How to calculate integral of $\int_0^\sqrt[3]4\!\sqrt\frac{x}{4-x^{3/2}}\,\mathrm{d}x$ Given the following integral:
$$\int_0^\sqrt[3]4\!\sqrt\frac{x}{4-x^{3/2}}\,\mathrm{d}x$$
How to solve it? I thought it may be possible to substitute it, but I didn't find anything to substitute. I tried to solve it with Maple, but the CAS didn't get it therefore I don't know how to carry on with this. Can you give me some hints?
 A: I think it is worth mentioning the case of the integration of the differential binomials.
The expression of the form
$$x^m(a+bx^n)^pdx$$
where $m,n,p,a,b$ are constant is called a differential binomial.
THEOREM. (Piskunov)
The integral
$$\int x^m(a+bx^n)^pdx$$
can be reduced if $m,n,p$ are rational numbers, to the integral of a rational function, and can thus be expressed in terms of elementary functions if:
$1.$ $p$ is an integer.
$2.$ $\dfrac{m+1}{n}$ is an integer.
$3.$ $\dfrac{m+1}{n}+p$ is an integer.
PROOF
We transform the integral writing $x^n = z$ so $dx = \frac 1 n z^{\frac 1 n -1}$. Then:
$$\int {{x^m}} {(a + b{x^n})^p}dx = \int {{z^{{{m + 1} \over n} - 1}}} {(a + bz)^p}dz = \int {{z^q}} {(a + bz)^p}dz$$
$1.$ Let $p$ be an integer. Being $q$ a rational number, let it be $\dfrac r s$. This integral then takes the form $$\int {R\left( {{z^{q/s}},z} \right)dz} $$
which can be reduced by substituting $z=t^s$.
$2.$ If $\dfrac{m+1}{n}$ is an integer. then $q=\dfrac{m+1}{n}-1$ is an integer. $p$ is rational $=\dfrac \lambda \mu$. The integral is reduced to $$\int {R\left( {{z^q},{{\left( {a + bz} \right)}^{{\lambda  \over \mu }}}} \right)dz} $$
which can be reduced substituting $a+bz=t^\mu$
$3.$ If $\dfrac{m+1}{n}+p$ is an integer then $\dfrac{m+1}{n}+p-1=q+p$ is an integer. We tranform the integral into
$$\int {{z^{q + p}}{{\left( {{{a + bz} \over z}} \right)}^p}dz} $$
where $q+p$ is an integer and $p=\dfrac \lambda \mu$ is rational. The integral is then
$$\int {R\left[ {z,{{\left( {{{a + bz} \over z}} \right)}^{{\lambda  \over \mu }}}} \right]dz} $$
which can be reduced using 
$${{a + bz} \over z} = {t^\mu }$$
Note. P.L. Chebyshev, a russian mathematician, proved the integrals just analysed can't be expressed in terms of elementary functions if it isn't the case $1$ , $2$ or $3$. 
A: $$\int\sqrt{\frac{x}{4-x^{3/2}}}\,\mathrm dx = -\frac{4\sqrt{x}}{3\sqrt{-\frac{x}{x^{3/2}-4}}}+\mathrm{constant}$$
where you can find the integration steps here by clicking on the button 'Show steps' next to the result.
In your particular case $x\geq 0$ over the whole integration domain such that we may simplify to
$$-\frac{4}{3}\sqrt{4-x^{3/2}}$$
evaluating at $x=0$ and $x=4^{1/3}$ gives the result
$$\int_0^{4^{1/3}}\sqrt{\frac{x}{4-x^{3/2}}}\,\mathrm dx = \frac{4}{3} (2-\sqrt{2})$$
