Optimal. Leaf size=307 \[ \frac {4 a b \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (a^2 (39 A+50 C)+4 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac {\left (5 a^2 (5 A+6 C)+12 A b^2\right ) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{120 d}+\frac {\left (a^4 (5 A+6 C)+12 a^2 b^2 (3 A+4 C)+8 b^4 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)+24 A b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^4}{6 d}+\frac {2 A b \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{15 d} \]
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Rubi [A] time = 1.12, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3048, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {4 a b \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 b^4 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a b \left (a^2 (39 A+50 C)+4 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac {\left (10 a^2 b^2 (49 A+66 C)+15 a^4 (5 A+6 C)+24 A b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {\left (5 a^2 (5 A+6 C)+12 A b^2\right ) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{120 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^4}{6 d}+\frac {2 A b \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{15 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 3021
Rule 3031
Rule 3047
Rule 3048
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+b \cos (c+d x))^3 \left (4 A b+a (5 A+6 C) \cos (c+d x)+b (A+6 C) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac {2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{30} \int (a+b \cos (c+d x))^2 \left (12 A b^2+5 a^2 (5 A+6 C)+2 a b (23 A+30 C) \cos (c+d x)+3 b^2 (3 A+10 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{120} \int (a+b \cos (c+d x)) \left (6 b \left (4 A b^2+a^2 (39 A+50 C)\right )+a \left (15 a^2 (5 A+6 C)+8 b^2 (32 A+45 C)\right ) \cos (c+d x)+b \left (24 b^2 (2 A+5 C)+5 a^2 (5 A+6 C)\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{360} \int \left (-3 \left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right )-96 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)-3 b^2 \left (24 b^2 (2 A+5 C)+5 a^2 (5 A+6 C)\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{720} \int \left (-192 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )-45 \left (8 b^4 (A+2 C)+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{15} \left (4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{16} \left (-8 b^4 (A+2 C)-12 a^2 b^2 (3 A+4 C)-a^4 (5 A+6 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (8 b^4 (A+2 C)+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {\left (4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {\left (8 b^4 (A+2 C)+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {\left (12 A b^2+5 a^2 (5 A+6 C)\right ) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}
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Mathematica [A] time = 4.79, size = 204, normalized size = 0.66 \[ \frac {15 \left (a^4 (5 A+6 C)+12 a^2 b^2 (3 A+4 C)+8 b^4 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (40 a^4 A \sec ^5(c+d x)+64 a b \left (5 \left (a^2 (2 A+C)+A b^2\right ) \tan ^2(c+d x)+15 \left (a^2+b^2\right ) (A+C)+3 a^2 A \tan ^4(c+d x)\right )+10 a^2 \left (a^2 (5 A+6 C)+36 A b^2\right ) \sec ^3(c+d x)+15 \left (a^4 (5 A+6 C)+12 a^2 b^2 (3 A+4 C)+8 A b^4\right ) \sec (c+d x)\right )}{240 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.93, size = 297, normalized size = 0.97 \[ \frac {15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 12 \, {\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \, {\left (A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 12 \, {\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \, {\left (A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (192 \, A a^{3} b \cos \left (d x + c\right ) + 64 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{3} b + 5 \, {\left (2 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} + 40 \, A a^{4} + 15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 12 \, {\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 64 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} b + 5 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 36 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.75, size = 1100, normalized size = 3.58 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.44, size = 511, normalized size = 1.66 \[ \frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {32 A \,a^{3} b \tan \left (d x +c \right )}{15 d}+\frac {4 A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {16 A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {8 a^{3} b C \tan \left (d x +c \right )}{3 d}+\frac {4 a^{3} b C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {3 A \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {9 A \,a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{4 d}+\frac {9 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {3 C \,a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {3 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {8 a A \,b^{3} \tan \left (d x +c \right )}{3 d}+\frac {4 a A \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {4 C a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {A \,b^{4} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {C \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 466, normalized size = 1.52 \[ \frac {128 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} b + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} b + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{3} - 5 \, A a^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, C a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a^{2} b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 720 \, C a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, A b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, C b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 1920 \, C a b^{3} \tan \left (d x + c\right )}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.97, size = 690, normalized size = 2.25 \[ \frac {\left (\frac {11\,A\,a^4}{8}+A\,b^4+\frac {5\,C\,a^4}{4}+\frac {15\,A\,a^2\,b^2}{2}+6\,C\,a^2\,b^2-8\,A\,a\,b^3-8\,A\,a^3\,b-8\,C\,a\,b^3-8\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {5\,A\,a^4}{24}-3\,A\,b^4-\frac {7\,C\,a^4}{4}-\frac {21\,A\,a^2\,b^2}{2}-18\,C\,a^2\,b^2+\frac {88\,A\,a\,b^3}{3}+\frac {56\,A\,a^3\,b}{3}+40\,C\,a\,b^3+\frac {88\,C\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {15\,A\,a^4}{4}+2\,A\,b^4+\frac {C\,a^4}{2}+3\,A\,a^2\,b^2+12\,C\,a^2\,b^2-48\,A\,a\,b^3-\frac {208\,A\,a^3\,b}{5}-80\,C\,a\,b^3-48\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {15\,A\,a^4}{4}+2\,A\,b^4+\frac {C\,a^4}{2}+3\,A\,a^2\,b^2+12\,C\,a^2\,b^2+48\,A\,a\,b^3+\frac {208\,A\,a^3\,b}{5}+80\,C\,a\,b^3+48\,C\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,A\,a^4}{24}-3\,A\,b^4-\frac {7\,C\,a^4}{4}-\frac {21\,A\,a^2\,b^2}{2}-18\,C\,a^2\,b^2-\frac {88\,A\,a\,b^3}{3}-\frac {56\,A\,a^3\,b}{3}-40\,C\,a\,b^3-\frac {88\,C\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {11\,A\,a^4}{8}+A\,b^4+\frac {5\,C\,a^4}{4}+\frac {15\,A\,a^2\,b^2}{2}+6\,C\,a^2\,b^2+8\,A\,a\,b^3+8\,A\,a^3\,b+8\,C\,a\,b^3+8\,C\,a^3\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,A\,a^4}{16}+\frac {A\,b^4}{2}+\frac {3\,C\,a^4}{8}+C\,b^4+\frac {9\,A\,a^2\,b^2}{4}+3\,C\,a^2\,b^2\right )}{\frac {5\,A\,a^4}{4}+2\,A\,b^4+\frac {3\,C\,a^4}{2}+4\,C\,b^4+9\,A\,a^2\,b^2+12\,C\,a^2\,b^2}\right )\,\left (\frac {5\,A\,a^4}{8}+A\,b^4+\frac {3\,C\,a^4}{4}+2\,C\,b^4+\frac {9\,A\,a^2\,b^2}{2}+6\,C\,a^2\,b^2\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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