Indefinite Integration

If $a > 0, b > 0$ and $\int\dfrac{1}{ax^2 + b}dx = \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$, then $\int\dfrac{1}{bx^2 + a}dx = \ldots$