How does one approach integrating:
$f(x)=\sin(\sqrt{ax^4+bx^3+cx^2+dx+e})$?
I have tried with substitution, which yields the following. I am quite new to this complexity of integration, so please tell me if there are any mistakes.
$\displaystyle\int f(x)dx=\displaystyle\int \sin(\sqrt{ax^4+bx^3+cx^2+dx+e})dx$
$u=\sqrt{ax^4+bx^3+cx^2+dx+e}$
$\displaystyle\int f(x)dx=\displaystyle\int \sin(u)dx$
Next, try to write $dx$ as a function $du$ and $u$.
$du=\dfrac{4ax^3+3bx^2+2cx+d}{\sqrt{ax^4+bx^3+cx^2+dx+e}}dx$
$du=\dfrac{4ax^3+3bx^2+2cx+d}{u}dx$
Here I get stuck, because I don't know what to do with the top part of the fraction to get rid of the remaining x's. I see that:
$4ax^3+3bx^2+2cx+d=\dfrac{d}{dx}u^2$,
but I am not sure what I can do with that.
Guidance is really appreciated.
How I got to this integral
For context, I am creating a model of a quad rotor drone. In this case simplified to a single continuously running motor. I will leave most of the physics formulas out of this and try to make things concise.
Angle of the drone under a single running motor in axis-angle representation.
$\underrightarrow{\theta}=\theta \underrightarrow{e},\lvert e\rvert=1$
$\underrightarrow{\theta}(t)=\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}+t\dot{\underrightarrow{\theta}}(0)+\underrightarrow{\theta}(0)$, with $\ddot{\underrightarrow{\theta}}$ constant.
$\theta(t)=\sqrt{(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_x+t\dot{\underrightarrow{\theta}}_x(0)+\underrightarrow{\theta}_x(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_y+t\dot{\underrightarrow{\theta}}_y(0)+\underrightarrow{\theta}_y(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_z+t\dot{\underrightarrow{\theta}}_z(0)+\underrightarrow{\theta}_z(0))^2}$
$\underrightarrow{e}(t)=\dfrac{\underrightarrow{\theta}(t)}{\theta(t)}$
Next, velocity of the drone, and where the complicated integral comes in.
Acceleration under the (constant) force of one rotor
$\underrightarrow{\alpha}=\dfrac{1}{m}\underrightarrow{F^{'}}=\dfrac{1}{m}(\cos(\theta)\underrightarrow{F}+\sin(\theta)(\underrightarrow{e}\times\underrightarrow{F})+\underrightarrow{e}(1-\cos(\theta))(\underrightarrow{e}\cdot\underrightarrow{F}))$
with $F$ the force applied by the rotor, and $F^{'}$ that force after it is rotated by $\underrightarrow{\theta}$.
Integrating this to obtain velocity
$\underrightarrow{v}=\dfrac{1}{m}\displaystyle\int\underrightarrow{a}(t)\underrightarrow{F}+\underrightarrow{b}(t)+\underrightarrow{c}(t)dt$, with
$\underrightarrow{a}(t)=\cos(\theta(t))$
$\underrightarrow{b}(t)=\sin(\theta(t))(\underrightarrow{e}(t)\times\underrightarrow{F})$
$\underrightarrow{c}(t)=\underrightarrow{e}(t)(1-\cos(\theta(t)))(\underrightarrow{e}(t)\cdot\underrightarrow{F})$
I approached the integral by splitting into the above three parts. This question is about integrating part a. In all its large form it looks like
$\displaystyle\int\underrightarrow{a}(t)dt=\displaystyle\int\cos(\sqrt{(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_x+t\dot{\underrightarrow{\theta}}_x(0)+\underrightarrow{\theta}_x(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_y+t\dot{\underrightarrow{\theta}}_y(0)+\underrightarrow{\theta}_y(0))^2+(\dfrac{1}{2}t^2\ddot{\underrightarrow{\theta}}_z+t\dot{\underrightarrow{\theta}}_z(0)+\underrightarrow{\theta}_z(0))^2})dt$
Which I simplified to the original equation given in this question. (I now see that it's actually $\cos$ and not $\sin$, but it's all the same)
comment
For example, consider $\int\sin\sqrt{x^4+1}\;dx$. Change variables using $y^2 = x^4+1$ to get $$ \int\frac{y\sin y}{2(y^2-1)^{3/4}}\;dy $$ This integral probably is not an elementary function. It is not known to Maple. I did not find it in Gradshteyn & Ryzhik.