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I want to calculate the DTFT of $$x[n] = \cos (\frac \pi 3 n) (n^4u[n+4] + n^2 u[n+2] - n^4 u[n-2] - n^2 u[n-4])$$

My first thought was to convert it to convolution since cosine decomposes to delta functions in the frequency domain $\Omega$ and use the property which states that $$n^m u[n + a] \leftrightarrow j^m \frac {d^m ( e^{j\Omega a} \mathcal {DTFT \{ u[n], n \to \Omega \})}} {d \Omega^m} $$ which turns out to be very frustrating. So I want a faster way to do it.

Thanks in advance

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    What do you mean with frustrating ? The multiplication of $x(n)$ by $n$ means differentating $X(\omega)$, and the multiplication of $x(n)$ by $e^{i b n}$ means shifting $X(\omega)$. So $X(\omega)$ is a sum of shifts and derivatives of $U(\omega)$ the DTFT of $u(n)$. Now the big question : what is the DTFT of $u(n)$ ? Do you know the Z-transform ? – 2017-01-04
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    Yes but I know how to do everything but it has a lot of calculations so I am searching for a smarter way. Of course, I know the Z transform :) – 2017-01-04
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    Ok, so what is $U(z)$ the Z-transform of $u(n)$ ? Is it complicated to shift and differentiate ? – 2017-01-04

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$n^4[u(n+4)-u(n-2)]$ is zero for $n<-4$ and $n>1$.

Similarly, $n^2[u(n+2)-u(n-4)]$ is zero for $n<-2$ and $n>3$.

Hence, the signal is composed of seven nonzero values. Therefore, the Fourier transform can be written as sum of seven appropriately shifted and weighted complex exponentials.