I would like to solve the following indefinite integral containing complex numbers

$$ \int \frac{1}{\sqrt{(x-a)(x-b)(x-c-di)(x-c+di)}}dx $$

where $a,b,c,d>0$ are real numbers and i is the known $i=\sqrt{-1}$. It implies that $d$ is the imaginary part of the complex number, respectively, $-c-di$ and $-c+di$. Roots are actually $(-a, -b, -c-di, -c+di)$. It comes from a 4th degree polynomial for which I found these roots. However, two of them are complex numbers as you can see. Is there any possibility to solve it through an exact or approximated existing method? Your support would be much appreciated. Thank you in advance. M.

  • $\begingroup$ Partial fractions? $\endgroup$
    – kipf
    Jul 11 at 16:26
  • $\begingroup$ Firstly the roots you have mentioned is incorrect instead they are a, b, c+di and c-di . It can be anticipated that the closed form would not have elementary functions, my guess is you would need to make more conditions on the coefficients. Could you state what is the motivation behind this question? $\endgroup$
    – ayan
    Jul 11 at 16:54
  • $\begingroup$ This can be thought of as the integral of the differential form $y^{-1}\,\mathrm{d}x$ on the algebraic curve given by $$y^2=(x-a)(x-b)((x-c)^2+d^2).$$ Since this curve is birationally equivalent to an elliptic curve, I believe that the integral essentially represents a closed-form in terms of an elliptic function. $\endgroup$ Jul 11 at 17:43
  • $\begingroup$ I do not see the problem (have a look at my answer to your previous question). $\endgroup$ Jul 12 at 3:39

1 Answer 1


In the Byrd-Friedman book ``Handbook of Elliptic Integrals'', 2nd edition, the integral 260.00 is $$ \int_a^y \frac{dt}{\sqrt{(t-a)(t-b)(t-c)(t-\bar c)}} = g F(\varphi,k) $$ where $\bar c$ is the complex-conjugated of $c$, $F$ the Incomplete Elliptic Integral of the First Kind, and $$ b_1=\frac{c+\bar c}{2}, $$ $$ a_1^2=-\frac{(c-\bar c)^2}{4}, $$ $$ A^2=(a-b_1)^2+a_1^2, \quad B^2=(b-b_1)^2+a_1^2 $$ $$ k^2=\frac{(A+B)^2-(a-b)^2}{4AB} $$ $$ \varphi = arccos \frac{(A-B)y+aB-bA}{(A+B)y-aB-bA} $$ $$ g=\frac{1}{\sqrt{AB}} $$

  • $\begingroup$ Before to close the answer (btw thank you so much!!!! great support!), I am just wondering: it is clear that the integration extremes are "a" and "y" in your answer as well as in the book. However, the lower intergation extreme "a" is same as the first root in (x-a). Therefore, in the equations that you have written, where do I apply the term "a" (meant as lower integr. extreme) and/or the term "a" (meant as the first root)? On this regard, I noticed you put (x-1) to the first root in the intergal. Is it to show me that the term "a" in A,k and φ refers only to the lower integration extreme? $\endgroup$
    – Mark Int
    Jul 12 at 16:50
  • $\begingroup$ could you pleae kindly answer back? Thank you! $\endgroup$
    – Mark Int
    Jul 13 at 18:58
  • 1
    $\begingroup$ All the definite integrals in the Byrd-Friedman book have one of the polynomial's roots as the lower or upper limit. If your integrals have both limits in between roots, you need to subtract two integrals of that kind. Apart from that, the term a is "applied" in the intermediate definition of A^2 as well as of k^2. $\endgroup$ Jul 13 at 20:27

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