Analytical solution for a separable scalar nonlinear ODE

I am considering the scalar ODE $$ \dot x(t) = -x(t)\sqrt{1+x(t)^4} \quad \text{with} \quad x(0) = x_0\in\mathbb{R}$$ for all $t\geq 0$. This ODE is separable: $$ -\int \frac{{\rm d}x}{x\sqrt{1+x^4}} = \int^{t}_{0}{\rm d}s$$ so we get the implicit equation $$ \frac{1}{2}{\rm arctanh}\left({\sqrt{1+x^4}}\right) + \mathbb{c} = t.$$ We proceed with the manipulation to get $$ x(t) = \left( \tanh^{2}(2t-2\mathbb{c}) -1 \right)^{1/4}.$$ At this point, I realize I have an issue... the argument $\tanh^2(y)-1=-{\rm sech}^2(y)$ is negative, no matter what $\mathbb{c}$ is so I get a complex solution. I am looking for a real solution for all real initial conditions $x_0$. Is this possible?

Edit

Wolfram Alpha gives me four different solutions which all have a similar negative argument in $(\cdot)^{1/4}$ which is puzzling.

Moreover, isn't the right-hand side at least Lipschitz continuous so uniqueness is assured?

Following the observations in the comments, here is my proposed answer. First, for $x\neq 0$ we have $$ -\int\frac{{\rm d}x}{x \sqrt{1+x^4}} = \frac{1}{2}{\rm arccoth}\left( \sqrt{x^4+1}\right) + \mathbb{c} = \int_{0}^{t} {\rm d}s = t.$$

Using the initial condition $x(0)=x_{0}\neq 0$ and solving for $\mathbb{c}$ gives $$ \mathbb{c} = g(x_{0}) = \frac{1}{2}{\rm arccoth}^2\left(\sqrt{1+x_{0^4}}\right).$$

Solving for $x$ and ignoring the imaginary roots yields $$ x = \pm\left(\coth^{2}(2\mathbb{c}-2t)-1\right)^\frac{1}{4} = \pm ({\rm csch}^{2}\left( 2g(x_0)-2t \right))^\frac{1}{4}$$.

Which leads to the solution of the ODE for all $x_{0}\in\mathbb{R}$ being $$x(t) = {\rm sign}(x_0)({\rm csch}^{2}\left( 2g(x_0)-2t \right))^\frac{1}{4}.$$