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.
[Math] Integrate $\int\sqrt{1+9t^4}\:dt$
integration
Best Answer
Rescaling the integrand to $\sqrt{1+x^4}$ is a simple matter so let's start there. First, some trickery with integration by parts can reduce the desired integral to one closer to the form of an elliptic integral of the first kind:
$$\begin{align} \int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x &=\int_{0}^{a}\frac{1+x^4}{\sqrt{1+x^4}}\,\mathrm{d}x\\ &=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\int_{0}^{a}\frac{x^4}{\sqrt{1+x^4}}\,\mathrm{d}x\\ &=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\int_{0}^{a}x\cdot\frac{x^3}{\sqrt{1+x^4}}\,\mathrm{d}x\\ &=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\left[\frac12x\sqrt{1+x^4}\right]_{0}^{a}-\frac12\int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x\\ &=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\frac12a\sqrt{1+a^4}-\frac12\int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x\\ \implies \frac32\int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x &=\frac{a}{2}\sqrt{1+a^4}+\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}\\ \implies \int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x &=\frac{a}{3}\sqrt{1+a^4}+\frac23\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}.\\ \end{align}$$
Focusing now on the integral $\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}$, applying a Landen transformation of the form $x=\frac{1-y}{1+y}$ yields,
$$\begin{align} \int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}} &=\int_{1}^{\frac{1-a}{1+a}}\frac{(1+y)^2}{\sqrt{2(1+6y^2+y^4)}}\cdot\frac{(-2)\,\mathrm{d}y}{(1+y)^2}\\ &=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{1+6y^2+y^4}}\\ &=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{\left(3+2\sqrt{2}+y^2\right)\left(3-2\sqrt{2}+y^2\right)}}.\\ \end{align}$$
Note that the constant terms in the last line above have the useful properties
$$(3+2\sqrt{2})^{-1}=3-2\sqrt{2};\\ \sqrt{3+2\sqrt{2}}=1+\sqrt{2}.$$
Scaling the integral by substituting $(\sqrt{2}+1)y=t$,
$$\begin{align} \int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}} &=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{\left[(1+\sqrt{2})^2+y^2\right]\left[(1-\sqrt{2})^2+y^2\right]}} \\ &=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{(\sqrt{2}+1)(\sqrt{2}-1)\sqrt{\left[1+(\sqrt{2}-1)^2y^2\right]\left[1+(\sqrt{2}+1)^2y^2\right]}} \\ &=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{\left[1+(\sqrt{2}-1)^2y^2\right]\left[1+(\sqrt{2}+1)^2y^2\right]}} \\ &=(2-\sqrt{2})\int_{(\sqrt{2}+1)\frac{1-a}{1+a}}^{\sqrt{2}+1}\frac{\mathrm{d}t}{\sqrt{\left[1+(\sqrt{2}-1)^4t^2\right]\left(1+t^2\right)}} \\ &=(2-\sqrt{2})\int_{(\sqrt{2}+1)\frac{1-a}{1+a}}^{\sqrt{2}+1}\frac{\mathrm{d}t}{\sqrt{1+t^2}\sqrt{1+(\sqrt{2}-1)^4t^2}}. \\ \end{align}$$
Now it's time for trigonometric substitution. For compactness of notation, write $(\sqrt{2}-1)^2=b$. Using $t=\tan{\theta}$,
$$\begin{align} \int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}} &=(2-\sqrt{2})\int_{(\sqrt{2}+1)\frac{1-a}{1+a}}^{\sqrt{2}+1}\frac{\mathrm{d}t}{\sqrt{1+t^2}\sqrt{1+b^2t^2}} \\ &=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\sec^2{\theta}\,\mathrm{d}\theta}{\sqrt{\sec^2{\theta}}\sqrt{1+b^2\tan^2}{\theta}} \\ &=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\sec^2{\theta}\,\mathrm{d}\theta}{\sqrt{\sec^4{\theta}}\sqrt{\cos^2{\theta}+b^2\sin^2}{\theta}} \\ &=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\mathrm{d}\theta}{\sqrt{\cos^2{\theta}+b^2\sin^2}{\theta}} \\ &=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\mathrm{d}\theta}{\sqrt{1-(1-b^2)\sin^2}{\theta}} \\ &=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\mathrm{d}\theta}{\sqrt{1-b^{\prime\,2}\sin^2}{\theta}}. \\ \end{align}$$
And presto chango, an elliptic integral of kind numero uno! I presume I can safely leave the remaining details to you, but let me know if should make anything clearer.