<s>[INST] $ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$$
doing the substitution
$$ x= \frac{t^2}{1-2 t}$$
according to the rule.
So the integral becomes:
$$  \int \frac{-2t^6}{(1-2t)^4}\, dt= \int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}\frac{-80t^3+90t^2-36t+5}{(1-2t)^4})\, dt=\int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}(\frac{10}{1-2t}-\frac{15}{2} \frac{1}{(1-2t)^2}+\frac{3}{(1-2t)^3}-\frac{1}{2} \frac{1}{(1-2t)^4}))\, dt=-\frac{1}{24}t^3-\frac{1}{8}t^2-\frac{5}{16}t-\frac{5}{16}\cdot \ln|1-2t|
-\frac{15}{64}\frac{1}{1-2t}+\frac{3}{64} \frac{1}{(1-2t)^2}-\frac{1}{16 \cdot 12} \frac{1}{(1-2t)^3}+cost $$
with $t=-x+ \sqrt{x^2+x}$.
The final result according to my book is instead $(\frac{1}{3}x^2-\frac{5}{12}x+\frac{15}{24})\sqrt{x^2+x}-\frac{5}{16}\ln( x+\frac{1}{2}+ \sqrt{x^2+x})$
And trying to obtain the same solution putting t in the formulas I'm definitely lost in the calculation...
I don't understant why this difference in the complexity of the solution...
Can someone show me where I'm making mistakes?
 [/INST] Just for your curiosity since you already received @hamam_Abdallah's answer.
You can compute this integral musch faster with a different change of variable.
Because of the denominator, let $x=\sinh^2(t)$
$$\int \frac{x^3}{\sqrt{x^2+x}}\, dx=2 \int \sinh ^6(t)\,dt$$
Now, using the multiple angle formulae
$$\sinh ^6(t)=\frac{15}{32} \cosh (2 t)-\frac{3}{16} \cosh (4 t)+\frac{1}{32} \cosh (6
   t)-\frac{5}{16}$$ and everything becomes simple.
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