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# reduction of elliptic integrals to standard form

Any integral of the form $\int R(x,\sqrt{P(x)})\,dx$, where $R$ is a rational
function and $P$ is a polynomial^{} of degree 3 or 4 can be expressed as a linear
combination of elementary functions and elliptic integrals of the first, second,
and third kinds.

To begin, we will assume that $P$ has no repeated roots. Were this not the case, we could simply pull the repeated factor out of the radical and be left with a polynomial of degree of 1 or 2 inside the square root and express the integral in terms of inverse trigonometric functions.

Make a change of variables $z=(ax+b)/(cx+d)$. By choosing the coefficients $a,b,c,d$ suitably, one can cast P into either Jacobi’s normal form $P(z)=(1-z^{2})(1-k^{2}z^{2})$ or Weierstrass’ normal form $P(z)=4z^{3}-g_{2}z-g_{3}$.

Note that

$R(z,\sqrt{P(z)})=\frac{A(z)+B(z)\sqrt{P(z)}}{C(z)+D(z)\sqrt{P(z)}}$ |

for suitable polynomials $A,B,C,D$. We can rationalize the denominator like so:

$\frac{A(z)+B(z)\sqrt{P(z)}}{C(z)+D(z)\sqrt{P(z)}}\times\frac{C(z)-D(z)\sqrt{P(% z)}}{C(z)-D(z)\sqrt{P(z)}}=F(z)+G(z)\sqrt{P(z)}$ |

The rational functions $F$ and $G$ appearing in the foregoing equation are defined like so:

$\displaystyle F(z)$ | $\displaystyle=$ | $\displaystyle\frac{A(z)C(z)-B(z)D(z)P(z)}{C^{2}(z)-D^{2}(z)P(z)}$ | ||

$\displaystyle G(z)$ | $\displaystyle=$ | $\displaystyle 2\frac{B(z)C(z)-A(z)D(z)}{C^{2}(z)-D^{2}(z)P(z)}$ |

Since $\int F(z)\,dz$ may be expressed in terms of elementary functions,
we shall focus our attention on the remaining piece, $\int G(z)\sqrt{P(z)}\,dz$,
which we shall write as $\int H(z)/\sqrt{P(z)}\,dz$, where $H=PG$..
Because we may decompose $H$ into partial fractions^{}, it suffices to consider
the following cases, which we shall all $A_{n}$ and $B_{n}$:

$A_{n}(z)=\int\frac{z^{n}}{\sqrt{P(z)}}\,dz$ |

$B_{n}(z,r)=\int\frac{1}{(z-r)^{n}\sqrt{P(z)}}\,dz$ |

Here, $n$ is a non-negative integer and $r$ is a complex number^{}.

We will reduce thes further using integration by parts. Taking antiderivatives, we have:

$\int\frac{z^{{n-1}}(zP^{{\prime}}(z)+2nP(z))}{2\sqrt{P(z)}}\,dz=z^{n}\sqrt{P(z% )}+C$ |

$\int\frac{(z-r)P^{{\prime}}(z)-2nP(z)}{2(z-r)^{{n+1}}\sqrt{P(z)}}\,dz=\frac{% \sqrt{P(z)}}{(z-r)^{n}}+C$ |

These identities will allow us to express $A_{n}$’s and $B_{n}$’s with large $n$ in terms of ones with smaller $n$’s.

At this point, it is convenient to employ the specific form of the polynominal $P$. We will first conside the Weierstrass normal form and then the Jacobi normal form.

Substituting into our identities and collecting terms, we find

$4(2n+3)A_{{n+2}}=(2n+1)g_{2}A_{n}+2ng_{3}A_{{n-1}}+z^{n}\sqrt{4z^{3}-g_{2}x-g_% {3}}+C$ |

$2n(4r^{3}-g_{2}r-g_{3})B_{{n+1}}+(2n-1)(12r^{2}-g_{2})B_{n}+24(n-1)rB_{{n-1}}+% 4(2n-3)B_{{n-2}}+\frac{\sqrt{4z^{3}-g_{2}x-g_{3}}}{(z-r)^{n}}+C=0$ |

Note that there are some cases which can be integrated in elementary terms. Namely, suppose that the power is odd:

$\int z^{{2m+1}}\sqrt{(1-z^{2})(1-k^{2}z^{2})}\,dz$ |

Then we may make a change of variables $y=z^{2}$ to obtain

$\frac{1}{2}\int y^{{2m}}\sqrt{(1-y)(1-k^{2}y)}\,dy,$ |

which may be integrated using elementary functions.

Next, we derive some identities using integration by parts. Since

$d\left((1-z^{2})(1-k^{2}z^{2})\sqrt{(1-z^{2})(1-k^{2}z^{2})}\right)=\left(% \frac{9}{2}k^{2}z^{3}-3(1+k^{2})z\right)\sqrt{(1-z^{2})(1-k^{2}z^{2})}\,dz,$ |

we have

$\displaystyle(2m+1)$ | $\displaystyle\int z^{{2m}}(1-z^{2})(1-k^{2}z^{2})\sqrt{(1-z^{2})(1-k^{2}z^{2})% }\,dz$ | |||

$\displaystyle+$ | $\displaystyle\int z^{{2m+1}}\left(\frac{9}{2}k^{2}z^{3}-3(1+k^{2})z\right)% \sqrt{(1-z^{2})(1-k^{2}z^{2})}\,dz$ | |||

$\displaystyle=$ | $\displaystyle z^{{2m+1}}(1-z^{2})(1-k^{2}z^{2})\sqrt{(1-z^{2})(1-k^{2}z^{2})}+C$ |

By colecting terms, this identity may be rewritten as follows:

$\displaystyle\left(1+2m+\frac{9}{2}k^{2}\right)$ | $\displaystyle\int z^{{2m+4}}\sqrt{(1-z^{2})(1-k^{2}z^{2})}\,dz-$ | |||

$\displaystyle(4+2m)(1+k^{2})$ | $\displaystyle\int z^{{2m+2}}\sqrt{(1-z^{2})(1-k^{2}z^{2})}\,dz+$ | |||

$\displaystyle\int z^{{2m}}\sqrt{(1-z^{2})(1-k^{2}z^{2})}=$ | ||||

$\displaystyle x^{{2k+1}}(1-z^{2})(1-k^{2}z^{2})\sqrt{(1-z^{2})(1-k^{2}z^{2})}+C$ |

By repeated use of this identity, we may express any integral of the form $\int z^{{2m}}\sqrt{P(z)}\,dz$ as the sum of a linear combination of $\int z^{2}\sqrt{P(z)}\,dz$ and $\int\sqrt{P(z)}\,dz$ and the product of a polyomial and $\sqrt{P(z)}$.

Likewise, we can use integration by parts to simplify integrals of the form

$\int\frac{\sqrt{P(z)}}{(z-r)^{n}}\,dz$ |

*Will finish later — saving in case of computer crash.*

## Mathematics Subject Classification

33E05*no label found*

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