Let $N$ be a positive integer, and $$ F(N) = \sum_{n=1}^{N} \frac{1-\cos\left(\frac{(2n-1)\pi}{2N}\right)} {\left[1+\cos\left(\frac{(2n-1)\pi}{2N}\right)\right]\left[5+3\cos\left(\frac{(2n-1)\pi}{2N}\right)\right]^2} $$ I want to konw the closed form of $F(n)$. In order to make an attack on it, I calculate some values, i.e.
$N=16$, $F(n)=117$; $N=32$, $F(n)=490$; $N=64$, $F(n)=2004$; $N=128$, $F(n)=8104$; $N=256$, $F(n)=32592$; $N=512$, $F(n)=130720$, etc.
I can not conclude the general expression of $F(n)$, and oeis does not seem to help a lot.
EDIT: My numerical effort shows that $F(N)=\frac{N(8N-11)}{16}$.
EDIT2:
$N=17$, $F1(N)=132.8125000000000187306828$, $F2(N)=132.8125$;
$N=18$, $F1(N)=149.6250000000000023235341$, $F2(N)=149.625$;
$N=50$, $F1(N)=1215.6250000000000000000000000000000000000000000050$, $F2(N)=1215.625$.
Let $\omega = e^{\frac{\pi}{2N}i}$ be the primitive $4N^{th}$ root of unity and $$\Lambda = \bigg\{\; \omega^{2k-1} : -N < k \le N \;\bigg\} = \bigg\{\; \omega^{\pm(2k-1)} : 1 \le k \le N \;\bigg\}$$ be the set of roots for the polynomial $z^{2N} + 1 = 0$.
For any integer $k$, let $c_k = \cos\left(\frac{(2k-1)\pi}{2N}\right) = \frac{\omega^{2k-1} + \omega^{1-2k}}{2}$. Notice $$\frac{1-z}{(1+z)(5+3z)^2} = \frac12\left(\frac{1}{1+z}\right) - \frac12\left(\frac{1}{\frac53 + z}\right) -\frac49\left(\frac{1}{\frac53 + z}\right)^2 $$ We can simplify our sum as $$F(N) = \frac12 f(1) - \frac12 f\left(\frac53\right) + \frac49 f'\left(\frac53\right) \quad\text{ where }\quad f(\alpha) = \sum\limits_{k=1}^N \frac{1}{\alpha + c_k} $$ For any $\alpha > 1$, let $\beta = \alpha + \sqrt{\alpha^2 - 1}$, we have $\displaystyle\;\alpha = \frac{\beta + \beta^{-1}}{2}$. Since $c_k = c_{1-k}$, we have
$$\begin{align} f(\alpha) &= \sum_{k=1}^{N} \frac{1}{\alpha + c_k} = \frac12 \sum_{k=-N+1}^N \frac{1}{\alpha + c_k} = \frac12\sum_{\lambda \in \Lambda}\frac{1}{\frac{\beta + \beta^{-1}}{2} + \frac{\lambda + \lambda^{-1}}{2}}\\ &= \sum_{\lambda \in \Lambda}\frac{\lambda}{(\lambda+\beta)(\lambda+\beta^{-1})} = \frac{1}{\beta-\beta^{-1}}\sum_{\lambda\in\Lambda}\left(\frac{\beta}{\lambda+\beta} - \frac{\beta^{-1}}{\lambda+\beta^{-1}}\right) \end{align} $$ Notice $\lambda \in \Lambda \iff -\lambda \in \Lambda$ and taking logarithm derivatives, we have $$\prod_{\lambda\in\Lambda} (z + \lambda) = \prod_{\lambda\in\lambda}(z - \lambda) = z^{2N} + 1 \quad\implies\quad \sum_{\lambda\in\Lambda} \frac{z}{\lambda + z} = \frac{2N z^{2N}}{z^{2N}+1}$$ This leads to $$f(\alpha) = \frac{2N}{\beta-\beta^{-1}}\left(\frac{\beta^{2N} - 1}{\beta^{2N} + 1}\right) $$ Notice $f(\alpha)$ is continuous at $\alpha = 1$. We find
$$f(1) = \lim_{\alpha\to 1}f(\alpha) = \lim_{\beta\to 1}f(\alpha) = \frac{2N}{2}\left(\frac{2N}{2}\right) = N^2$$
This expression for $f'(\alpha)$ is pretty messy. If I didn't make any mistake, it is $$\begin{align} f'(\alpha) &= \frac{d\beta}{d\alpha}\frac{df(\alpha)}{d\beta} = \frac{\beta}{\sqrt{\alpha^2-1}} \frac{d\log f(\alpha)}{d\beta} f(\alpha)\\ &= \frac{4N}{(\beta-\beta^{-1})^2}\left(-\frac{\beta+\beta^{-1}}{\beta - \beta^{-1}} + \frac{4N\beta^{2N}}{\beta^{4N}-1}\right)\left(\frac{\beta^{2N} - 1}{\beta^{2N} + 1}\right)\end{align}$$
For the problem at hand, we need $f(\alpha)$ and $f'(\alpha)$ at $\alpha = \frac53$ which is equivalent to $\beta = 3$. Throwing the function $f(\alpha)$ and $f'(\alpha)$ to a CAS and ask it to simplify the mess at $\beta = 3$, we get
$$\begin{align} F(N) &= \frac12 N^2 - \frac{3N}{8}\left(\frac{9^N-1}{9^N+1}\right) -\frac{N \left( 5\cdot 9^{2N}-16N\cdot 9^N-5\right)}{16\left( 9^N+1\right)^2}\\ &= \frac{(8N-11)N}{16}+\frac{N((8N+11)9^N + 11)}{8(9^N+1)^2} \end{align} $$ In particular, the first few values of $F(N)$ are $$\frac{1}{25},\frac{1188}{1681},\frac{327129}{133225},\frac{56551312}{0764961},\frac{316019361}{34869025},\frac{979687020612}{70607649841},\ldots$$
As a double check, I have computed the values of these $F(N)$ numerically using original expression and they match up to $50$ decimal places.