scipy.special.elliprc(x, y, out=None) = <ufunc 'elliprc'>#

Degenerate symmetric elliptic integral.

The function RC is defined as [1]

\[R_{\mathrm{C}}(x, y) = \frac{1}{2} \int_0^{+\infty} (t + x)^{-1/2} (t + y)^{-1} dt = R_{\mathrm{F}}(x, y, y)\]
x, yarray_like

Real or complex input parameters. x can be any number in the complex plane cut along the negative real axis. y must be non-zero.

outndarray, optional

Optional output array for the function values

Rscalar or ndarray

Value of the integral. If y is real and negative, the Cauchy principal value is returned. If both of x and y are real, the return value is real. Otherwise, the return value is complex.

See also


Completely-symmetric elliptic integral of the first kind.


Symmetric elliptic integral of the second kind.


Completely-symmetric elliptic integral of the second kind.


Symmetric elliptic integral of the third kind.


RC is a degenerate case of the symmetric integral RF: elliprc(x, y) == elliprf(x, y, y). It is an elementary function rather than an elliptic integral.

The code implements Carlson’s algorithm based on the duplication theorems and series expansion up to the 7th order. [2]

New in version 1.8.0.



B. C. Carlson, ed., Chapter 19 in “Digital Library of Mathematical Functions,” NIST, US Dept. of Commerce.


B. C. Carlson, “Numerical computation of real or complex elliptic integrals,” Numer. Algorithm, vol. 10, no. 1, pp. 13-26, 1995.


Basic homogeneity property:

>>> import numpy as np
>>> from scipy.special import elliprc
>>> x = 1.2 + 3.4j
>>> y = 5.
>>> scale = 0.3 + 0.4j
>>> elliprc(scale*x, scale*y)
>>> elliprc(x, y)/np.sqrt(scale)

When the two arguments coincide, the integral is particularly simple:

>>> x = 1.2 + 3.4j
>>> elliprc(x, x)
>>> 1/np.sqrt(x)

Another simple case: the first argument vanishes:

>>> y = 1.2 + 3.4j
>>> elliprc(0, y)
>>> np.pi/2/np.sqrt(y)

When x and y are both positive, we can express \(R_C(x,y)\) in terms of more elementary functions. For the case \(0 \le x < y\),

>>> x = 3.2
>>> y = 6.
>>> elliprc(x, y)
>>> np.arctan(np.sqrt((y-x)/x))/np.sqrt(y-x)

And for the case \(0 \le y < x\),

>>> x = 6.
>>> y = 3.2
>>> elliprc(x,y)
>>> np.log((np.sqrt(x)+np.sqrt(x-y))/np.sqrt(y))/np.sqrt(x-y)