Integration of the square root of a quadratic I am in the tricky situation of trying to integrate the following. $$\sqrt{4 a^2 (y-b)^2+c^4}$$ $a, b$ and $c$ are all known constants. Can anybody provide insight as to how to do this? I have tried to rearrange to fit the form: $$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$ But do not seem able to do so. Maybe excessive toiling has hidden an obvious answer from my eyes. Thanks for the help.

$$\sqrt{4 a^2 (y-b)^2+c^4}=\dfrac1{2|a|}\sqrt{(y-b)^2+\left(\dfrac{c^2}{2a}\right)^2}$$

Using Trigonometric substitutions , set $y-b=\dfrac{c^2}{2a}\cdot\tan u$

and use How to integrate $\sec^3 x \, dx$?

Or Indefinite integral of secant cubed