1. *The Kepler problem*

This problem is explained above. We ran it for five different eccentricities (i.e., *ecc* = 0, 0.2, 0.4, 0.6, 0.8), while we recorded the efficiency measures using the endpoint errors for *xend* = 10*π* and *xend* = 20*π*.

The efficiency measure ratios of DLMP6(5) vs. NEW6(5) for Kepler are presented in Tables 3 and 4.


**Table 3.** Efficiency measure ratios of DLMP6(5) vs. NEW6(5) for Kepler in [0, 10*π*].

**Table 4.** Efficiency measure ratios of DLMP6(5) vs. NEW6(5) for Kepler in [0, 20*π*].

