Faster way to get a rational integral $\int_2^3 \frac{x^{100}+1}{x^3+1} \, dx$

Well...maybe the division isn't that bad. Do you know about geometric series? What's

$$ 1 + r + r^2 + \ldots + r^{32}? $$ It's $$ \frac{r^{33} - 1}{r-1} $$ If you apply this to $r = -x^3$, you get $$ \frac{-x^{99} - 1}{-(x^3)-1} = \frac{x^{99} + 1}{x^3 + 1} $$ I know that's not what you wanted, but if you do a little algebra, you can do this: \begin{align} \frac{x^{100} + 1}{x^3 + 1} &= \frac{x^{100} + x - x + 1}{x^3 + 1}\\ &= \frac{x^{100} + x}{x^3 + 1} + \frac{-x + 1}{x^3 + 1}\\ &= x\frac{x^{99} + 1}{x^3 + 1} + \frac{-x + 1}{x^3 + 1}. \end{align} Now integrating the first fraction is easy because of the geometric series, and all you have to deal with is the last one.

We have $$ J=\int_{2}^{3}\frac{x-1}{x^3+1}\,dx = \frac{1}{3}\log\left(\frac{21}{16}\right)\tag{1}$$ by partial fraction decomposition, and that is the only annoying part, since $$ I = \int_{2}^{3}\frac{x^{100}+1}{x^3+1}\,dx = -J+\int_{2}^{3}x\cdot\frac{x^{99}+1}{x^3+1}\,dx \tag{2}$$ and $$ \frac{x^{99}+1}{x^3+1} = 1-x^3+x^6-\ldots+x^{96} \tag{3} $$ so: $$ \boxed{\int_{2}^{3}\frac{x^{100}+1}{x^3+1}\,dx} =-J+ \int_{2}^{3}\sum_{k=0}^{32}(-1)^k x^{3k+1}dx=\boxed{-\frac{1}{3}\log\left(\frac{21}{16}\right)+\sum_{k=0}^{32}(-1)^k\frac{3^{3k+2}-2^{3k+2}}{3k+2}}\tag{4} $$

It is not so bad with long division since $$\frac{x^{100}+1}{x^3+1}=\sum_{n=0}^{32}(-1)^nx^{3n+1}+\frac{1-x}{x^3+1}$$ and $$\frac{1-x}{x^3+1}=\frac{1-2 x}{3 \left(x^2-x+1\right)}+\frac{2}{3 (x+1)}$$ So, the antiderivative is quite simple but, for the definite integral, I hope and wish that you are very patient !