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.
 A: Let's write $\sqrt{1+9t^4}$ as
$$
\sqrt{1+9t^4} = \frac{3t^2\sqrt{1+9t^4}}{3t^2}=3t^2\sqrt{1/(9t^4)+1}
$$
Now we can use the fractional power binomial expansion.
\begin{align}
(1/(9t^4)+1)^{1/2}&= 1 + \frac{1}{18t^4}-\frac{1}{5184t^8} +\cdots\\
3t^2(1/(9t^4)+1)^{1/2}&= 3t^2 + \frac{1}{6t^2}-\frac{1}{1728t^6} +\cdots
\end{align}
Now let's integrate
\begin{align}
\int\sqrt{1+9t^4}dt &= t^3 - \frac{1}{6t} + \frac{1}{1080t^5} -\frac{1}{34992t^9} + \cdots\\ 
&= 3\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!t^{3-4n}}{(4n-3)(2n-1)36^n(n!)^2}
\end{align}
A: Hint: for this part$$\int\frac{\sqrt{1+u}\:du}{\sqrt{u}}$$
Let $v=\sqrt\frac{1+u}{u}$ or $u=\frac{1}{v^2-1}$ and thus $du=-\frac{2vdv}{(v^2-1)^2}$
Then it comes to solving
$$-\int \frac{2v^2dv}{(v^2-1)^2}$$
Gross as it looks, it is at least rational and therefore "doable". 
However when I make Maple calculate $\int\sqrt{1+9t^4}\:dt$ it does not give a closed-form result. So I'm afraid there must be some mistake when you simplified it to $(4)$.
A: $$\int\sqrt{1+9t^4}dt = \int\sqrt{1+(3t^2)^2}dt$$
put $$3t^2 = \sinh x \implies 6tdt = \cosh xdx \implies dt = \frac{\cosh x}{2\sqrt{3}\sqrt{\sinh x}}dx$$
$$\int\sqrt{1+\sinh^2x}\frac{\cosh x}{2\sqrt{3}\sqrt{\sinh x}}dx$$
$$\frac{1}{2\sqrt{3}}\int\sqrt{\cosh^2x}\frac{\cosh x}{\sqrt{\sinh x}}dx$$
$$\frac{1}{2\sqrt{3}}\int \cosh x\frac{\cosh x}{\sqrt{\sinh x}}dx$$
Taking $\cosh x$ as 1st function and $\frac{\cosh x}{\sqrt{\sinh x}}$ as 2nd function then by parts intigeration we get
$$\frac{1}{2\sqrt{3}}[\cosh x \frac{(\sinh x)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}-\int \sinh x\frac{(\sinh x)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}dx]$$
$$\frac{1}{2\sqrt{3}}[\cosh x\sinh x-\int \sinh^2xdx]$$
$$\frac{1}{2\sqrt{3}}[\cosh x\sinh x-\int \frac{\cosh 2x-1}{2}dx]$$
$$\frac{1}{2\sqrt{3}}[\cosh x\sinh x- \frac{1}{2} \int (\cosh 2x-1)dx]$$
$$\frac{1}{2\sqrt{3}}[\cosh x\sinh x- \frac{1}{2} (\frac{\sinh 2x}{2}-x)]$$
as $\sinh 2x = 2\sinh x\cosh x = 2(3t^2)\sqrt{1+9t^4} = 6t^2\sqrt{1+9t^4}$
$$\frac{1}{2\sqrt{3}}[\sqrt{1+9t^4}(3t^2)- \frac{1}{2} (\frac{6t^2\sqrt{1+9t^4}}{2}-\sinh^{-1}(3t^2))]$$
A: 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.
A: $$
\begin{aligned} \displaystyle \int \sqrt{1+9 t^4} \, \mathrm{d}t =&\frac{1}{3} t \sqrt{1+9 t^4}+\frac{2}{3} \displaystyle \int \frac{1}{\sqrt{1+9 t^4}} \, \mathrm{d}t\\ =&\frac{1}{3} t \sqrt{1+9 t^4}+\frac{\left(1+3 t^2\right) \sqrt{\frac{1+9 t^4}{\left(1+3 t^2\right)^2}} F\left(2 \tan ^{-1}\left(\sqrt{3} t\right)|\frac{1}{2}\right)}{3 \sqrt{3} \sqrt{1+9 t^4}}\\ \end{aligned}
$$
