Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^2 d x=d y, y(0)=-2$. Let the maximum and minimum values of the function $y=y(x)$ in $\left[0, \frac{\pi}{3}\right]$ be $\alpha$ and $\beta$, respectively. If $(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}$, then $\gamma+\delta$ equals _________.

<p>$$\begin{aligned} & \frac{d y}{d x}=(x+y+z)^2 \\ & \text { Put } x+y+z=t \\ & \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \\ & \text { Given DE } \Rightarrow \frac{d t}{d x}-1=t^2 \\ & \Rightarrow \frac{d t}{1+t^2}=d x \Rightarrow \tan ^{-1} t=x+c \\ & \Rightarrow x+y+z=\tan (x+c) \\ & \Rightarrow y(x)=\tan (x+c)-x-2 \\ & \because y(0)=-2 \Rightarrow-2=\tan c-0-2 \\ & \qquad \Rightarrow c=0 \\ & \Rightarrow y(x)=\tan x-x-2 \\ & \frac{d y}{d x}=\sec ^2 x-1 \geq 0 \end{aligned}$$</p> <p>$\Rightarrow y(x)$ is increasing if $x \in\left(0, \frac{\pi}{3}\right)$</p> <p>$$\begin{aligned} & \Rightarrow \alpha=y\left(\frac{\pi}{3}\right), \beta=y(0) \\ & \Rightarrow \alpha=-\frac{\pi}{3}-2+\sqrt{3} \text { and } \beta=-2 \end{aligned}$$</p> <p>Now, $(3 \alpha+\pi)^2+\beta^2=(6+3 \sqrt{3})^2+(-2)^2$</p> <p>$=67-36 \sqrt{3}=y+\delta \sqrt{3}$.</p> <p>$\Rightarrow \gamma+\delta=31$</p>