Let $y=f\left(x\right)$ be a thrice differentiable function in $(-5,5)$ . Let the tangents to the curve $y=f\left(x\right)$  at $(1,f\left(1\right))$ and  $(3,f\left(3\right))$ make angles  $\pi /6$ and  $\pi /4$, respectively with positive  $x-$axis. If  $27{\int }_1^3({(f\text{'}\left(t\right))}^2+1)f"\left(t\right)dt=\alpha +\beta \sqrt{3}$ where  $\alpha ,\beta$ are integers, then the value of $\alpha +\beta$  equals 

$$\begin{gathered} y=f\left(x\right)\Rightarrow \frac{dy}{dx}=f\text{'}\left(x\right) \\ {\frac{dy}{dx})}_{(1,f\left(1\right))}=f\text{'}\left(1\right)=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\Rightarrow f\text{'}\left(1\right)=\frac{1}{\sqrt{3}} \\ {\frac{dy}{dx})}_{(3,f\left(3\right))}=f\text{'}\left(3\right)=\tan \frac{\pi }{4}=1\Rightarrow f\text{'}\left(3\right)=1 \\ 27\underset{1}{\overset{3}{\int }}({(f\text{'}\left(t\right))}^2+1)f"\left(t\right)dt=\alpha +\beta \sqrt{3} \\ I=\underset{1}{\overset{3}{\int }}({(f\text{'}\left(t\right))}^2+1)f"\left(t\right)dt \\ f\text{'}\left(t\right)=z\Rightarrow f\text{'}\left(t\right)dt=dz \\ z=f\text{'}\left(3\right)=1 \\ z=f\left(1\right)=\frac{1}{\sqrt{3}} I=\underset{1/\sqrt{3}}{\overset{1}{\int }}(z^2+1)dx={(\frac{z^3}{3}+z)}_{1/\sqrt{3}}^1 \\ =(\frac{1}{3}+1)-(\frac{1}{3}.\frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}}) =\frac{4}{3}-\frac{10}{9\sqrt{3}}=\frac{4}{3}-\frac{10}{27}\sqrt{3} \\ \alpha +\beta \sqrt{3}=27(\frac{4}{3}-\frac{10}{27}\sqrt{3})=36-10\sqrt{3} \\ \alpha =36,\beta =-10 \\ \alpha +\beta =36-10=26 \end{gathered}$$