Integrate $\int\sqrt{1+9t^4}\:dt$

Question

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.

Answer 1

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.

Answer 2

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}

Answer 3

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".

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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 $(4)$.

Answer 4

$$\int\sqrt{1+9t^4}dt = \int\sqrt{1+(3t^2)^2}dt$$ 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 $\sinh 2x = 2\sinh x\cosh x = 2(3t^2)\sqrt{1+9t^4} = 6t^2\sqrt{1+9t^4}$ $$\frac{1}{2\sqrt{3}}[\sqrt{1+9t^4}(3t^2)- \frac{1}{2} (\frac{6t^2\sqrt{1+9t^4}}{2}-\sinh^{-1}(3t^2))]$$

Answer 5

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.

Answer 6

$$ \begin{aligned} \displaystyle \int \sqrt{1+9 t^4} \, \mathrm{d}t =&\frac{1}{3} t \sqrt{1+9 t^4}+\frac{2}{3} \displaystyle \int \frac{1}{\sqrt{1+9 t^4}} \, \mathrm{d}t\\ =&\frac{1}{3} t \sqrt{1+9 t^4}+\frac{\left(1+3 t^2\right) \sqrt{\frac{1+9 t^4}{\left(1+3 t^2\right)^2}} F\left(2 \tan ^{-1}\left(\sqrt{3} t\right)|\frac{1}{2}\right)}{3 \sqrt{3} \sqrt{1+9 t^4}}\\ \end{aligned} $$