If $\alpha and \beta$ are the roots of the equation $x^2+5x+1=0$, then the value of ${\left(\frac{\alpha }{1+\beta }\right)}^2+{\left(\frac{\beta }{\alpha +1}\right)}^2=$

$$\begin{gathered} x^2+5x+1=0 \\ \alpha +\beta =-5, \alpha \beta =1 \\ {\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta =25-2=23 \\ (1+\alpha )(1+\beta )=1+(\alpha +\beta )+\alpha \beta =1-5+1=-3 \\ {\left(\frac{\alpha }{1+\beta }\right)}^2+{\left(\frac{\beta }{1+\alpha }\right)}^2={(\frac{\alpha }{1+\beta }+\frac{\beta }{1+\alpha })}^2-2\left(\frac{\alpha }{1+\beta }\right)\left(\frac{\beta }{1+\alpha }\right) \\ {\left(\frac{{\alpha }^2+{\beta }^2+\alpha +\beta }{(1+\beta )(1+\alpha )}\right)}^2-2\left(\frac{1}{-3}\right)={\left(\frac{23-5}{-3}\right)}^2+\frac{2}{3}=36+\frac{2}{3}=\frac{110}{3} \end{gathered}$$