You've probably seen this integral before if you paid attention

$N$ is the smallest angle, in radians, of a triangle with side lengths $4,$ $5,$ and $\sqrt{41-8\sqrt{10}}.$

The following integral $I$ is equal to $\dfrac{\pi\sqrt{\alpha}+\beta\sqrt{\gamma}}{\delta},$ for positive square-free integers $\alpha$ and $\gamma$ and positive integers $\beta$ and $\delta.$ What is $\alpha+\beta+\gamma+\delta?$

$I=\int_0^N\dfrac{\sec^4\theta}{\sec^4\theta-\tan^4\theta}\text{ }d\theta$

$\textbf{Details and Assumptions}$

There is exactly $\textbf{one}$ place in this problem where you may need to use a four-function calculator.