Mutant Ellipsoid

By symmetry,

$\LARGE{\start{equation}V=8\int _{ 0 }^{ \frac { 1 }{ 4 } } \int _{ 0 }^{ \frac { 1 }{ 3 } (1-16x^{ 2 })^{ 1/4 } } \frac { 1 }{ 2 } (1-16x^{ 2 }-81y^{ 4 })^{ 1/6 }dydx}$

$\LARGE{=4\int _{ 0 }^{ \frac { 1 }{ 4 } } (1-16x^{2})^{1/6}dx\int _{ 0 }^{ \frac { 1 }{ 3 } (1-16x^{ 2 })^{ 1/4 } } (1-\frac{81y^{ 4 }}{1-16x^{2}})^{ 1/6 }dy}$

A substitution to get the integral into the form of the beta function.

$\LARGE{u = \frac{81y^{4}}{1-16x^{2}}}$ --> $\LARGE{du = \frac{324y^{3}}{1-16x^{2}}}dy$

$\LARGE{V=\frac{1}{3}\int_{0}^{1/4}(1-16x^{2})^{7/6-3/4}dx\int_{0}^{1}u^{1/4-1}(1-u)^{7/6-1}du}$

$\LARGE{=\frac{1}{3}B(\frac{1}{4},\frac{7}{6})\int_{0}^{1/4}(1-16x^{2})^{5/12}dx}$

And again:

$\LARGE{u = 16x^{2}}$ --> $\LARGE{du = 32x dx}$

$\LARGE{V=\frac{1}{96}B(\frac{1}{4},\frac{7}{6})\int_{0}^{1}4u^{1/2-1}(1-u^{17/12-1})du}$

$\LARGE{=\frac{1}{24}B(\frac{1}{4},\frac{7}{6})B(\frac{1}{2},\frac{17}{12})}$

$\LARGE{=\frac{\Gamma (\frac{1}{4})\Gamma (\frac{7}{6})\Gamma (\frac{1}{2})\Gamma (\frac{17}{12})}{24\times \Gamma (\frac{17}{12})\Gamma (\frac{23}{12})}=\frac{\Gamma(\frac{1}{4})\Gamma(\frac{7}{6})}{24\times \Gamma(\frac{23}{12})}\sqrt{\pi }=\frac{\frac{1}{6}\Gamma(\frac{1}{4})\Gamma(\frac{1}{6})}{24\times \frac{11}{12}\Gamma(\frac{11}{12})}\sqrt{\pi }=\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{6})}{132\times\Gamma(\frac{11}{12})}\sqrt{\pi }}$

$\LARGE{A+B+C+D+E+F+G=\color{#3D99F6}{1+4+1+6+132+11+12}=\color{#D61F06}{\boxed{167}}}$