Optimal. Leaf size=253 \[ \frac {a^4 (72 A+83 B+100 C) \tan (c+d x)}{15 d}+\frac {7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (417 A+488 B+550 C) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {(43 A+52 B+50 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{60 d}+\frac {(37 A+48 B+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{120 d}+\frac {a (2 A+3 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{15 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^4}{6 d} \]
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Rubi [A] time = 0.83, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3043, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac {a^4 (72 A+83 B+100 C) \tan (c+d x)}{15 d}+\frac {7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (417 A+488 B+550 C) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {(37 A+48 B+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{120 d}+\frac {(43 A+52 B+50 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{60 d}+\frac {a (2 A+3 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{15 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^4}{6 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2968
Rule 2975
Rule 3021
Rule 3043
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^4 (2 a (2 A+3 B)+a (A+6 C) \cos (c+d x)) \sec ^6(c+d x) \, dx}{6 a}\\ &=\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^3 \left (a^2 (37 A+48 B+30 C)+3 a^2 (3 A+2 B+10 C) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx}{30 a}\\ &=\frac {(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^2 \left (6 a^3 (43 A+52 B+50 C)+a^3 (73 A+72 B+150 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{120 a}\\ &=\frac {(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x)) \left (3 a^4 (417 A+488 B+550 C)+3 a^4 (159 A+176 B+250 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{360 a}\\ &=\frac {(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int \left (3 a^5 (417 A+488 B+550 C)+\left (3 a^5 (159 A+176 B+250 C)+3 a^5 (417 A+488 B+550 C)\right ) \cos (c+d x)+3 a^5 (159 A+176 B+250 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{360 a}\\ &=\frac {a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int \left (48 a^5 (72 A+83 B+100 C)+315 a^5 (7 A+8 B+10 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{720 a}\\ &=\frac {a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{16} \left (7 a^4 (7 A+8 B+10 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{15} \left (a^4 (72 A+83 B+100 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {\left (a^4 (72 A+83 B+100 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (72 A+83 B+100 C) \tan (c+d x)}{15 d}+\frac {a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \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 = 2.09, size = 265, normalized size = 1.05 \[ -\frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (840 (7 A+8 B+10 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sin (c+d x) (16 (672 A+643 B+620 C) \cos (c+d x)+20 (229 A+216 B+174 C) \cos (2 (c+d x))+4032 A \cos (3 (c+d x))+735 A \cos (4 (c+d x))+576 A \cos (5 (c+d x))+4165 A+4408 B \cos (3 (c+d x))+840 B \cos (4 (c+d x))+664 B \cos (5 (c+d x))+3480 B+4640 C \cos (3 (c+d x))+810 C \cos (4 (c+d x))+800 C \cos (5 (c+d x))+2670 C)\right )}{30720 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 203, normalized size = 0.80 \[ \frac {105 \, {\left (7 \, A + 8 \, B + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (7 \, A + 8 \, B + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (72 \, A + 83 \, B + 100 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 15 \, {\left (49 \, A + 56 \, B + 54 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 32 \, {\left (18 \, A + 17 \, B + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 10 \, {\left (41 \, A + 24 \, B + 6 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 48 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 40 \, A a^{4}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.78, size = 392, normalized size = 1.55 \[ \frac {105 \, {\left (7 \, A a^{4} + 8 \, B a^{4} + 10 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (7 \, A a^{4} + 8 \, B a^{4} + 10 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (735 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 840 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1050 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 4165 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 4760 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 5950 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 9702 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 11088 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 13860 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 11802 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 13488 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16860 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7355 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9320 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10690 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3000 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2790 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.61, size = 385, normalized size = 1.52 \[ \frac {20 a^{4} C \tan \left (d x +c \right )}{3 d}+\frac {83 a^{4} B \tan \left (d x +c \right )}{15 d}+\frac {35 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {7 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {34 a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {24 A \,a^{4} \tan \left (d x +c \right )}{5 d}+\frac {12 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {49 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {49 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {7 a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {41 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {27 a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.36, size = 645, normalized size = 2.55 \[ \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^{4} + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 32 \, {\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 )} B a^{4} + 960 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 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 )} - 180 \, A 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 )} - 120 \, B 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 )} - 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 )} - 120 \, A a^{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 )} - 480 \, B a^{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 )} - 720 \, C a^{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 a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 480 \, B a^{4} \tan \left (d x + c\right ) + 1920 \, C a^{4} \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 = 4.71, size = 338, normalized size = 1.34 \[ \frac {7\,a^4\,\mathrm {atanh}\left (\frac {7\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,A+8\,B+10\,C\right )}{4\,\left (\frac {49\,A\,a^4}{4}+14\,B\,a^4+\frac {35\,C\,a^4}{2}\right )}\right )\,\left (7\,A+8\,B+10\,C\right )}{8\,d}-\frac {\left (\frac {49\,A\,a^4}{8}+7\,B\,a^4+\frac {35\,C\,a^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-\frac {833\,A\,a^4}{24}-\frac {119\,B\,a^4}{3}-\frac {595\,C\,a^4}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {1617\,A\,a^4}{20}+\frac {462\,B\,a^4}{5}+\frac {231\,C\,a^4}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {1967\,A\,a^4}{20}-\frac {562\,B\,a^4}{5}-\frac {281\,C\,a^4}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {1471\,A\,a^4}{24}+\frac {233\,B\,a^4}{3}+\frac {1069\,C\,a^4}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {207\,A\,a^4}{8}-25\,B\,a^4-\frac {93\,C\,a^4}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
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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