How to solve this integral/better way to approach? $$\int_{0}^{c} dy \sqrt{\frac{c-1/2y^2+1/3y^3}{1+2y}}$$ where c is a constant. This is coming from trying to find the area $$\int_{U \le c} dq_1dq_2$$ where $$U=\frac{1}{2}(q_1^2+q_2^2)-\frac{1}{3}q_2^3+q_1^2q_2$$ bounded by energy $c=U(q_1,q_2)$.
 A: (Not an answer, just a comment that was too long). 
You can prove that the value of the integral is $\frac{c^2}{2\sqrt{6}} + O(c)$ with the following algebraic simplifications. First note that the integral can be written as
$$ I = \frac{1}{\sqrt{6}}\int_0^c \sqrt{ (y-1)^2 + \frac{6c-1}{2y+1}} \ dy. $$
It follows that 
$$I > \frac{1}{\sqrt{6}} \int_0^c (y-1) \ dy = \frac{c(c-2)}{2\sqrt{6}}.$$ 
Similarly, using the fact that $\sqrt{a+b} < \sqrt{a} + \sqrt{b}$ (which does not quite hold in some regions of the domain but seems to be insignificant for large $c$), we have
$$ I < \frac{1}{\sqrt{6}} \int_0^c (y-1) \ dy + \frac{1}{\sqrt{6}}\int_0^c \sqrt{\frac{6c-1}{2y+1}} \ dy  = \frac{c^2}{2\sqrt{6}} + O(c).$$
Thus we can conclude that $I = \frac{c^2}{2\sqrt{6}} + O(c).$ I think what could possibly help you is the following:


*

*If you want just a numerical answer, the function you are integrating is very smooth and convex so getting high precision values is doable.

*If you want a more accurate answer, you need to specify what regions of $c$ you are interested in. My answer holds for $c \rightarrow \infty$ but there are certainly more accurate answers for other cases such as $c << 1$.

A: The Appell-Lauricella function is defined by the series
$$
F[\{a,c\};\{b_1,b_2,\dots,b_n\};\{x_1,x_2,\ldots,x_n\}]:=
$$
$$
=\sum_{i_1,i_2,\ldots,i_n\geq 0}\frac{(a)_{i_1+i_2+\ldots+i_n}(b_1)_{i_1}(b_2)_{i_2}\ldots(b_n)_{i_n}}{(c)_{i_1+i_2+\ldots+i_n}i_1!i_2!\ldots i_n!}x_1^{i_1}x_2^{i_2}\ldots x_n^{i_n},
$$
where $n\geq2$, $a,c,b_1,b_2,\ldots,b_n\in\textbf{C}$ and $|x_1|<1,|x_2|<1,\ldots,|x_n|<1$.
Then holds the following
THEOREM.
For $Re(c)>Re(a)>0$ and $|x_1|<1,|x_2|<1,\ldots,|x_n|<1$, we have
$$
F[\{a,c\};\{b_1,b_2,\dots,b_n\};\{x_1,x_2,\ldots,x_n\}]=
$$
$$
=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int^{1}_{0}t^{a-1}(1-t)^{c-a-1}(1-x_1t)^{-b_1}(1-x_2t)^{-b_2}\ldots (1-x_nt)^{-b_n}dt.
$$
Using the above theorem I will prove that
$$
\int^{c}_{0}\sqrt{\frac{c-y^2/2+y^3/3}{1+2y}}dy=
\frac{c\sqrt{4-l}}{2\sqrt{6}}|l-1|\times
$$
$$
\times F\left[\{1,2\};\{\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\};\{-2c,\frac{2c}{l-1},\frac{4c}{4-l-\sqrt{3}\sqrt{(4-l)l}},\frac{4c}{4-l+\sqrt{3}\sqrt{(4-l)l}}\}\right],
$$
where $c=\frac{1}{24}(4-9l+6l^2-l^3)$.
For to prove the above evaluation make the change of variable $y\rightarrow -y$ to get
$$
\int^{c}_{0}\sqrt{\frac{c-y^2/2+y^3/3}{2y+1}}dy=i\int^{-c}_{0}\sqrt{\frac{y^2/2+y^3/3-c}{-2y+1}}dy,
$$
then $y\rightarrow \frac{1-w}{2}$ to get
$$
i\int^{-c}_{0}\sqrt{\frac{y^2/2+y^3/3-c}{-2y+1}}dy=\frac{\sqrt{c}}{4\sqrt{6}}\int^{2c+1}_{1}\sqrt{\frac{24+1/c(w-4)(w-1)^2}{w}}dw.
$$
Now if $c=\frac{1}{24}(4-9l+6l^2-l^3)$ we can write
$$
24+(-4+w)(-1+w)^2/c=\frac{24(l-w)(9-6l+l^2-6w+lw+w^2)}{(l-4)(l-1)^2}.
$$
Hence we can write the last integral in the form of theorem and use it to get the result, which is the Appell-Lauricella function.
