Applicare la formula: $\int\sqrt{1+\theta ^3}dx$$=\frac{2}{5\sqrt{\theta ^3+1}}\left(\theta ^4+\sqrt[6]{-1}\sqrt[4]{\left(3\right)^{3}}\sqrt{\sqrt[6]{-1}\left(\theta +1\right)}\sqrt{\theta ^2-\theta +1}F\left(\arcsin\left(\frac{\sqrt{-\sqrt[6]{\left(-1\right)^{5}}\left(\theta +1\right)}}{\sqrt[4]{3}}\right)\Big\vert \sqrt[3]{-1}\right)+\theta \right)+C$, dove $1+x^3=1+t^3$, $1/2=\frac{1}{2}$, $dx=dt$, $x=t$ e $x^3=t^3$