closed form of $\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$ closed form of $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$$
for $a,b>0$
n=1 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}$$
n=2 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{a^3b^3}$$
but what the integral for n ????
I hope to be there two different solution one by contour integration and the other by real analysis and thanks for all

Here is a different solution using differentiation under integral sign. 
We firstly look at the the case $n=1$, then
\begin{align*}
I_1(\alpha,\beta) & = \int_{0}^{2\pi}\frac{dx}{\alpha \cos^2x+\beta \sin^2x} \\
                  & = \int_{0}^{2\pi} \frac{\sec^2x}{\alpha + \beta \tan^2x} \,\mathrm{d}x\\
                  & = \frac{1}{\beta} \int_0^{2\pi} \frac{1}{\left(\sqrt{\frac{\alpha}{\beta}}\right)^2 + \tan^2 (x)}\; \mathrm{d}(\tan x)\,\\
& = \frac{1}{\sqrt{\alpha \beta}} \left[ \left(\tan^{-1}\left(\sqrt{\frac{\beta}{\alpha}}\tan (x)\right)\right) \right]_0^{2\pi} = \frac{2\pi}{\sqrt{\alpha\beta}}
\end{align*}
Where $a^2 = \alpha$ and $b^2 = \beta$.
Differentiating $I_n$ with respect to $\alpha$ and $\beta$ yields
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}\alpha} I_n & = - \int_0^{2\pi} \frac{n \cos^2x}{(\alpha \cos^2x + \beta \sin^2x)^{n+1}}  \\
\frac{\mathrm{d}}{\mathrm{d}\beta} I_n & = - \int_0^{2\pi} \frac{n \sin^2x}{(\alpha \cos^2x + \beta \sin^2x)^{n+1}} 
\end{align*}
Addition gives that
\begin{align}
\nabla I_n = \left( \frac{\mathrm{d}}{\mathrm{d}\alpha} + 
             \frac{\mathrm{d}}{\mathrm{d}\beta} \right) I_1
           = -n \int_0^{2\pi} \frac{(\sin^2x + \cos^2x)\mathrm{d}x}{(\alpha \cos^2x + \beta \sin^2x)^2} 
           = -n I_{n+1}
\end{align}
Where $\nabla$ is the nabla-operator. This gives us the recurrence relation
\begin{align}
\nabla I_{n-1} = (1-n) I_n \tag{1}
\end{align}
Which can be used to evaluate each $n \in \mathbb{N} / \{ 1 \}$. 
This is a fine answer, but if one wants an explicit expression it can
be shown by induction that 
$$ \boxed{ \displaystyle
   I_{n+1} = \frac{2\pi(-1)^{n}}{n!} \left( \frac{\partial }{\partial \alpha} + \frac{\partial}{\partial \beta} \right)^n \frac{1}{\sqrt{\alpha\beta}} 
}
$$
Where the last part is the binomial expansion of the partial derivatives, and this holds for all $n \in \mathbb{N}$.
Example for $n=3$ we have
$$
\left( \frac{\partial }{\partial \alpha} + \frac{\partial}{\partial \beta} \right)^2 f(\alpha,\beta) = \left(\frac{\partial^2}{\partial \alpha^2}+2\frac{\partial^2}{\partial \alpha \, \partial \beta}+\frac{\partial^2}{\partial \beta^2}\right)f(\alpha,\beta)
$$
We can also express the partial derivatives as a sum like 
$$
\left( \frac{\partial }{\partial \alpha} + \frac{\partial}{\partial \beta} \right)^n 
                               =
\sum_{k=0}^b \binom{n}{k} \frac{\partial^{n-k} }{\partial \alpha^{n-k} } 
                  \frac{\partial^{k} }{\partial \beta^{k} } 
                               =
\sum_{k=0}^\infty \binom{n}{k} \frac{\partial^{n} }{\partial \alpha^{n-k} \,\partial^k \beta} 
$$
without that doing us much good. For actually computing explicit values of $n$ the recurrence relation is more convenient. 
EDIT: To give an intuition of the "closed" expression one can first rewrite 
$(1)$ to 
$$
I_n = - \frac{1}{n-1} \nabla I_{n-1}
$$
Direct computation of the first values gives

One can also show that the expression satisfies $(1)$ directly by insertion, and the use of the chain rule. 
