Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?

Lengthy but not impossible. We have for sure
$$\begin{eqnarray*} f(x)&=&\frac{1}{(x+3)(x+4)^2(x+5)^3}\\&=&\frac{A}{x+3}+\frac{B}{x+4}+\frac{C}{x+5}+\frac{D}{(x+4)^2}+\frac{E}{(x+5)^2}+\frac{F}{(x+5)^3}\end{eqnarray*}\tag{1}$$
where
$$ F=\lim_{x\to -5}(x+5)^3 f(x) = \lim_{x\to -5}\frac{1}{(x+3)(x+4)^2} = -\frac{1}{2},$$
$$ D=\lim_{x\to -4}(x+4)^2 f(x) = \lim_{x\to -4}\frac{1}{(x+3)(x+5)^3} = -1,$$
$$ A = \lim_{x\to -3}(x+3) f(x)=\lim_{x\to -3}\frac{1}{(x+4)^2 (x+5)^3} = \frac{1}{8}.\tag{2} $$
By setting $g(x)=f(x)-\frac{F}{(x+5)^3}-\frac{D}{(x+4)^2}-\frac{A}{(x+3)}$ it follows that:
$$ \begin{eqnarray*}g(x)&=&\frac{20-3x-x^2}{8(x+4)(x+5)^2}\\&=&\frac{B}{x+4}+\frac{C}{x+5}+\frac{E}{(x+5)^2}\end{eqnarray*}\tag{3}$$
where
$$ E = \lim_{x\to -5}g(x)(x+5)^2 = -\frac{5}{4}, $$
$$ B = \lim_{x\to -4}g(x)(x+4) = 2\tag{3} $$
and $C=-\frac{17}{8}$ follows from computing $g(x)-\frac{E}{(x+5)^2}-\frac{B}{x+4}$, or from a general property of the partial fraction decomposition of "rapidly decaying" meromorphic functions, that in this case grants $C=-(A+B)$, since $A,B,C$ are the residues of $f(x)$ at its poles. You may deduce the same by noticing that $\lim_{x\to +\infty} x\cdot f(x)=0$.
It follows that:
$$ \int\frac{dx}{(x+3)(x+4)^2(x+5)^3}\\=C+\frac{1}{8} \left(\frac{8}{x+4}+\frac{2}{(x+5)^2}+\frac{10}{x+5}+\log(x+3)+16\log(x+4)-17\log(x+5)\right).$$
