^{s-1}})\zeta(s)$

we have:-

$\large = 120 \cdot \frac{15}{16}\zeta(5)$

$\large = \frac{225}{2}\zeta(5)$

I would request @Mark Hennings to check if I have missed anything which might cause convergence issues.