Is there anyway to solve $\displaystyle \int t \frac{\sin \left(\frac{t}{2} \sqrt{ a \left(t+ \frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}}\right) }{ \sqrt{ a \left(t+ \frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}}} \operatorname{d}t \tag8$ either by analytical method or from geometrical method with out using numerical methods? means looking for a closed form with out infinite series expansion in the result
NB: Main issue is the lack of general form
Observe that the radical can be simplified to $\sqrt{at^2+bt+c}$
Now Integrate by Parts $\displaystyle\int t\dfrac{\sin\left(\dfrac{t\sqrt{at^2+bt+c}}2\right)}{\sqrt{at^2+bt+c}}\ dt$
$\displaystyle=t\int\dfrac{\sin\left(\dfrac{t\sqrt{at^2+bt+c}}2\right)}{\sqrt{at^2+bt+c}}\ dt-\int\left[\frac{d(t)}{dt}\int\dfrac{\sin\left(\dfrac{t\sqrt{at^2+bt+c}}2\right)}{\sqrt{at^2+bt+c}}\ dt\right]dt \tag 1$ is the result of Integration by Parts
and for $\displaystyle\int\dfrac{\sin\left(\dfrac{t\sqrt{at^2+bt+c}}2\right)}{\sqrt{at^2+bt+c}}\ dt,$ set $t\sqrt{at^2+bt+c}=2u$ to find something really interesting