<s>[INST] Fastest way to integrate $\int_0^1 x^{2}\sqrt{x+x^2} \hspace{2mm}dx $ This integral looks simple, but it appears that its not so.
All Ideas are welcome, no Idea is bad, it may not work in this problem, but may be useful in some other case some other day ! :)
 [/INST] Note that 
$$x^2\sqrt{x+x^2}=\frac{x^3+x^4}{\sqrt{x(1+x)}}$$
So, let us look for a polynomial $P(x)=a x^3+bx^2+cx+d$ such that the derivative
$(P(x)\sqrt{x(1+x)})^\prime$ is as close as we can to this function. An easy calculation shows that
$$
\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{4 a x^4+(\frac{7 a }{2}+3 b)
   x^3+(\frac{5 b }{2}+2 c)
   x^2+(\frac{3 c }{2}+d )x+\frac{d}{2}}{\sqrt{x(1+x)}}
$$
So, choosing $a=\frac{1}{4}$, $b=\frac{1}{24}$, $c=-\frac{5}{96}$ and $d=\frac{5}{64}$ we see that
$$
\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{  x^4+ 
   x^3}{\sqrt{x(1+x)}}+\frac{5}{64}\cdot\frac{1}{2\sqrt{x(1+x)}}
$$
This reduces the considered integral to a simple one:
$$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}\Big]_0^1-\frac{5}{64}\int_0^1\frac{dx}{2\sqrt{x(1+x)}}$$
The last integral is easy since $\log(\sqrt{x}+\sqrt{1+x})$ is a primitive of the integrand.
Thus
$$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}-\frac{5}{64}\log(\sqrt{x}+\sqrt{1+x})\Big]_0^1$$
Finally,
$$\int_0^1x^2\sqrt{x(1+x)}dx=
\frac{61}{96 \sqrt{2}}-\frac{5}{64}
   \log \left(1+\sqrt{2}\right).
$$
</s>