Hilbert Transform of a sine Function with Quadratic Argument $\sin(At^2 + Bt + \pi/4)$
Directly calculating this seems difficult. 

My argumentation is the following. For a signal $s(t)$, the so-called analytic signal $s_{\mathrm{an}}(t)$ can be obtained by 
\begin{equation}
s_{\mathrm{an}}(t) = s(t)+\mathrm{j} \mathcal{H}\{s(t)\}\,\,, \text{where}\,S_{\mathrm{an}}(\omega) = 0\,\forall\, \omega &lt; 0 
\end{equation}
The analytic signal essentially corresponds to the spectral content of $s(t)$ in the positive frequencies only.

For your first example of a simple sinusoid, you can also come to the result for the Hilbert transform if you consider the analytic signal. The real sinusoid consists of the frequency components $\pm B$. The analytic signal should then be the component at $B$, which is then obviously a complex exponential, yielding the result for the Hilbert transform.

Now for your chirp signal $x(t)$, the situation is somewhat more complicated. If we think about a virtual "instantaneous frequency course" of the signal, it is
\begin{equation}
\omega_{x}(t) = \pm (2At+B)\,.
\end{equation}
Now this is somewhat strange, corresponding to two linearly changing components of opposite slope, crossing the zero-frequency point at $\omega_{x}(t=-\frac{B}{2A})=0$.

Now the analytic signal would have to represent the part of this frequency course above the zero-line of the $\omega-t$ plane (I might add some plots later). This means it would have to first have a negative slope, go down to frequency zero, and then abruptly change to a positive slope!

This means that the analytic signal would have to look something like
\begin{equation}
x_{\mathrm{an}}(t) = c_1 \exp{(-\mathrm{j} (At^2 + Bt + \frac{\pi}{4}))}\,\forall\,t &lt; -\frac{B}{2A}
\end{equation}
and
\begin{equation}
x_{\mathrm{an}}(t) = c_2 \exp{(\mathrm{j} (At^2 + Bt + \frac{\pi}{4}))}\,\forall\,t \geq -\frac{B}{2A}\,\,,
\end{equation}
where the $c$ are some constant with $\vert c \vert = 1$.

Now we can determine the Hilbert transform of $x(t)$ by observing and checking in the equation for the analytic signal. This yields
\begin{equation}
\mathcal{H}\{x(t)\} = \cos{(At^2 + Bt + \frac{\pi}{4})}\,\forall\,t &lt; -\frac{B}{2A}\,,\,\text{with}\,c_1 = -\mathrm{j}\,,
\end{equation}
and
\begin{equation}
\mathcal{H}\{x(t)\} = -\cos{(At^2 + Bt + \frac{\pi}{4})}\,\forall\,t \geq -\frac{B}{2A}\,,\,\text{with}\,c_2 = \mathrm{j}\,.
\end{equation}

One can probably also write these as one equation with the absolute value function. In any case, the point is the Hilbert transform seems to contain a discontinuity, which is what makes this particularly confusing to calculate I suspect.

I know it is somewhat "handwaivy", but I think the general idea/result is correct, so hope this helps!