<s>[INST] How to integrate $\frac{1}{(x+1)(x+2)^2(x+3)^3}$? I tried to solve it with partial fraction decomposition but the expression becomes way too difficult to solve. I could only solve three of six(A-F) expressions of the partial fraction expansion.
 [/INST] When partial fraction decomposition becomes a bit overwhelming, you can apply the Horowitz-Ostrogradsky algorithm ! [Manuel Bronstein - Symbolic Integration I]
It is very mechanical, only the calculation of $H$ is tedious, but the rest is quite easy.
So we start with $$\frac AD=\frac{1}{(x+1)(x+2)^2(x+3)^3}$$
$A=1$
$D=(x+1)(x+2)^2(x+3)^3$
$D\,'=(x+2)(x+3)^2(6x^2+22x+18)$
$D^-=\gcd(D,D\,')=(x+2)(x+3)^2$
$D^*=D/D^-=(x+1)(x+2)(x+3)$
$B=\sum_{i=0}^{\deg(D^-)-1}b_ix^i=b_0+b_1x+b_2x^2$
$C=\sum_{i=0}^{\deg(D^*)-1}c_ix^i=c_0+c_1x+c_2x^2$

And let's identify to the null polynomial $$\forall x:\quad H(x)=A-B\,'D^*+BD^*{D^{-}}'/D^--CD^-=0$$

$H(x)=-c_2x^5+(b_2-c_1-8c_2)x^4+(2b_1-2b_2-c_0-8c_1-21c_2)x^3+(3b_0+4b_1-15b_2-8c_0-21c_1-18c_2)x^2+(10b_0-4b_1-12b_2-21c_0-18c_1)x+(7b_0-6b_1-18c_0+1)$

This system solves to $\begin{cases}B=\frac 94x^2+\frac{25}{2}x+17\\C=\frac 94x+\frac 52\end{cases}$
And the formula says :
$$\int \frac AD\mathop{dx}=\frac{B}{D^-}+\int\frac{C}{D^*}\mathop{dx}=\frac{9x^2+50x+68}{4(x+2)(x+3)^2}+\int\frac{9x+10}{4(x+1)(x+2)(x+3)}\mathop{dx}$$

The last part is still solved by partial fraction decomposition, but is much simpler:
$\int=-\frac{17}8\ln(x+3)+2\ln(x+2)+\frac 18\ln(x+1)$
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