integral $\int \sqrt{x^4+x^3} \, dx$ $$
\int \sqrt{x^4+x^3} \, dx?
$$
Using the binomial method and by setting $\frac{1}{x}+1=t$, I get to solve 
$$
\int \frac{-t^{\frac{3}{2}}}{(t^2-1)^4} dt? 
$$
It is on degree $\frac 32$. How to solve this it seems a bit hard and impossible.

$$\int \sqrt{x^4 + x^3} \;dx = \int x\sqrt{x^2 + x}\;dx = \frac{1}{2} \int (2x+1)\sqrt{x^2+x}\;dx - \frac{1}{2} \int \sqrt{x^2 + x} \;dx \\
= \frac{1}{2} \int (2x+1)\sqrt{x^2+x}\;dx - \frac{1}{2} \int \sqrt{\left( x + \frac{1}{2} \right)^2-\frac{1}{4}} \;dx $$
For the first term, substitute $t = x^2 + x$. For the second term, substitute $x + \frac{1}{2} = \frac{1}{2}\cosh u$
