5. As shown in Figure 1, square $ABCD$ is inscribed in $\odot O$, point $P$ is on the arc $\overparen{AB}$, and $DP$ intersects $AC$ at point $Q$. If $QP=QO$, then the value of $\frac{QC}{QA}$ is ( ).
(A) $2 \sqrt{3}-1$
(B) $2 \sqrt{3}$
(C) $\sqrt{3}+\sqrt{2}$
(D) $\sqrt{3}+2$
5.D.

As shown in Figure 4, let the radius of $\odot O$ be $r$, $QO = m$, then $QP = m$, $QC = r + m$, $QA = r - m$.

In $\odot O$, according to the intersecting chords theorem, we have
$$
QA \cdot QC = QP \cdot QD,
$$

which is $(r-m)(r+m) = m \cdot QD$.
Therefore, $QD = \frac{r^2 - m^2}{m}$.
Connecting $DO$, by the Pythagorean theorem, we get
$$
QD^2 = DO^2 + QO^2.
$$

That is $\left(\frac{r^2 - m^2}{m}\right)^2 = r^2 + m^2$.
Solving this, we get $m = \frac{\sqrt{3}}{3} r$.

Therefore, $\frac{QC}{QA} = \frac{r + m}{r - m} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \sqrt{3} + 2$.