In $4$d Euclidean space, the Maldacena-Wilson line is defined as:
$$\mathcal{W}(C) = \frac{1}{N} \text{Tr} \left\lbrace \mathcal{P} \exp \int_C d\tau \left( i \dot{x}_\mu A_\mu^a(x) + \left| \dot{x} \right|\theta_i \phi^{i,a}(x) \right) \right\rbrace \tag{1}$$
where $\phi$ and $A_\mu$ are scalar and gluon fields respectively, $\tau$ is the $4$th component of $x$, $\dot{x}=(0,0,0,1)$, $\mathcal{P}$ refers to the path-ordering operator, $C$ is an infinite line on the $\tau$-direction, and the trace is meant to act on the color indices $a$, $b$ ($i$, $j$ are SO($6$) indices). When expanding the exponential to $2$nd order, we get for example the following term:
$$\frac{1}{2! N} \text{Tr}\ \mathcal{P} \int_C d\tau_1\ d\tau_2\ \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{2}$$
One can get rid of the path ordering by realizing the following relation:
$$\begin{align} \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^{\tau_1} d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} &= \text{Tr} \int_{-\infty}^\infty d\tau_2 \int_{\tau_2}^{\infty} d\tau_1 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \\ &= \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{\tau_1}^{\infty} d\tau_2 \left| \dot{x}_2 \right| \left| \dot{x}_1 \right| \theta_i \theta_j \phi_2^{i,a} \phi_1^{j,b} \\ &= \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{\tau_1}^{\infty} d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{3} \end{align}$$
where in the last line I used the cyclicity of the trace. This implies:
$$\text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^{\tau_1} d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} = \frac{1}{2} \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^\infty d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{4}$$
and thus
$$\frac{1}{2! N} \text{Tr}\ \mathcal{P} \int_C d\tau_1\ d\tau_2\ \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} = \frac{1}{2! N} \frac{1}{2} \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^\infty d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{5}$$
Now let us expand the Wilson line to $3$rd order. Is there a convenient formula such as $(5)$ to get rid of the path ordering at this order? Since the trace acts now on $3$ generators, I imagine it would be necessary to have something antisymmetric like $\epsilon (\tau_1 \tau_2 \tau_3)$ in the integral in order to change the sign depending on the path ordering. But how to set the limit of integration to $-\infty$ and $+\infty$ for all integrals?
Specifically I am searching for a formula for the following "mixed term":
$$\frac{1}{2! 1! N} \text{Tr}\ \mathcal{P} \int_C d\tau_1\ d\tau_2\ d\tau_3 \left| \dot{x}_1 \right| \dot{x}_{2\mu} \left| \dot{x}_3 \right| \theta_i \theta_j\ \phi_1^{i,a} A_{2\mu}^b \phi_3^{j,c} \tag{6}$$
As a response to the comment, a useful reference is probably this one, where an expression for $(6)$ involving a $\epsilon(\tau_1 \tau_2 \tau_3)$ is used in eq. $(13)$ without explaining how. I would like to know if I can always do that or if this is a special case.
