I am looking for a faster way to find the appropriate Euler-Lagrange equations for a variational problem of the form $$\mathcal{L}[x,u] = \int_a^b L\left(x(t),u(t),\frac{u_t}{x_t},\frac{1}{x_t}\frac{d}{dt}\left[\frac{u_t}{x_t}\right]\right)x_t\,dt$$ obtained by parameterizing the standard second-order Lagrangian $\mathcal{L}[u] = \int_{x_0}^{x_1}L(x,u,u_x,u_{xx})\,dx$ in terms of a dummy variable $t$. Specifically, I would like a faster way of doing this than the approach that you generally take for the computation of the first-order EL equations and first integral formulas. If we consider the first-order case $$\mathcal{L}[x,u] = \int_a^b L\left(x(t),u(t),\frac{u_t}{x_t}\right)x_t\,dt$$ and define a path $\gamma:(-1,1)\rightarrow G$ with $\gamma(0) = e$ then we may define the group infinitesimals by $$\left.\frac{d}{d\varepsilon}\right|_{\varepsilon = 0}\gamma(\varepsilon)\cdot x := \theta\qquad \left.\frac{d}{d\varepsilon}\right|_{\varepsilon = 0}\gamma(\varepsilon)\cdot u := \psi\qquad \left.\frac{d}{d\varepsilon}\right|_{\varepsilon = 0}\gamma(\varepsilon)\cdot u_I := \psi_I$$ where $I = xx..x$. Taking the standard approach to finding the EL equations, we can then compute $$\begin{align*}0&=\int_a^b \left.\frac{d}{d\varepsilon}\right|_{\varepsilon = 0}L\left(\gamma(\varepsilon)\cdot x(t),\gamma(\varepsilon)\cdot u(t),\gamma(\varepsilon)\cdot \frac{u_t}{x_t}\right)\gamma(\varepsilon)\cdot x_t\,dt\\ &=\int_a^b \frac{\partial L}{\partial x} x_t \theta + \frac{\partial L}{\partial u}x_t \psi + \frac{\partial L}{\partial \frac{u_t}{x_t}}\left(\frac{x_t\psi_t - u_t\theta_t}{x_t}\right) + L\theta_t\, dt \end{align*}$$ which, when you try to apply integration by parts, yields the following EL equations $$\begin{align*}E^x(L) &= \frac{\partial L}{\partial x}x_t - \frac{d}{dt}\left[L - \frac{u_t}{x_t}\frac{\partial L}{\partial \frac{u_t}{x_t}}\right]\\ E^u(L)&= \frac{\partial L}{\partial u}x_t - \frac{d}{dt}\left[\frac{\partial L}{\partial \frac{u_t}{x_t}}\right]\end{align*}$$ from the equation $$\int_a^b E^x(L)\theta + E^u(L)\psi + \frac{d}{dt}\left[\left(L - \frac{u_t}{x_t}\frac{\partial L}{\partial \frac{u_t}{x_t}}\right)\theta + \frac{\partial L}{\partial \frac{u_t}{x_t}}\psi\right]\,dt.$$ Now, we could just repeat this computation for the EL equations in the second-order case, but that gets pretty nasty considering the quotient rules and makes it difficult to generalize completely, so I was wondering if anyone knows of a better way to go about this computation. My first thought was to try to use Noether's Theorem for multiple dependent variables, but when I try to do that I do not seem to get the same equations out. If anyone can shed some light on where I've gone wrong or provide a general formula, I would greatly appreciate it. Thank you!
Update: If it is helpful to anyone, here are the EL equations for the second and third-order cases.
For the second-order: $$\begin{align*}E^x(L) &= \frac{\partial L}{\partial x} x_t - \frac{d}{dt}\left[ L - \frac{u_t}{x_t}D_3(L) + \frac{-2u_{tt}x_t + 3u_tx_{tt}}{x_t^3} D_4(L)\right] + \frac{d^2}{dt^2}\left[\frac{-u_t}{x_t^2}D_4(L)\right]\\ E^u(L)&= \frac{\partial L}{\partial u}x_t - \frac{d}{dt}\left[D_3(L) - \frac{x_{tt}}{x_t^2}D_4(L)\right] + \frac{d^2}{dt^2}\left[\frac{1}{x_t}D_4(L)\right]\end{align*}$$
For the third-order: $${\scriptsize \begin{align*}E^x(L) &= \frac{\partial L}{\partial x} x_t - \frac{d}{dt}\left[ L - \frac{u_t}{x_t}D_3(L) + \frac{-2u_{tt}x_t + 3u_tx_{tt}}{x_t^3} D_4(L) + \frac{4u_t x_{ttt} x_t^5 - 15u_tx_{tt}^2x_t^4 + 12u_{tt}x_{tt}x_{t}^5 - 3u_{ttt}x_{t}^6}{x_t^9}D_5(L)\right] + \frac{d^2}{dt^2}\left[\frac{-u_t}{x_t^2}D_4(L) + \frac{-3u_{tt}x_{t} + 6u_tx_{tt}}{x_t^4}D_5(L)\right] - \frac{d^3}{dt^3}\left[\frac{-u_t}{x_t^3}D_5(L)\right]\\ E^u(L)&= \frac{\partial L}{\partial u}x_t - \frac{d}{dt}\left[D_3(L) - \frac{x_{tt}}{x_t^2}D_4(L) + \frac{-x_t^6x_{ttt} + 3x_{t}^5x_{tt}^2}{x_{t}^9}D_5(L)\right] + \frac{d^2}{dt^2}\left[\frac{1}{x_t}D_4(L)+\frac{-3x_t^6x_{tt}}{x_t^9}D_5(L)\right] - \frac{d^3}{dt^3}\left[\frac{1}{x_t^2}D_5(L)\right]\end{align*}}$$
Where $D_i(L)$ indicates taking the derivative with respect to the $i$-th coordinate of $L$.