If the solution curve $y = y ( x )$ of the differential equation $\left( 1 + y ^ { 2 } \right) \left( 1 + \log _ { e } x \right) d x + x d y = 0 , x > 0$ passes through the point $( 1,1 )$ and $y ( e ) = \frac { \alpha - \tan \left( \frac { 3 } { 2 } \right) } { \beta + \tan \left( \frac { 3 } { 2 } \right) }$, then $\alpha + 2 \beta$ is