mmetric elliptic integral of the second kind.
    elliprj : Symmetric elliptic integral of the third kind.

    Notes
    -----
    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]_

    .. versionadded:: 1.8.0

    References
    ----------
    .. [1] B. C. Carlson, ed., Chapter 19 in "Digital Library of Mathematical
           Functions," NIST, US Dept. of Commerce.
           https://dlmf.nist.gov/19.16.E6
    .. [2] B. C. Carlson, "Numerical computation of real or complex elliptic
           integrals," Numer. Algorithm, vol. 10, no. 1, pp. 13-26, 1995.
           https://arxiv.org/abs/math/9409227
           https://doi.org/10.1007/BF02198293

    Examples
    --------
    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)
    (0.5484493976710874-0.4169557678995833j)

    >>> elliprc(x, y)/np.sqrt(scale)
    (0.5484493976710874-0.41695576789958333j)

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

    >>> x = 1.2 + 3.4j
    >>> elliprc(x, x)
    (0.4299173120614631-0.3041729818745595j)

    >>> 1/np.sqrt(x)
    (0.4299173120614631-0.30417298187455954j)

    Another simple case: the first argument vanishes:

    >>> y = 1.2 + 3.4j
    >>> elliprc(0, y)
    (0.6753125346116815-0.47779380263880866j)

    >>> np.pi/2/np.sqrt(y)
    (0.6753125346116815-0.4777938026388088j)

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

    >>> x = 3.2
    >>> y = 6.
    >>> elliprc(x, y)
    0.44942991498453444

    >>> np.arctan(np.sqrt((y-x)/x))/np.sqrt(y-x)
    0.44942991498453433

    And for the case :math:`0 \le y < x`,

    >>> x = 6.
    >>> y = 3.2
    >>> elliprc(x,y)
    0.4989837501576147

    >>> np.log((np.sqrt(x)+np.sqrt(x-y))/np.sqrt(y))/np.sqrt(x-y)
    0.49898375015761476

    Ú