Tricky Area

Firstly, we use the substitution $y = -x^3$ to show that the area is equal to $A \; = \; \int_{-1}^0 \sqrt{1 + x^3}\,dx \; = \; \tfrac13\int_0^1 y^{-\frac23}(1-y)^{\frac12}\,dy \; = \; \tfrac13B(\tfrac13,\tfrac32) \;= \; \frac{\Gamma(\tfrac13)\Gamma(\tfrac32)}{\Gamma(\tfrac{11}{6})} \; = \; \frac{\sqrt{\pi}}{5}\,\frac{\Gamma(\tfrac13)}{\Gamma(\tfrac{5}{6})}$ using the identity $\Gamma(z+1) = z\Gamma(z)$ a couple of times, and the fact that $\Gamma(\tfrac12)= \sqrt{\pi}$ . The duplication formula $\sqrt{2\pi}\Gamma(2z) = 2^{2z-\frac12}\Gamma(z)\Gamma(z+\tfrac12)$ gives us that $A \; = \; \frac{\sqrt{\pi}}{5}\, \frac{\Gamma(\tfrac13)^2 2^{\frac13}}{\sqrt{2\pi}\Gamma(\tfrac23)} \; = \; \frac{\Gamma(\tfrac13)^2}{5\times 2^{\frac13}\Gamma(\tfrac23)}$ and then the relfection formula $\Gamma(z)\Gamma(1-z) = \pi\mathrm{cosec}\,\pi z$ tells us that $A \; = \; \frac{\Gamma(\tfrac13)^3\sin\tfrac13\pi}{5\pi2^{\frac13}} \; = \; \frac{\sqrt{3}}{5\pi2^{\frac43}}\Gamma(\tfrac13)^3 \; = \; \frac{\sqrt{3}\sqrt[3]{4}}{20\pi}\big(-\tfrac23\,!\big)^3$ making the answer $4+3+20-2 = \boxed{25}$