Evaluate $\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$

Question

$$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$

My approaches so far has been using substitution with $\tan x = t$ and $\tan \frac x2 = t$ but the calculations has been harder than I think they should.

I've also tried using ordinary polynom integration to simplify the integral but I'm having problems with factorizing the denominator.

Answer 1

Choose $t=\tan x$ and then $dt=(1+\tan^2x)dx$.

So your integral is: $$\int \frac{t}{t^3+3 t^2+2 t+6} dt=\int \frac{t}{(t+3)(t^2+2)} dt.$$

We have: $$\frac{1}{(t+3)(t^2+2)}=\frac{At+B}{t^2+2}+\frac{C}{t+3}=\frac{(A+C)t^2+(3A+B)t+(3B+2C)}{(t+3)(t^2+2)}.$$

We easily get: $A=-C=-\dfrac{1}{11}$ and $B=\dfrac{3}{11}$.

Therefore:

$$\begin{equation}\begin{split}\int \frac{t}{t^3+3 t^2+2 t+6} dt&=\dfrac{1}{11}\int t\left(\frac{3-t}{t^2+2}+\frac{1}{t+3} \right)dt,\\&=\dfrac{1}{11}\int \frac{3t-t^2}{t^2+2}+\frac{t}{t+3}dt\\&=\dfrac{1}{11}\int \frac{3t-t^2}{t^2+2}dt+\dfrac{1}{11}\int\frac{t}{t+3}dt\\&=-\dfrac{1}{11}\int \frac{-3t+t^2+2-2}{t^2+2}dt+\dfrac{1}{11}\int\frac{t+3-3}{t+3}dt\\&=\cdots\end{split}\end{equation}$$

Answer 2

$$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$ $$\int \frac{\tan x(\tan^2(x)+1)}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$ $$\int \frac{\tan x(\sec^2(x))}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$

Frome here, let $u=\tan(x)$, $\dfrac{du}{dx} = \sec^2(x)$., thus,

$$\int \frac{u}{u^3 + 3u^2 + 2u + 6}$$

From there, use partial fractions.

Answer 3

Given

$$ \int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx. $$

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Consider the substitution

$$ \boxed{\color{red}{\tan(x) = \sqrt{2} \tan(y)}}, $$

so

$$ \Big( \tan^2(x) + 1 \Big) dx = \sqrt{2} \Big( \tan^2(y) + 1 \Big) dy. $$

so we obtain

$$ \begin{eqnarray} \int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx &=& \int \frac{\tan(x) \Big( \tan^2(x) + 1 \Big)} { \Big( \tan(x) + 3 \Big) \Big( \tan^2(x) + 2 \Big) } dx\\ &=& \color{blue}{\int \frac{ \tan(y)}{ \sqrt{2} \tan(y) + 3 } dy}. \end{eqnarray} $$

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Now

$$ \begin{eqnarray} \int \frac{ \sqrt{2} - 3 \tan(y)}{ \sqrt{2} \tan(y) + 3 } dy &=& \int \frac{ \sqrt{2} \cos(y) - 3 \sin(y) }{ \sqrt{2} \sin(y) + 3 \cos(y) } dy\\ &=& \int \frac{1}{ \sqrt{2} \sin(y) + 3 \cos(y) } d \Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big)\\ &=& \ln\Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big), \end{eqnarray} $$

and

$$ \begin{eqnarray} \int \frac{ \sqrt{2} \tan(y) + 3}{ \sqrt{2} \tan(y) + 3 } dy &=& y. \end{eqnarray} $$

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We also have

$$ \begin{eqnarray} \frac{\sqrt{2}}{11} \frac{ \sqrt{2} \tan(y) + 3}{ \sqrt{2} \tan(y) + 3 } - \frac{3}{11} \frac{ \sqrt{2} - 3 \tan(y)}{ \sqrt{2} \tan(y) + 3 } &=& \frac{\tan(y)}{ \sqrt{2} \tan(y) + 3}. \end{eqnarray} $$

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So we obtain

$$ \begin{eqnarray} \color{blue}{\int \frac{\tan(y)}{ \sqrt{2} \tan(y) + 3} dy} &=& \color{darkgreen}{\frac{\sqrt{2}}{11} y - \frac{3}{11} \ln\Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big)}. \end{eqnarray} $$

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The final result is

$$ \int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx = \color{darkgreen}{\frac{\sqrt{2}}{11} y - \frac{3}{11} \ln\Big( \sqrt{2} \sin(y) + 3 \cos(y) \Big)}, $$

where

$$ \color{red}{y = \arctan\left( \frac{ \tan(x) }{ \sqrt{2} } \right)}. $$

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Using the properties

$$ \begin{eqnarray} \sin(\arctan(z)) &=& \frac{z}{\sqrt{1+z^2}},\\ \cos(\arctan(z)) &=& \frac{1}{\sqrt{1+z^2}},\\ \end{eqnarray} $$

we obtain

$$ \boxed{ \begin{eqnarray} \int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx &=& \frac{\sqrt{2}}{11} \arctan\left( \frac{\tan(x)}{\sqrt{2}} \right)\\ && \hspace{1em} + \frac{3}{11} \ln\left( \frac{ \tan(x) + 3 }{\sqrt{ 1 + \tan^2(x)/2 }} \right). \end{eqnarray}} $$