Need some hints to solve $ \int \frac 1 { \sqrt{x+\sqrt{x^3}} } dx $ (from $ \int \frac {x} { \sqrt{ 1 + x^2+\sqrt{(1 + x^2)^3}} } dx $)

Question

I'm working on the indefinite integral for $ \int \frac {x} { \sqrt{ 1 + x^2+\sqrt{(1 + x^2)^3}} } dx $ and after u-substitution with $ u=1+x^2 $ and $ \frac {du}{dx}=2x $ I get $ \int \frac 1 { \sqrt{u+\sqrt{u^3}} } du $.

Not sure where to go from here.

Answer 1

$$\int \frac{1}{\sqrt{x+\sqrt{x^3}}}dx$$ $\color{green}{\sqrt{x^3}=x^{1.5},\:\quad \mathrm{assuming}\:x\ge 0}$ $$=\int \frac{1}{\sqrt{x+x^{1.5}}}dx =\int \frac{1}{\sqrt{\left(\sqrt{x}+1\right)x}}dx$$

Then Apply Integral Substitution:

$u=\left(\sqrt{x}+1\right)\quad \:du=\frac{1}{2\sqrt{x}}dx$

$\color{green}{\Rightarrow \:du=\frac{1}{2\sqrt{x}}dx}$

$\color{green}{\Rightarrow \:dx=2\sqrt{x}du}$

$=\int \frac{1}{\sqrt{ux}}2\sqrt{x}du$

$=\int \frac{2\sqrt{x}}{\sqrt{xu}}du$

$\color{green}{u=\left(\sqrt{x}+1\right)\quad \Rightarrow \quad \:x=u^2-2u+1}$

$$\Rightarrow=\int \frac{2\sqrt{u^2-2u+1}}{\sqrt{\left(u^2-2u+1\right)u}}du=2\int \frac{\sqrt{u^2-2u+1}}{\sqrt{\left(u^2-2u+1\right)u}}du=2\int \frac{\sqrt{\left(u-1\right)^2}}{\sqrt{\left(u-1\right)^2u}}du$$

Note $\color{green}{\sqrt{\left(u-1\right)^2u}=\sqrt{\left(u-1\right)^2}\sqrt{u}}$

$$=2\int \frac{\sqrt{\left(u-1\right)^2}}{\sqrt{\left(u-1\right)^2}\sqrt{u}}du=2\int \frac{1}{\sqrt{u}}du=2\int \:u^{-0.5}du$$ So $\color{green}{\mathrm{Substitute\:back}\:u=\left(\sqrt{x}+1\right)}$

$$\int \frac{1}{\sqrt{x+\sqrt{x^3}}}dx=\color{red}{4\sqrt{\sqrt{x}+1}+C}$$

Answer 2

Set $\sqrt{1+x^2}=u,1+x^2=u^2\implies x\ dx=u\ du$

$$\int\dfrac{udu}{\sqrt{u^2+u^3}}=\int\dfrac{du}{\sqrt{1+u}}=?$$ as $\sqrt{u^2}=|u|=+u$ as $u\ge1>0$