By the linearity of integration, trace, and matrix multiplication, we can pull $\int$ inside:
$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx=\mathrm{tr}\left( \int (Ax+B)^{-1}dx\, C\right).$
I originally thought $\log(Ax+B)\;A^{-1}$ would be the antiderivative for this, but oenamen helpfully points out this can't be said in full generality ($\log XY=\log X+\log Y$ does not necessarily hold if the matrices $X$ and $Y$ don't commute!); instead, we need a special $B^{-1}$ factor. We have:
$\frac{d}{dx}\log(I+xU) =\quad U(I+xU)^{-1} =(I+xU)^{-1}U.$
Hence, applying the above alongside $X^{-1}Y^{-1}=(YX)^{-1}$:
$\begin{array}{c l} \frac{d}{dx}\log \big(B^{-1}(Ax+B)\big) & =\frac{d}{dx}\log(I+xB^{-1}A) \\ & = (I+xB^{-1}A)^{-1}B^{-1}A \\ & = \big(B\cdot(I+B^{-1}Ax)\big)^{-1}A \\ & = (Ax+B)^{-1}A. \end{array}$
Therefore we conclude
$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx =\mathrm{tr}\left(\log\big(I+xB^{-1}A)\; A^{-1}C\right)+const. $