How to integrate- $\int \sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}}}}dx$?

Question

How to integrate $$\int \sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}}}}dx$$? I am trying to evaluate this integral by substituting $x=(\tan\theta)^{\frac{2}{3}}$. Therefore, $dx=\frac{2}{3}(\cot\theta)^{\frac{1}{3}}(\sec^{2}\theta)d\theta$. Therefore, the above integral will become $$\frac{2}{3}\int (\sqrt{(\tan\theta)^{\frac{2}{3}}+\sqrt{(\tan\theta)^{\frac{4}{3}}+|\tan\theta|}})(\cot\theta)^{\frac{1}{3}}(\sec^{2}\theta)d\theta$$. Now here $3$ cases will arise. the $1^{st}$ case is $|\tan\theta|=\tan\theta, when(\tan\theta>0)$. The $2^{nd}$ case is $|\tan\theta|=-\tan\theta, when(\tan\theta<0)$. And lastly the $3^{rd}$ case is $|\tan\theta|=0, when(\tan\theta=0)$. But here I am very much confused whether I can take the case of $|\tan\theta|=0, when(\tan\theta)=0$. Because if $\tan\theta=0$, then $\cot\theta=\infty$. Therefore I think the above integral will be undefined in that case. So, please help me out with this question.

Answer 1

It would be interesting to consider the infinite series $$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}}}}=\sum_{n=0}^\infty \frac{a_n}{b_n}\,x^{\frac{2n+3}{8} }$$ Coefficients $a_n$ and $b_n$ form respectively sequences $A002596$ and $A046161$ in $OEIS$ and they have a closed form. So, we know all of them and then $$\int\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}}}}=\sum_{n=0}^\infty \frac{8\,a_n}{(2n+11)\,b_n}\,x^{\frac{2n+11}{8} }$$

Edit

@JJacquelin having shown a solution, letting $$x=\frac{1}{t^4 \left(t^2-2\right)^2}$$ the integrale becomes $$-8\int \frac{t^2-1}{t^6 \left(t^2-2\right)^4}\,dt$$ Using partial fraction decomposition, the integrand is $$\frac{1}{2 t^6}+\frac{1}{2 t^4}+\frac{1}{4 t^2}-\frac{1}{4 \left(t^2-2\right)}+\frac{1}{2 \left(t^2-2\right)^3}-\frac{1}{\left(t^2-2\right)^4}$$ and the integral is $$\large\color{blue}{\frac{21 }{256 \sqrt{2}}\log \left(\frac{\sqrt{2}-t}{\sqrt{2}+t}\right)-}$$ $$\large\color{blue}{\frac{315 t^{10}-1680 t^8+2772 t^6-1152 t^4-256 t^2-1536}{1920_, t^5\,\left(t^2-2\right)^3}}$$

Answer 2

$$I=\int \sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}}}}dx$$ $$I=\int \sqrt{x+x\sqrt{1+x^{-1/2}}}dx=\int \sqrt{x}\sqrt{1+\sqrt{1+x^{-1/2}}}dx$$ $t=x^{-1/2}\quad\implies\quad x=t^{-2}\quad\implies\quad dx=-2t^{-3}dt$ $$I=-2\int t^{-4}\sqrt{1+\sqrt{1+t}}\:dt$$ Erom WolframAlpha :