If $W_t$ is a standard Brownian motion, I know that $$\int_0^t \operatorname{sign}(W_s) \, {\rm d}W_s$$ is also a standard Brownian motion, where $\operatorname{sign}(x) = \begin{cases} +1 &\text{if } x \geq 0\\ -1 &\text{otherwise}\end{cases}$
Thanks to Lévy's theorem, when $X_t = a + b t + W_t$, we know that
$$Z_t = \frac1{b} \int_0^t \text{sign}(X_s) \, {\rm d}X_s$$
is a Brownian motion. I wonder if we can get a simple formula for it expected value. In others words, because of the martingale property of the stochastic integral, I would like to compute
$$ \mathbb{E} \left(\int_0^t \operatorname{sign}(X_s) \,{\rm d}s \right) $$
I get expressions involving
$$\Phi(x) = \int_{x}^{\infty} e^{-u^2/2} \frac{du}{\sqrt{2 \pi}}$$
but no simple formula.