Given
My approach:
since $$r=1+cos(\theta)$$ we get $$\begin{cases} x=cos(\theta)+cos^2(\theta) \\ y=sin(\theta)+\frac{sin(2\theta)}{2} \end{cases}$$
therefore substituting in the integral gives
$$-\int _0^{2\pi }\:\left(\left(e^{cos\left(\theta \:\:\right)+cos^2\left(\theta \:\:\right)}+arctan\left(sin\left(\theta \:\:\right)+\frac{sin\left(2\theta \:\:\right)}{2}\right)\right)\left(sin\left(\theta \:\:\right)+sin\left(2\theta \:\:\right)\right)d\theta \:\:+\left(3\cdot \:\:\left(cos\left(\theta \:\:\right)+cos^2\left(\theta \:\:\right)\right)\cdot \:\:\left(sin\left(\theta \:\:\right)+\frac{sin\left(2\theta \:\:\right)}{2}\right)-\frac{cos\left(\theta \:\:\right)+cos^2\left(\theta \:\:\right)}{1+\left(sin\left(\theta \:\:\right)+\frac{sin\left(2\theta \:\:\right)}{2}\right)^2}\right)\cdot \:\left(cos\left(\theta \:\right)+cos\left(2\theta \:\right)\right)d\theta \:\right)$$
There is no way this is the method I should use or do I really have to integrate this?
In my opinion the easiest way to do this problem is to use Green's theorem (later, you might refer to this as a special case of Stokes' theorem). Namely, if $D$ is the region bounded by $C$ and $P$ and $Q$ are functions of $x$ and $y$ defined on an open region containing $D$ with continuous partial derivatives, then $$\oint_C (P~dx+Q~dy)=\iint_D \left(Q_x-P_y\right)~dy~dx.$$ With $P=e^x+\arctan y$ and $Q=\frac{x}{1+y^2}-3xy$, one gets $Q_x=\frac{1}{1+y^2}-3y$ and $P_y=\frac{1}{1+y^2}$. Hence one has that $$\oint_C (e^x+\arctan y)~dx+\left(\frac{x}{1+y^2}-3xy\right)~dy=-3 \iint_D y~dy~dx.$$ Switching to polar coordinates results in the computation of the integral $$\oint_C (e^x+\arctan y)~dx+\left(\frac{x}{1+y^2}-3xy\right)~dy=-3\int_0^{2\pi}\int_0^{1+\cos(\theta)} r^2\sin(\theta)~dr~d\theta.$$ This double integral evaluates very nicely (exercise - use symmetry).