Integration At Its Best

$\displaystyle \left( \int _{ 0 }^{ 1 }{ \dfrac { { x }^{ 2 }\cdot dx }{ \sqrt { 1-{ x }^{ 4 } } } } \right) \cdot \left( \int _{ 0 }^{ 1 }{ \dfrac { dx }{ \sqrt { 1+{ x }^{ 4 } } } } \right)$

The value of the expression above can be expressed in the form of $\dfrac { \pi }{ \beta \sqrt { \gamma } } .$ Find ${ \beta }^{ \gamma }.$