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.

Before applying Inverse Trigonometric Substitution, I prefer to preprocess the expression as follows
$$\begin{array}{lll}
\int\sqrt{4a^2(y-b)^2+c^4}dy &=& \displaystyle\int\sqrt{\frac{4a^2c^4(y-b)^2}{c^4}+c^4}dy\\
&=& \displaystyle \int c^2\sqrt{\frac{4a^2(y-b)^2}{c^4}+1}dy\\
&=& \displaystyle \int c^2\sqrt{\frac{(2ay-2ab)^2}{c^4}+1}dy\\
&=& \displaystyle \int c^2\sqrt{\bigg(\frac{2ay-2ab}{c^2}\bigg)^2+1}dy\\
\end{array}$$
Next, making the substitution
$$u=\frac{2ay-2ab}{c^2}\implies u'(y) = \frac{2a}{c^2}$$
we have
$$\int\frac{c^2\sqrt{\bigg(\frac{2ay-2ab}{c^2}\bigg)^2+1}du(y)}{\frac{2a}{c^2}} = \frac{c^4}{2a}\int \sqrt{u^2+1}du=\dots$$
