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.

Let  $$\displaystyle I =  \int \sqrt{4a^2(y-b)^2+c^4}dy\;,$$ Let $(y-b) = t\;,$ Then $dy = dt$
So Integral $$\displaystyle I = \int \sqrt{4a^2t^2+c^4}dt = 2a\underbrace{\int \sqrt{t^2+k^2}dt}_{J}\;,$$ Where $\displaystyle k= \frac{c^2}{2a}$
For calculation of Integral $J\;,$ We Use Integration by parts.
Now Let $$\displaystyle J = \int \sqrt{t^2+k^2}\cdot tdt = \sqrt{t^2+k^2}\cdot t-\int\frac{t^2}{\sqrt{t^2+k^2}}dt$$
So $$\displaystyle J= \sqrt{t^2+k^2}\cdot t-\int\frac{(t^2+k^2)-k^2}{\sqrt{t^2+k^2}}dt = \sqrt{t^2+k^2}\cdot t-J+k^2\ln |t+\sqrt{t^2+k^2}|$$
So we Get $$\displaystyle J = \frac{1}{2}\sqrt{t^2+k^2}\cdot t+\frac{k^2}{2}\ln|t+\sqrt{t^2+k^2}|$$
So we get $$\displaystyle I = 2a\cdot J = a\cdot \sqrt{t^2+k^2}\cdot t+ak^2\cdot \ln|t+\sqrt{t^2+k^2}|$$
Where $\displaystyle k = \frac{c^2}{2a}$
