homework and exercises - Derivation of equations of motion in Nordstrom's theory of scalar gravity?

Nordstrom's theory of a particle moving in the presence of a scalar field $\varphi (x)$ is given by

$$S = -m\int e^{\varphi (x)}\sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}}d\lambda ,$$ where $\lambda$ is the parametrization of the worldline of the particle, ignoring the free field part $\int \eta_{\alpha \beta}\partial^{\alpha}\varphi \partial^{\beta} \varphi d^{4}x$.

How does one derive the equations of motion in terms of the parameter

$$d\tau = \sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}} d\lambda.$$ $u^{\alpha} = \frac{dx^{\alpha}}{d\tau} \Rightarrow u_{\alpha}u_{\beta}\eta^{\alpha \beta} = 1$?

My attempt:

$$\delta S = 0 \Rightarrow \int \left( \frac{\partial (e^{\varphi } \sqrt{...})}{\partial x^{\alpha}}\delta x^{\alpha} +\frac{\partial (e^{\varphi } \sqrt{...})}{\partial \left( \frac{d x^{\alpha}}{d \lambda } \right)}\frac{d}{d\lambda} \delta x^{\alpha} \right)d\lambda = |d\tau = \sqrt{...}d\lambda | =$$ $$= \int \left(\sqrt{...}e^{\varphi}\partial_{\alpha}\varphi - \frac{d}{d \lambda} \left(\frac{d x_{\alpha}}{d\tau}e^{\varphi}\right) \right)\delta x^{\alpha}d \lambda =$$ $$=\int \left( \partial_{\alpha}\varphi - \frac{d^{2}x_{\alpha}}{d \tau^{2}} - \frac{dx_{\alpha}}{d\tau} \frac{d \varphi }{d\tau }\right) \delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda =$$ $$=\int \left( \partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi \right)\delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda \Rightarrow$$ $$\partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi = 0 \Rightarrow \partial_{\alpha} \varphi = e^{-\varphi}\frac{d }{d \tau}\left( e^{\varphi } u_{\alpha}\right).$$ Unfortunately, this equation doesn't look like the equation from Wikipedia, $$\frac{d (\varphi u_{\alpha})}{d \tau} = -\partial_{\alpha } \varphi.$$ I can explain the part of differences by renaming the function, $e^{\varphi } \to \varphi$, in the expression for action (then my equation reduces to the form $\partial_{\alpha} \varphi = \frac{d }{d \tau}\left( \varphi u_{\alpha}\right)$), but I can't explain why my equation has the wrong sign.