= Solution
It is helpful first to separate an exact <gradient>:
$$
\mathbf F=\nabla\bigl(\alpha x^2+\alpha^2xy+\alpha xz\bigr)+(0,\beta z,\beta y^2).
$$
The <fundamental theorem for line integrals> makes the gradient contribution $2\alpha+\alpha^2$ for either path. On the curved path, $dy=2t\,dt$, $dz=3t^2\,dt$, $y=t^2$ and $z=t^3$, so the remaining contribution is
$$
\beta\int_0^1(z\,dy+y^2\,dz)=\beta\int_0^1(2t^4+3t^6)\,dt=\frac{29\beta}{35}.
$$
Thus
$$
\boxed{\int\mathbf F\cdot d\mathbf r=\alpha^2+2\alpha+\frac{29\beta}{35}}.
$$
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