Find $\int\frac{\tan^2(x)}{\sqrt{x}} \, dx$

Question

I can not find the next antiderivative

$$\displaystyle\int\dfrac{\tan^2(x)}{\sqrt{x}}\,dx$$

I tried with integration by parts the second integral, please help me.

My try

$$ \begin{aligned} \int \dfrac{\sec^2(x)-1}{\sqrt{x}}\,dx =\int\dfrac{\sec^2(x)}{\sqrt{x}}\,dx-\int\dfrac{dx}{\sqrt{x}}=\int\dfrac{\sec^2(x)}{\sqrt{x}}\,dx-2\cdot\sqrt{x}+C\end{aligned} $$ Thank you so much.

Answer 1

I don't think that there is a closed form solution to your question(as said also in the comments).This is my argument:

Let A=$ ∫ \frac{\tan ^2\left(x\right)}{\sqrt x}dx$.Because I want an integration by parts,I write it as:$∫ \frac{\tan ^2\left(x\right)}{\sqrt x} x^\prime dx=\frac{x\tan ^2\left(x\right)}{\sqrt x}-2 ∫ \frac{x\tan \left(x\right)}{\sqrt x \cos ^2\left(x\right)}dx+\frac{1}{2}A \Leftrightarrow \frac{1}{2}A=\frac{x\tan ^2\left(x\right)}{\sqrt x}-2∫ \frac{x\tan \left(x\right)}{\sqrt x \cos ^2\left(x\right)}dx$

I want to focus now on $∫ \frac{x\tan \left(x\right)}{\sqrt x \cos ^2\left(x\right)}dx$. With the substitution $\tan \left(x\right)=t$ we have that $∫ \frac{x\tan \left(x\right)}{\sqrt x \cos ^2\left(x\right)}dx=∫ \sqrt {\tan ^{-1}} ⋅ tdt$ or in other words $∫ \sqrt{x} ⋅ \tan \left(x\right) dx$.

But when you plug the last integral into Wolfram Alpha it says: " no result found in terms of standard mathematical functions"

I know that maybe the "answer" it's long but I wanted to show you why this has no closed form solution.

Answer 2

$\int\dfrac{\tan^2x}{\sqrt x}~dx$

$=\int\dfrac{\sec^2x-1}{\sqrt x}~dx$

$=\int\dfrac{\sec^2x}{\sqrt x}~dx-\int\dfrac{1}{\sqrt x}~dx$

$=\int\dfrac{1}{\sqrt x}~d(\tan x)-2\sqrt x$

$=\dfrac{\tan x}{\sqrt x}-\int\tan x~d\left(\dfrac{1}{\sqrt x}\right)-2\sqrt x$

$=\dfrac{\tan x}{\sqrt x}-2\sqrt x-\int\tan u^2~d\left(\dfrac{1}{u}\right)$ (Let $u=\sqrt x$)

$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\dfrac{\tan u^2}{u^2}~du$

$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{8}{(2n+1)^2\pi^2-4u^4}~du$ (use Mittag-Leffler Expansion of tangent)

$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)\pi+2u^2)((2n+1)\pi-2u^2)}~du$

$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{4}{(2n+1)\pi((2n+1)\pi+2u^2)}~du+\int\sum\limits_{n=0}^\infty\dfrac{4}{(2n+1)\pi((2n+1)\pi-2u^2)}~du$

$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tan^{-1}\dfrac{\sqrt2u}{\sqrt{(2n+1)\pi}}+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tanh^{-1}\dfrac{\sqrt2u}{\sqrt{(2n+1)\pi}}+C$

$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tan^{-1}\dfrac{\sqrt{2x}}{\sqrt{(2n+1)\pi}}+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tanh^{-1}\dfrac{\sqrt{2x}}{\sqrt{(2n+1)\pi}}+C$

Answer 3

Only a note.

$\displaystyle \int\frac{(\tan x)^2}{\sqrt{x}}dx = \int\frac{1}{\sqrt{x}}(\frac{d}{dx}\tan x)\,dx - \int\frac{1}{\sqrt{x}} = \frac{\tan x}{\sqrt{x}}-\frac{\sqrt{x}}{2}+ \frac{1}{2}\int\frac{\tan x}{\sqrt{x}^3} dx $

with $\enspace\displaystyle\frac{1}{2}\int\frac{\tan x}{\sqrt{x}^3} dx=\int\frac{\tan (t^2)}{t^2} dt $

The question has changed to find a formula for $\enspace\displaystyle\int\frac{\tan (x^2)}{x^2} dx$

for which the Taylor series around $\,0\,$ is $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}\frac{x^{4n-3}}{4n-3}\,$ .

I've never seen a closed form for $\enspace\displaystyle\int\frac{\tan (x^2)}{x^2} dx\,$ , maybe it doesn't exist with conventional functions.