I want to perform a definite integral
$ \int_{t_1}^{t_2} \frac{dt}{\sqrt{a t^3+ b t^2+ct+d}}, $ and will be happy to get the answer in terms of elliptic functions. Can somebody please guide me on how to achieve this using the exhaustive list of functions and results at DLMF (NIST) website?
On
Hint
I think that for such an integral, I should rewrite first $$a t^3+ b t^2+ct+d=a(t-\alpha)(t-\beta)(t-\gamma)$$ This would give for the antiderivative $$\int \frac{dt}{\sqrt{a t^3+ b t^2+ct+d}}=-\frac{2 F\left(\sin ^{-1}\left(\frac{\sqrt{\beta -\alpha }}{\sqrt{t-\alpha }}\right)|\frac{\alpha -\gamma }{\alpha -\beta }\right)}{ \sqrt{a(\beta -\alpha)}}$$ where appears the elliptic integral of the first kind.
The above result shows the change of variable you would need to use, namely $$t=\alpha +(\beta -\alpha ) \csc ^2(u)$$
The way to derive the integral really depends on the characteristics of the variables, $a$,$b$,$c$,and $d$, and also on the region where the integral takes place. Let us first assume here $a>0$ and \begin{equation} \begin{split} \sqrt{at^3+bt^2+ct+d} =\sqrt{a}\sqrt{t^3+\frac{b}{a}t^2+\frac{c}{a}t+\frac{d}{a}}\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \end{split} \end{equation} Now setting $(1)=0$ and find the solution of this equation. In one case you may get three real solutions, $a_1$, $a_2$, $a_3$ and we assume here $a_1>a_2>a_3$. The other case is that you may get one real solution, $a_1$ and two conjugated complex solutions, $b_1$ and $b_2$. Here, I will give the derivation in the case of the integral region from $a_1$ to $t$ In the three real solutions case, transform the variable $t$ to $u$ using the relation,
\begin{equation} \begin{split} sn^2[u,k]=\frac{t-a_1}{t-a_2} \,\,\,\,\,\,\,\,\,\,\,\,\,(2) \end{split} \end{equation} where $sn[u]$ is the Jacobian sn function. It is the inversed function of elliptic integral shown below. \begin{equation} \begin{split} u=\int^{x}_0\frac{dx}{\sqrt{(1-x^2)(1-k^2 x^2)}} \,\,\,\,\,\,\,\,\,\,\,\,\,(3) \end{split} \end{equation} and $x=sn[u,k]$, where k is the modulus. In the case of (2), the specific value of modulus k will be derived in the following process. From (2), \begin{equation} \begin{split} t=\frac{a_1-a_2 sn^2[u,k]}{cn^2[u,k]} \,\,\,\,\,\,\,\,\,\,\,\,\,(4) \end{split} \end{equation} where, \begin{equation} \begin{split} cn^2[u,k]=1-sn^2[u,k] \end{split} \end{equation} and is the Jacobian cn function. The derivative of $t$ in (4) yields, \begin{equation} \begin{split} dt=\frac{2(a-b)sn[u,k]dn[u,k]}{cn^3[u,k]}du \,\,\,\,\,\,\,\,\,\,\,\,\,(5) \end{split} \end{equation} Here $dn[u,k]$ is the Jacobian dn function and has the relation, \begin{equation} \begin{split} dn^2[u,k]=1-k^2sn^2[u,k] \end{split} \end{equation} Also, the derivatives of sn, cn, dn functions with regard to u used in (5) can be easily derived from the relation (3) as, \begin{equation} \begin{split} &\frac{d}{du}sn[u,k]=cn[u,k]dn[u,k] \\ &\frac{d}{du}cn[u,k]=-sn[u,k]dn[u,k] \\ &\frac{d}{du}dn[u,k]=-k^2sn[u,k]cn[u,k] \\ \end{split} \end{equation} Inserting (4) and (5) in the integral \begin{equation} \begin{split} \frac{1}{\sqrt{a}}\int^t_{a_1}\frac{dt}{\sqrt{t^3+\frac{b}{a}t^2+\frac{c}{a}t+\frac{d}{a}}} =\frac{1}{\sqrt{a}}\int^t_{a_1}\frac{dt}{(t-a_1)(t-a_2)(t-a_3)} \,\,\,\,\,\,\,\,\,\,\,\,\,(6) \end{split} \end{equation} yields \begin{equation} \begin{split} &\frac{1}{\sqrt{a}}\int^t_{a_1}\frac{dt}{(t-a_1)(t-a_2)(t-a_3)} \\ &=\frac{1}{\sqrt{a}}\int^{u_1}_0\frac{2(a_1-a_2)sn\,dn\,du} {\sqrt{(a_1-a_3cn^2-a_2sn^2)[-a_1(1-cn^2)+a_2sn^2][-a_1+a_2(sn^2+cn^2)]}} \\ &=\frac{1}{\sqrt{a}}\int^{u_1}_0\frac{2(a_1-a_2)sn\,dn\,du} {\sqrt{(a_1-a_3)[1-(a_2-a_3)/(a_1-a_3)sn^2][(a_2-a_1)sn^2](a_2-a_1)}} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,(7) \end{split} \end{equation} where $\sin u_1=\sqrt{(t-a_1)/(t-a_2)}.$From now on, I will just write $sn$, $cn$, and $dn$ and omit using the bracket $[u,k]$. By putting $k^2=(a_2-a_3)/(a_1-a_3)$, (7) is simplified as \begin{equation} \begin{split} &\frac{1}{\sqrt{a}}\int^{u_1}_0\frac{2(a_1-a_2)sn\,dn\,du} {\sqrt{(a_1-a_3)[1-(a_2-a_3)/(a_1-a_3)\,sn^2][(a_2-a_1)\,sn^2](a_2-a_1)}} \\ &\frac{1}{\sqrt{a(a_1-a_3)}}\int^t_{a_1}\frac{2(a_1-a_2)sn\,dn\,du} {\sqrt{(1-k^2\,sn^2)[(a_2-a_1)\,sn^2](a_2-a_1)}} \\ &=\frac{2}{\sqrt{a(a_1-a_3)}}\int^{u_1}_0\,du =\frac{2}{\sqrt{a(a_1-a_3)}}\,F\left[\sin^{-1}\sqrt{\frac{t-a_1}{t-a_2}},k\right] \,\,\,\,\,\,\,\,\,\,\,\,\,(8) \end{split} \end{equation} where $F$ is the Jacobian elliptic integral of the first kind. In the case where the solutions of $(1)=0$ are one real value, $a_1$, and two conjugated complex values $b_1$, $b_2$, we introduce new parameters $\beta$, $\alpha$, and $A$ as, \begin{equation} \begin{split} &\beta=\frac{b_1+b_2}{2} \\ &\alpha^2=-\frac{(b_1-b_2)^2}{4}>0 \\ &A^2=(\beta-a_1)^2+\alpha^2 \end{split} \end{equation} This time the transformation of the variable from t to u is done by assuming the relation, \begin{equation} \begin{split} cn[u,k]=\frac{A+a_1-t}{A-a_1+t} \,\,\,\,\,\,\,\,\,\,\,\,\,(9) \end{split} \end{equation} Therefore, in the same manner as derived above \begin{equation} \begin{split} &t=\frac{a_1+A+(a_1-A)\,cn}{1+cn} \\ &dt=\frac{2\,A\,sn\,dn}{(1+cn)^2}\,du \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,(10) \end{split} \end{equation} Inserting (10) to (6) yields \begin{equation} \begin{split} &\frac{1}{\sqrt{a}}\int^{u_1}_0 2A\sqrt{\frac{(1+cn)^3}{2A^2(1-cn)[2A-(A+\beta-a_1)\,sn^2]}} \frac{sn\,dn}{(1+cn)^2}\,du \\ &=\frac{1}{\sqrt{a}}\int^{u_1}_0 \frac{sn\,dn\,du}{\sqrt{A(1+cn)(1-cn)[1-(A+\beta-a_1)/(2A)\,sn^2]}} \\ =&\frac{1}{\sqrt{a\,A}}\int^{u_1}_0du =\frac{1}{\sqrt{a\,A}}F\left[\cos^{-1}\frac{A+a_1-t}{A-a_1+t},k\right] \end{split} \end{equation} where $k$ in this case is $k^2=(A+\beta-a_1)/(2A)$, and $\cos u_1=(A+a_1-t)/(A-a_1+t)$. There are other ways of transforming the integral depending on the regions of the integral and the sign of $a$. I will not refer to it, since the principle is the same as mentioned so far.