How to integrate $\sqrt{1+x\sqrt{1+x}}$?

Question

How to integrate $$\int \sqrt{1+x\sqrt{1+x}}dx$$ ? I am trying to integrate this expression by substituting $(1+x)=u$. Now $dx=du$. So, $x=(u-1)$. Now substituting this value of $x$ in the actual expression we get $$\int \sqrt{1+(u-1)\sqrt{u}}du$$. So, now after simplifying I am getting $$\int \sqrt{1+u^{\frac{3}{2}}-u^{\frac{1}{2}}}du$$. Now, I am thinking of substituting $u^{\frac{1}{2}}=t$ Then $\frac{1}{2\sqrt{u}}du=dt$. Therefore, $du=2\sqrt{u}dt$$=$$2tdt$. So, the final integral is becoming $$2×\int t×\sqrt{1+t^{3}-t}dt$$. But after this step I can't approach further. Please help me out. Also, I haven't understood the solution of Wolfram Alpha.

Answer 1

$$I=\int \sqrt{1+x\sqrt{1+x}}\,dx$$ $$x=t^2-1 \quad \implies \quad I=2\int t \sqrt{1-t+t^3}\,dt$$

Write $$\sqrt{1-t+t^3}=\sqrt{(t-a)(t-b)(t-c)}$$ where $(a,b,c)$ are the roots of the cubic equation (one is real and the two other are complex conjugate). The real one is $$a=-\frac{2}{\sqrt{3}} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{3 \sqrt{3}}{2}\right)\right)$$

You face a very complex result in terms of elliptic integrals of the first and second kinds.

Edit

For a series solution, you could write $$ t \sqrt{1-t+t^3}=\sum_{n=0}^\infty a_n\, t^{n+1}$$ with $$a_n=\frac 1{2n}\big((2n-3)a_{n-1}-(2n-9) a_{n-3} \big)$$ with $$a_0=1 \qquad \qquad a_1=-\frac 12\qquad \qquad a_2=-\frac 18$$