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Find the sum closed form $$\sum_{k=0}^{n}\arctan{\dfrac{k^4+6k^3+10k^2-k-9}{(k+1)(k+2)(k+3)(k^3+7k^2+15k+8)}}$$

For problems involving sums, the idea is to use trigonometricidentities to write the sum in the form $$\sum_{k=1}^{n}[g(k)-g(k-1)]$$ and I initially considered pairing every two terms up to use the $arctanx+arctany $ trick, but it doesn't work because each arctanarctan term has a different coefficient.

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    The first few values for $n=0,1,\dots$ of $\tan\left(\sum_{k=0}^n \arctan(\dots)\right)$ are -3/16, -8/45, -5/32, -24/175, -35/288, -16/147, -63/640, -80/891, -33/400, -120/1573... – 2017-01-31
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    $\arctan f(k)-\arctan f(k+1)$ is also a telescopic term, that equals $\arctan\frac{f(k)-f(k+1)}{1+f(k)\,f(k+1)}$. Can you write such horrible fraction as $\frac{f(k)-f(k+1)}{1+f(k)\,f(k+1)}$? – 2017-01-31
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    You may also consider that the denominator plus $i$ times the numerator factors as $$\left(3+(7+2 i) k+(5+i) k^2+k^3\right) \left((16-3 i)+(20-4 i) k+(8-i) k^2+k^3\right)$$ – 2017-01-31
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    i.e. as $f(k)\;\overline{f(k+1)}$. – 2017-01-31

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By setting $f(k)=3+(7+2i)k+(5+i)k^2+k^3$ we are dealing with:

$$ \sum_{k=1}^{n}\text{arg}\left(f(k)\;\overline{f(k+1)}\right) =\text{arg }f(1)-\text{arg }f(n+1).\tag{1}$$ In order to notice that, I factored $(k+1)(k+2)(k+3)(k^3+7k^2+15k+8)+i(k^4+6k^3+10k^2-k-9)$, getting: $$\left(3+(7+2 i) k+(5+i) k^2+k^3\right) \left((16-3 i)+(20-4 i) k+(8-i) k^2+k^3\right)\tag{2}$$ and checked that $(2)$ has the $f(k)\,\overline{f(k+1)}$ structure.