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# scipy.interpolate.splantider

`scipy.interpolate.``splantider`(tck, n=1)

  Compute the spline for the antiderivative (integral) of a given spline.

  Parameters: | tck : BSpline instance or a tuple of (t, c, k) Spline whose antiderivative to compute n : int, optional Order of antiderivative to evaluate. Default: 1
  Returns: | BSpline instance or a tuple of (t2, c2, k2) Spline of order k2=k+n representing the antiderivative of the input spline. A tuple is returned iff the input argument tck is a tuple, otherwise a BSpline object is constructed and returned.

  See also

  `splder`, `splev`, `spalde`, `BSpline`

  Notes

  The `splder` function is the inverse operation of this function. Namely, `splder(splantider(tck))` is identical to tck, modulo rounding error.

  New in version 0.13.0.

  Examples

      >>> from scipy.interpolate import splrep, splder, splantider, splev
      >>> x = np.linspace(0, np.pi/2, 70)
      >>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2)
      >>> spl = splrep(x, y)

  The derivative is the inverse operation of the antiderivative, although some floating point error accumulates:

      >>> splev(1.7, spl), splev(1.7, splder(splantider(spl)))
      (array(2.1565429877197317), array(2.1565429877201865))

  Antiderivative can be used to evaluate definite integrals:

      >>> ispl = splantider(spl)
      >>> splev(np.pi/2, ispl) - splev(0, ispl)
      2.2572053588768486

  This is indeed an approximation to the complete elliptic integral \(K(m) = \int_0^{\pi/2} [1 - m\sin^2 x]^{-1/2} dx\):

      >>> from scipy.special import ellipk
      >>> ellipk(0.8)
      2.2572053268208538