Trig. Indefinite Integral $\int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$ $\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$
$\underline{\bf{My \; Try}}$:: Let $\tan x = t$. Then $\sec^2 xdx = dt\Rightarrow \displaystyle dx = \frac{1}{1+\tan^2 t}dt\Rightarrow dx = \frac{1}{1+t^2}dt$ 
So $\displaystyle \int\frac{t+t^3}{1+t^3}\cdot \frac{1}{1+t^2}dt = \int\frac{t}{1+t^3}dt$
Now My Question is can we solve the Given Integral without Using Partial fraction Method
If Yes How can I solve
plz Help me , Thanks 

Mathematica evaluates it as 
$$\frac{\tan ^{-1}\left(\frac{2 \tan (x)-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{1}{6} \log \left(\tan ^2(x)-\tan (x)+1\right)-\frac{1}{3} \log (\tan (x)+1)$$
This can be done via partial fractions.
\begin{align*}
\int \frac{t}{1 + t^3} \; dt &= \frac 1 3 \int \frac{t+1}{t^2 - t + 1} dt - \frac 1 3 \int \frac{1}{1+t} \; dt\\ 
 &= \frac 1 3 \int \frac{t+1}{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4} \; dt - \frac 1 3 \log(1+t)\\ 
 &=  \frac 1 3  \int \frac{t - \frac 1 2  + \frac 3 2 }{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4} \; dt - \frac 1 3 \log(1+t)\\ 
 &= \frac 1 6 \log \left(  \left ( t - \frac 1 2 \right )^2 + \frac 3 4 \right ) - \frac 1  3 \log(1+t) + \frac 1 2 \int \frac{1}{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4}\;  dt \\ 
 &= \frac 1 6 \log \left(  \left ( t - \frac 1 2 \right )^2 + \frac 3 4 \right ) - \frac 1  3 \log(1+t)  + \frac 1 {\sqrt{3}} \arctan \left( \frac{t - \frac 1 2 }{ \frac{\sqrt 3}{2 } } \right ) 
\end{align*}
