In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential:

$\u25fb{A}^{\alpha}={\mu}_{0}{J}^{\alpha}$

To get $A$ in terms of $J$, however, we have to use a considerably uglier formula; or, at least, this is the formula presented in textbooks:

${A}^{\alpha}(t,\mathbf{r})=\frac{{\mu}_{0}}{4\pi}\int \frac{{J}^{\alpha}(t-\frac{1}{c}\Vert \mathbf{r}-{\mathbf{r}}^{\prime}\Vert ,{\mathbf{r}}^{\prime})}{\Vert \mathbf{r}-{\mathbf{r}}^{\prime}\Vert}{d}^{3}{\mathbf{r}}^{\prime}$

The first equation is evidently Lorentz covariant. The second one, on the other hand, doesn't look covariant at all. The integrand has some messy dependence on $(t,\mathbf{r})$ and the integration goes over only the spatial dimensions.

Can we rewrite the second equation in a covariant form? If not, then why not?

$\u25fb{A}^{\alpha}={\mu}_{0}{J}^{\alpha}$

To get $A$ in terms of $J$, however, we have to use a considerably uglier formula; or, at least, this is the formula presented in textbooks:

${A}^{\alpha}(t,\mathbf{r})=\frac{{\mu}_{0}}{4\pi}\int \frac{{J}^{\alpha}(t-\frac{1}{c}\Vert \mathbf{r}-{\mathbf{r}}^{\prime}\Vert ,{\mathbf{r}}^{\prime})}{\Vert \mathbf{r}-{\mathbf{r}}^{\prime}\Vert}{d}^{3}{\mathbf{r}}^{\prime}$

The first equation is evidently Lorentz covariant. The second one, on the other hand, doesn't look covariant at all. The integrand has some messy dependence on $(t,\mathbf{r})$ and the integration goes over only the spatial dimensions.

Can we rewrite the second equation in a covariant form? If not, then why not?