So for the equation below I used long division first, and got $\int{(1+\frac{-3x^2+9x-27}{x^3-3x^2+9x-27}})dx$ (So my signs should be the other way around, thank you to Alijah Ahmed for pointing it out, see answer below). Now I'm not sure how to proceed, completing the square would give me something weird, I think maybe using substitution? $$\int{\frac{x^3}{x^3-3x^2+9x-27}}dx$$
Update: using partial fractions, $A=1.5, B=1.5, C=-4.5$ and I obtained $x+ \frac{3}{2}ln(x-3)+ \frac{3}{4}ln(x^2+9)-\frac{3}{2}arctan(\frac{x}{3})+C$ as an answer.
As the comments have mentioned, use partial fractions.
Also, note that the denominator $x^3-3x^2+9x-27$ factorises to $(x-3)(x^2+9)$ which helps, so you want to find $A$, $B$ and $C$ such that $$\frac{A}{x-3}+\frac{Bx+c}{x^2+9}=\frac{\color{red}{+}3x^2\color{red}{-}9x\color{red}{+}27}{(x-3)(x^2+9)}$$ (you had the signs for the numerator after long division wrong - corrections in red in the equation above, i.e. you should have had $\int{(1+\frac{3x^2-9x+27}{x^3-3x^2+9x-27}})dx$)