I am trying to derive the Adams Mouton 2 step method for $$y'(t)= f(t,y(t))$$ by interpolating $f$ at the nodes $(x,f_j)$, $(x+h,f_{j+1})$ and $(x+h+\vartheta h,f_{j+2})$. I get the following lagrange interpolating polynomial $$p\left(t\right)=f_j\frac{\left(t-x-h\right)\left(t-x-h-\vartheta \:h\right)}{h^2\left(1-\vartheta \right)}+f_{j+1}\frac{\left(t-x\right)\left(t-x-h-\vartheta \:h\right)}{-\vartheta \:h^2}\:+f_{j+2}\frac{\left(t-x\right)\left(t-x-h\right)}{\vartheta \:h^2\left(1+\vartheta \right)}$$
Now to get the approximation $$y(x+h+\vartheta h)=y(x+h)+\int_{x+h}^{x+h+\vartheta h}f(t,y(t))~dt\approx y(x+h)+\int_{x+h}^{x+h+\vartheta h}p(t)~dt$$
but I can't seem to be able to integrate the polynomial. Are there any tricks to integrating it?
I might get this wrong, but in my opinion there are no tricks in integrating $p$, it is just a standard integration: the integral can be decomposed in the sum of three different integrals, and for instance $\int^{x+h+\theta h}_{x+h}{f_{j+2}(t-x)(t-x-h)/(\theta h^2 (1+\theta))}dt=f_{j+2}/(\theta h^2 (1+\theta))\int^{x+h+\theta h}_{x+h}{(t-x)(t-x-h)}dt=f_{j+2}/(\theta h^2 (1+\theta))\int^{x+h+\theta h}_{x+h}{t^2-t(x+h)-xt+x(x+h)} dt$
Then you just need to be able to integrate $t$ and $t^2$ (and costants).