Since the acceleration is the derivative of velocity, you can write $ \frac{\mathrm{d} v}{\mathrm{d} t} = p_0 + p_1 v + p_2 v^2 + p_3 v^3 $ separating the variables we get the integral form $ \int \frac{\mathrm{d}v}{p_0 + p_1 v + p_2 v^2 + p_3 v^3} = \int \mathrm{d}t = t + c$
Which we can integrate using partial fractions (also see this page). To summarise the method:
Using the fundamental theorem of algebra we can factor the polynomial $ p_0 + p_1 v + p_2 v^2 + p_3 v^3 = p_3 (v + \alpha_1)(v + \alpha_2)(v + \alpha_3) $ where the $\alpha$s are the roots of the polynomial (assume they are distinct for now; repeated roots will require some additional work). Then we look for $\beta_1,\beta_2,\beta_3$ such that $ \sum \frac{\beta_i}{v+\alpha_i} = \frac{1}{(v+\alpha_1)(v+\alpha_2)(v+\alpha_3)} $ Expanding the sum you see that this requires
$\begin{align} \beta_1 + \beta_2 + \beta_3 &= 0 \\ \beta_1 (\alpha_2 + \alpha_3) + \beta_2(\alpha_1+\alpha_3) + \beta_3(\alpha_1 + \alpha_2) &= 0 \\ \beta_1 \alpha_2\alpha_3 + \beta_2\alpha_1\alpha_3 + \beta_3 \alpha_1\alpha_2 &= 1 \end{align}$ which is a linear system that can be solved.
This way we reduce our integral equation to $ t + c = \frac{1}{p_3}\int \frac{\beta_1}{v + \alpha_1} + \frac{\beta_2}{v+\alpha_2} + \frac{\beta_3}{v+\alpha_3} \mathrm{d}v $ where the $\alpha$ and $\beta$ coefficients are determined from the polynomial you started with. This gives us the implicit solution
$ p_3t + C = \beta_1 \ln (v+\alpha_1) + \beta_2 \ln(v+\alpha_2) + \beta_3 \ln(v+\alpha_3) $
or
$ e^{p_3 t + C} = (v+\alpha_1)^{\beta_1}(v+\alpha_2)^{\beta_2}(v+\alpha_3)^{\beta_3} \tag{*}$
However, this is generally where one gets stuck. To obtain $d$ from $v$ you have to integrate $v$ one more time. But now equation (*) may not have nice analytic representation for $v$, nevermind a simple integral for you to obtain $d$. In those cases the best you can do is probably ask Mathematica.
(Sometime you may get lucky. For example, if your polynomial is a perfect cube, then you have $ \int \frac{\mathrm{d}v}{p(v+q)^3} = -\frac{1}{2p(v+q)^2} + C $ then you get that $ v + q = \sqrt{2p t + C} $ which one can easily integrate to get $d = \int v~\mathrm{d}t$. But those depends on special form of the coefficients $p_i$ which you have not specified.)