Integrate $ \int \frac { e^{\arctan(x)}}{{(1+x^2)}^{\frac{3}{2}}} \ dx $ $y=arctanx$
$tany=x$
\begin{align}
\int \frac { e^{\Large\arctan(x)}}{{(1+x^2)}^{\Large\frac{3}{2}}} \ dx&=\int \frac {e^{\Large\arctan(\tan y)}}{{(1+\tan^2y)}^{\Large\frac{3}{2}}}dy\\
&=\int \frac {e^{y}}{\sec^3 y} dy\\
&= e^y \cos^3 y+ \int 3e^{y}\sin y\ \cos^2 y\ dy\\
\end{align}
Is this right so far or am I doing something wrong?
It's been quite a while since I've done integration with trig substitutions. Last time I did this integral I did not use trig subtitution and still got the correct answer, I can't find my solutions from then(over 2 years ago).

You may observe that
$$
\left(-\frac{1}{(1+x^2)^{1/2}}\right)'=\frac {x}{{(1+x^2)}^{3/2}},\quad \left(\frac{x}{(1+x^2)^{1/2}}\right)'=\frac {1}{{(1+x^2)}^{3/2}},
$$
then just perform two integrations by parts as follows.
$$
\begin{align}
\int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}}dx&=\frac {x\: e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}-\int \frac { x\:e^{\Large\arctan x}}{{(1+x^2)}^{3/2}} \ dx\\\\
&=\frac {x\: e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}-\left(-\frac { e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}+\int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}} \ dx\right)\\\\
&=\frac {(x+1)\: e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}-\int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}} \ dx
\end{align}$$ giving easily
$$\bbox[30px,border:1px solid #ff8000]{
\begin{align}
\int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}}dx=\frac {(x+1)\: e^{\Large\arctan x}}{2(1+x^2)^{1/2}}+C.
\end{align}}
$$
