<s>[INST] Integrate $\int\sqrt{1+9t^4}\:dt$ I need to find
\begin{align}
\int\sqrt{1+9t^4}\:dt.
\end{align}
What I have so far:
\begin{align}
\int\sqrt{1+9t^4}\:dt & =\int\sqrt{1+\left(3t^2\right)^2}\:dt,\tag{1}
\end{align}
now let $3t^2=\tan\left(\theta\right)\implies \displaystyle t=\sqrt{\frac{\tan\left(\theta\right)}{3}},\:\:dt=\frac{1}{2}\left(\frac{\tan\left(\theta\right)}{3}\right)^{-1/2}\left(\frac{\sec^{2}\left(\theta\right)}{3}\right)$, which gives me
\begin{align}
\int\left[1+\tan^2\left(\theta\right)\right]^{1/2}\frac{\sqrt{3}\sec^2\left(\theta\right)}{6\sqrt{\tan\left(\theta\right)}}\:d\theta\tag{2}&=\frac{\sqrt{3}}{6}\int\frac{\sec^3\left(\theta\right)}{\sqrt{\tan\left(\theta\right)}}\:d\theta\\
& = \frac{\sqrt{3}}{6}\int\frac{\sec^2\left(\theta\right)\sec\left(\theta\right)\:d\theta}{\sqrt{\tan\left(\theta\right)}}\tag{3},
\end{align}
now let $u=\tan\left(\theta\right),\:\:du=\sec^{2}\left(\theta\right)\:d\theta$, which gives us
\begin{align}
\frac{\sqrt{3}}{6}\int\frac{\sqrt{1+u^{\color{red}{2}}}\:du}{\sqrt{u}}\tag{4},
\end{align}
and where do I go from here? Or perhaps I'm just going in circles and haven't made any progress with this result.
 [/INST] Substituting $u=t\sqrt3$ and $v=\dfrac1{1+u^4}$ we will be able to express this integral in terms of the 
incomplete beta function of arguments $\dfrac14$ and $-\dfrac34$. Alternately, by expanding the integrand 
into its binomial series, and switching the order of summation and integration, we can rewrite 
the same integral in terms of the hypergeometric function.
</s>