Elliptic Integrals involving square roots of polynomial of fourth order

Question

Does any one know the solution of the integral \begin{equation} \int_{a}^{u}\frac{x^{2}}{\sqrt{x(x-a)(x-b)(x-c) }}.dx \end{equation}

where $u > a > b > 0 > c$

I believe it has one polynomial component and a linear combination of Elliptic Integrals of the First, Second and Third Kind. I believe [but not sure] it is of the form

\begin{equation} \frac{ \sqrt{x(x-a)(x-b)(x-c)} } {x-b}+\text{linear combination of Elliptic Integrals}\end{equation}

I believe the limits $u>a>b>0>c$ has a bearing on the form of the elliptic integrals.

Any help would be appreciated.

Answer 1

The tables list integrals of inverse square roots of polynomials of fourth order up to $$\int \frac{x \ dx}{\sqrt{(a-x)(b-x)(c-x)(d-x)}}$$ only eg. Gradhsteyn/Rhyzik 3.148 p242 or more special cases with double roots. So its worth to try Mathematica, since possibly its only complexity of the results, that makes it impossible to print the result. A bit of mathematical AI

   j=  Assuming[a > b > c > x > d > 0, 
       Simplify[\[Integral]x^2 ((x - a) (x - b) (x - c) (x - d))^(-1/ 2)
     [DifferentialD]x]
 
   (two pages of output)

The problem is the catalog of different square roots, tabulated in Gradhsteyn/Rhyzik for all integrals leading to elliptic integrals. This can be automated by searching all power forms and replacing them by greek characters

repl = ( Rule @@@ ({#,Take[CharacterRange["\[Alpha]", "\[Omega]"], 
 Length[#]]}\[Transpose]) &)[
Union@Quiet[
Cases[j,HoldPattern[ _*_*(_)^-1*(_)^-1] |  Power[_, 1/2 | -1/2], \[Infinity]]]];
TeXForm[TableForm[repl]]

$$\begin{array}{c} \frac{(a-c) (b-d)}{(a-d) (b-c)}\to \alpha \\ \frac{(a-b) (c-d)}{(a-c) (b-d)}\to \beta \\ \frac{(a-d) (b-x)}{(a-x) (b-d)}\to \gamma \\ \sqrt{\frac{(a-d) (b-x)}{(a-x) (b-d)}}\to \delta \\ \sqrt{(a-x) (b-c) (b-d) (b-x) (c-d) (c-x)}\to \varepsilon \\ \sqrt{(a-c) (c-b) (c-d) (d-x)}\to \zeta \\ \frac{1}{\sqrt{-((b-c) (b-d) (c-d) (d-x))}}\to \eta \\ \sqrt{(a-d) (d-b) (c-d) (d-x)}\to \theta \\ \sqrt{(a-d) (b-d) (d-c) (d-x)}\to \iota \\ \sqrt{-((d-a) (b-d) (d-c) (d-x))}\to \kappa \\ \sqrt{-((a-d) (d-b) (d-c) (d-x))}\to \lambda \\ \frac{(a-c) (x-d)}{(a-x) (c-d)}\to \mu \\ \sqrt{(a-d) (d-b) (d-c) (x-d)}\to \nu \\ \sqrt{\frac{(a-c) (x-d)}{(a-x) (c-d)}}\to \xi \\ \end{array}$$

Verification may need longer than I have time to wait

Assuming[ a > b > c > x > d > 0, Simplify[PowerExpand[D[j, x]^-2]]]

Answer 2

For the general case the Byrd-Friedman "Handbook of elliptic integrals for engineers and scientists" has [258.11] $$ \int_a^y \frac{t^m}{\sqrt{(t-a)(t-b)(t-c)(t-d)}}dt = a^m g\int_0^{u_1}\frac{(1-\alpha_1^2 sn^2 u)^m}{(1-\alpha^2 sn^2 u)^m} du see 340.04 $$ where $g=2/\sqrt{(a-c)(b-d)}$, $k^2=(b-c)(a-d)/[(a-c)(b-d)]$, $y>a>b>c>d$, where $\alpha_1^2=(a-d)b/[a(b-d)]$. [340.04] is $$ \int \frac{(1-\alpha_1^2 sn^2 u)^m}{(1-\alpha^2 sn^2 u)^m}du =(\frac{\alpha_1^2}{\alpha^2})^m\sum_{j=0}^m \binom{m}{j} (\frac{\alpha^2-\alpha_1^2}{\alpha_1^2})^j V_j $$ where the $V_j$ are given in [336] and in equation (77) in https://arxiv.org/abs/2203.17227 .

Answer 3

In my answer I assumed the normal elliptic integral with x between the smallest singular points. The condition, that x greater than the largest singular point and the smallest is zero, yields a managable exercise in steering the Mathematica engine clear of its defiiciencies with respect to elliptic problems.

Declare a complete list of conditions

   $Assumptions = 
      x > a && x > b && x > c && x >0 && a > b && a > c && b > c

$$\text{int1}=\int \frac{x^2}{\sqrt{x (x-a) (x-b) (x-c)}} \, dx$$

Get the basis of the elliptic integrals occuring

  base = Cases[int1, _EllipticF | _EllipticE | _EllipticPi, \[Infinity]]

$$\left\{E\left(\sin ^{-1}\left(\sqrt{\frac{(a-b) (x-c)}{(c-b) (x-a)}}\right)|\frac{a (c-b)}{(a-b) c}\right),F\left(\sin ^{-1}\left(\sqrt{\frac{(a-b) (x-c)}{(c-b) (x-a)}}\right)|\frac{a (c-b)}{(a-b) c}\right),\Pi \left(\frac{c-b}{a-b};\sin ^{-1}\left(\sqrt{\frac{(a-b) (x-c)}{(c-b) (x-a)}}\right)|\frac{a (c-b)}{(a-b) c}\right)\right\}$$

Get them out of the way

   repl = Rule @@@ ({base, {\[ScriptCapitalE], \[ScriptCapitalF], \
                     \[CapitalPi]}}\[Transpose])

and extract the coefficients and simpilfy the sparse array converted to Normal

     coeffs =  CoefficientArrays[  int1 /. repl // 
          ExpandAll,
            {\[ScriptCapitalE], \[ScriptCapitalF], \[CapitalPi]}] // 
              Together // Normal // FullSimplify

$$\left\{\sqrt{\frac{x (x-b) (x-c)}{x-a}},\left\{-i c \sqrt{\frac{b-a}{c}},i \sqrt{\frac{1}{c (b-a)}} \left(a^2+a (b+c)-b c\right),i (c-a) \sqrt{\frac{1}{c (b-a)}} (a+b+c)\right\}\right\}$$

Reconstruct the expression (in order to differentiate)

       int2 = First[coeffs] + Rest[coeffs] . base

$$i \left(a^2+a (b+c)-b c\right) \sqrt{\frac{1}{c (b-a)}} F\left(\sin ^{-1}\left(\sqrt{\frac{(a-b) (x-c)}{(c-b) (x-a)}}\right)|\frac{a (c-b)}{(a-b) c}\right)+\sqrt{\frac{x (x-b) (x-c)}{x-a}}-i c \sqrt{\frac{b-a}{c}} E\left(\sin ^{-1}\left(\sqrt{\frac{(a-b) (x-c)}{(c-b) (x-a)}}\right)|\frac{a (c-b)}{(a-b) c}\right)+i (c-a) (a+b+c) \sqrt{\frac{1}{c (b-a)}} \Pi \left(\frac{c-b}{a-b};\sin ^{-1}\left(\sqrt{\frac{(a-b) (x-c)}{(c-b) (x-a)}}\right)|\frac{a (c-b)}{(a-b) c}\right)$$

Compute the derivative squared for verification

    (D[int2, x] /. repl // FullSimplify)^2 // FullSimplify

$$\frac{x^3}{(x-a) (x-b) (x-c)}$$

The underlying problems with elliptic functions in Mathematica rest on the fact, that any internal call of any elliptic term runs through the complete test for special cases of the arguments (zeroes, poles, periods). Renaming the basis occurring in an expression, reduces the algebra to simple normal. The test of linear independence of the basis can be done in another run, eg with numerical values of the singular points.