find the power representation of $x^2 \arctan(x^3)$ Wondering what im doing wrong in this problem im ask to find the power series representation of 
$x^2 \arctan(x^3)$
now i know that arctan's power series representation is this
$$\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}  $$
i could have sworn that for solving for this i could just have use that formula and then distribute the $x^2$ but i'm getting the wrong answer here is the steps im taking.
1- plug in $x^3$ for $x$
$$x^2\sum_{n=0}^\infty (-1)^n \frac{(x^3)^{2n+1}}{2n+1}  $$
2- distribute the $2n+1$ to $x^3$
$$x^2\sum_{n=0}^\infty (-1)^n \frac{x^{6n+3}}{2n+1}  $$
3- distribute the $x^2$
$$\sum_{n=0}^\infty (-1)^n \frac{x^{6n+5}}{2n+1}  $$
however this is wrong according to my book the answer should be
$$\sum_{n=0}^\infty (-1)^n \frac{x^{3n+2}}{n}  $$
can someone please point out my mistake.  Please forgive me in advance for any mathematical blunders that i post. I'm truly sorry.
Thanks
Miguel

The book's answer would be right if it said
$$
\sum_{\text{odd }n\ge 0} (-1)^{(n-1)/2)} \frac{x^{3n+2}}{n}.
$$
That would be the same as your answer, i.e.
$$
\sum_{\text{odd }n\ge 0} (-1)^{(n-1)/2)} \frac{x^{3n+2}}{n} = \sum_{n\ge0} (-1)^n\frac{x^{6n+5}}{2n+1}.
$$
