Integral of $\sqrt{a^2+b^2t^2}$

I'm trying to calculate mass of some line and this is the integral needed to be solved. Wolfram shows me some way with the fuction sec and reduction methods and I don't know how to use these. is there any other way that I miss?

$$\int\sqrt{a^2+b^2t^2}dt$$

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Let $t=\dfrac ab\tan\theta\;\Rightarrow\;dt=\dfrac ab\sec^2\theta\ d\theta$, the integral turns out to be \begin{align} \int\sqrt{a^2+b^2t^2}\ dt&=\int\sqrt{a^2+a^2\tan^2\theta}\cdot\dfrac ab\sec^2\theta\ d\theta\\ &=\dfrac {a^2}b\int\sec^3\theta\ d\theta\\ &=\dfrac {a^2}b\int\frac{1}{\cos^3\theta}\cdot\frac{\cos\theta}{\cos\theta}\ d\theta\\ &=\dfrac {a^2}b\int\frac{\cos\theta}{\cos^4\theta}\ d\theta\\ &=\dfrac {a^2}b\int\frac{\cos\theta}{(1-\sin^2\theta)^2}\ d\theta. \end{align} Now, let $u=\sin\theta\;\Rightarrow\;du=\cos\theta\ d\theta$, then $$\dfrac {a^2}b\int\frac{\cos\theta}{(1-\sin^2\theta)^2}\ d\theta=\dfrac {a^2}b\int\frac{1}{(1-u^2)^2}\ du.$$ The last integral admits a decomposition by partial fractions: $$\dfrac {a^2}b\int\frac{1}{(1-u^2)^2}\ du=\dfrac {a^2}{4b}\int\left(\frac{1}{1-u}+\frac{1}{(1-u)^2}+\frac{1}{1+u}+\frac{1}{(1+u)^2}\right)\ du$$ Integrating term-by-term, we will obtain \begin{align} \int\sqrt{a^2+b^2t^2}\ dt&=\dfrac {a^2}{4b}\left(\ln\left|\frac{1+\sin \theta}{1-\sin \theta}\right|+2\sec \theta\tan \theta\right)+C\\ &=\dfrac {a^2}{4b}\left(\ln\left|\frac{\sqrt{a^2+b^2t^2}+bt}{\sqrt{a^2+b^2t^2}-bt}\right|+\frac{2bt\sqrt{a^2+b^2t^2}}{a^2}\right)+C\\ &=\dfrac {a^2}{4b}\left(\ln\left|\frac{\sqrt{a^2+b^2t^2}+bt}{\sqrt{a^2+b^2t^2}+bt}\cdot\frac{\sqrt{a^2+b^2t^2}+bt}{\sqrt{a^2+b^2t^2}-bt}\right|+\frac{2bt\sqrt{a^2+b^2t^2}}{a^2}\right)+C\\ &=\dfrac {a^2}{2b}\left(\ln\left|\frac{\sqrt{a^2+b^2t^2}+bt}{a}\right|+\frac{bt\sqrt{a^2+b^2t^2}}{a^2}\right)+C.\\ \end{align}
You can also use IBP to evaluate $\displaystyle\int\sec^3\theta$ by taking $u=\sec\theta$ and $dv=\sec^2\theta\ d\theta$. – Tunk-Fey May 19 '14 at 17:45
Hint: use a hyperbolic trig sub. Sub $\frac{bt}{a} = \sinh x.$
The circular trig sub $\frac{bt}{a} = \tan x$ would involve the integral of $\sec^3 x.$ Not that trivial, although doable.