What is the indefinite integral of $\sqrt[4] {\tan \left( x\right) }$

Question

$$\int\sqrt[4] {\tan \left( x\right) } dx$$

I'm really stuck right now with this integral, so any kind of advice would be appreciated.

Perhaps a pretty nifty substitution that can save me from partial fraction decomposition.

Answer 1

[Too long for a comment]

Mathematica gives this answer.

$\displaystyle\int \sqrt[4]{\tan (x)} \, dx=\\\cos \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}-\cos \left(\frac{\pi }{8}\right)\right)\right)+\\\cos \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}+\cos \left(\frac{\pi }{8}\right)\right)\right)-\\\sin \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}-\sin \left(\frac{\pi }{8}\right)\right)\right)-\\\sin \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}+\sin \left(\frac{\pi }{8}\right)\right)\right)+\\\frac{1}{2} \cos \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}+2 \sin \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)-\\\frac{1}{2} \cos \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}-2 \sin \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)-\\\frac{1}{2} \sin \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}+2 \cos \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)+\\\frac{1}{2} \sin \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}-2 \cos \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)+C$

I wouldn’t want to have to have worked it out by hand, but it doesn’t look impossible.

Answer 2

$$ \small \left( x^2 + \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 + \sqrt{2 - \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 - \sqrt 2} \; x + 1 \right) = x^8 + 1 $$

Answer 3

I usually factor a cyclotomic denominator like this into linear factors, expand into partial fractions, integrate to logarithms, and then convert those complex logarithms into logarithms and arctangents, but this time I thought I would try combining complex conjugate partial fractions before integrating and see how it went. First the substitution $z^4=\tan x$: $$\int\sqrt[4]{\tan x}dx=4\int\frac{z^4}{z^8+1}dz$$ And now $$\frac{4z^4}{z^8+1}=\frac{4z^4}{\prod_{k=0}^7(z-\omega_k)}=\sum_{k=0}^7\frac{A_k}{z-\omega_k}$$ Where $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, and $\theta_k=\frac{(2k+1)\pi}8$ And then as @Count Iblis remarked, it's easy to get the numerators by L'Hopital's rule $$\lim_{z\rightarrow\omega_k}\frac{4z^4}{z^8+1}=\lim_{z\rightarrow\omega_k}\frac{4z^4}{8z^7}=\frac1{2\omega_k^3}=-\frac12\omega_k^5$$ Combining complex conjugate fractions, $$\begin{align}\frac{4z^4}{z^8+1} & =\sum_{k=0}^7\frac{-\frac12\omega_k^5}{z-\omega_k}=\sum_{k=0}^3\frac{-\frac12\omega_k^5(z-\omega_{7-k})-\frac12\omega_{7-k}^5(z-\omega_k)}{(z-\omega_k)(z-\omega_{7-k})}\\ & =\sum_{k=0}^3\frac{-z\cos5\theta_k}{z^2-2z\cos\theta_k+1}=\sum_{k=0}^3\frac{(z-\cos\theta_k)\cos3\theta_k+\cos\theta_k\cos3\theta_k}{z^2-2z\cos\theta_k+1}\end{align}$$ So $$\begin{align}\int\sqrt[4]{\tan x}dx & =\sum_{k=0}^3\int\frac{(z-\cos\theta_k)\cos3\theta_k+\cos\theta_k\cos3\theta_k}{z^2-2z\cos\theta_k+1}d\theta\\ & =\sum_{k=0}^3\left[\frac12\cos3\theta_k\ln\left(z^2-2z\cos\theta_k+1\right)+\frac{\cos3\theta_k\cos\theta_k}{\sin\theta_k}\tan^{-1}\left(\frac{z-\cos\theta_k}{\sin\theta_k}\right)\right]+C\\ & =\sum_{k=0}^3\left[\frac12\cos3\theta_k\ln\left(\sqrt{\tan x}-2\sqrt[4]{\tan x}\cos\theta_k+1\right)+\frac{\cos3\theta_k\cos\theta_k}{\sin\theta_k}\tan^{-1}\left(\frac{\sqrt[4]{\tan x}-\cos\theta_k}{\sin\theta_k}\right)\right]+C\end{align}$$ Evaluating the trig functions, $$\begin{align}\int\sqrt[4]{\tan x}dx & =\frac14\sqrt{2-\sqrt2}\ln\left(\sqrt{\tan x}-\sqrt[4]{\tan x}\sqrt{2+\sqrt2}+1\right)+\\ & \frac12\sqrt{2+\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4+2\sqrt2}-\frac{2+\sqrt2}2\right)-\\ & \frac14\sqrt{2+\sqrt2}\ln\left(\sqrt{\tan x}-\sqrt[4]{\tan x}\sqrt{2-\sqrt2}+1\right)-\\ & \frac12\sqrt{2-\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4-2\sqrt2}-\frac{2-\sqrt2}2\right)+\\ & \frac14\sqrt{2+\sqrt2}\ln\left(\sqrt{\tan x}+\sqrt[4]{\tan x}\sqrt{2-\sqrt2}+1\right)-\\ & \frac12\sqrt{2-\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4-2\sqrt2}+\frac{2-\sqrt2}2\right)-\\ & \frac14\sqrt{2-\sqrt2}\ln\left(\sqrt{\tan x}+\sqrt[4]{\tan x}\sqrt{2+\sqrt2}+1\right)+\\ & \frac12\sqrt{2+\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4+2\sqrt2}+\frac{2+\sqrt2}2\right)+C\end{align}$$ All in all, combining complex conjugates before integrating seemed to work well in this case. It took much more time to typeset the solution than to solve the integral by hand. It agrees with the Mathematica solution already posted.