Optimal. Leaf size=156 \[ -\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {a^3 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^3 x (6 A+7 B+5 C)+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]
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Rubi [A] time = 0.51, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3043, 2976, 2968, 3023, 2735, 3770} \[ -\frac {(6 A-3 B-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {a^3 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^3 x (6 A+7 B+5 C)-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2968
Rule 2976
Rule 3023
Rule 3043
Rule 3770
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^3 (a (3 A+B)-a (3 A-C) \cos (c+d x)) \sec (c+d x) \, dx}{a}\\ &=-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^2 \left (3 a^2 (3 A+B)-a^2 (6 A-3 B-5 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{3 a}\\ &=-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-3 B-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x)) \left (6 a^3 (3 A+B)+15 a^3 (B+C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-3 B-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int \left (6 a^4 (3 A+B)+\left (6 a^4 (3 A+B)+15 a^4 (B+C)\right ) \cos (c+d x)+15 a^4 (B+C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-3 B-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int \left (6 a^4 (3 A+B)+3 a^4 (6 A+7 B+5 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {1}{2} a^3 (6 A+7 B+5 C) x+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-3 B-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\left (a^3 (3 A+B)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^3 (6 A+7 B+5 C) x+\frac {a^3 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-3 B-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end {align*}
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Mathematica [A] time = 0.98, size = 227, normalized size = 1.46 \[ \frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (6 (6 A+7 B+5 C) (c+d x)+3 (4 A+12 B+15 C) \sin (c+d x)-12 (3 A+B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 (3 A+B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 A \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+3 (B+3 C) \sin (2 (c+d x))+C \sin (3 (c+d x))\right )}{96 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 156, normalized size = 1.00 \[ \frac {3 \, {\left (6 \, A + 7 \, B + 5 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.83, size = 281, normalized size = 1.80 \[ -\frac {\frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (6 \, A a^{3} + 7 \, B a^{3} + 5 \, C a^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (3 \, A a^{3} + B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (3 \, A a^{3} + B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 221, normalized size = 1.42 \[ \frac {a^{3} A \sin \left (d x +c \right )}{d}+\frac {a^{3} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {7 a^{3} B x}{2}+\frac {7 a^{3} B c}{2 d}+\frac {C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {11 a^{3} C \sin \left (d x +c \right )}{3 d}+3 A x \,a^{3}+\frac {3 A \,a^{3} c}{d}+\frac {3 a^{3} B \sin \left (d x +c \right )}{d}+\frac {3 C \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a^{3} C x}{2}+\frac {5 C \,a^{3} c}{2 d}+\frac {3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 210, normalized size = 1.35 \[ \frac {36 \, {\left (d x + c\right )} A a^{3} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 36 \, {\left (d x + c\right )} B a^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} + 18 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, B a^{3} \sin \left (d x + c\right ) + 36 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.98, size = 290, normalized size = 1.86 \[ \frac {6\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+7\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+5\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}+\frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {23\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{12}+\frac {3\,C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{24}+A\,a^3\,\sin \left (c+d\,x\right )+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{8}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{8}}{d\,\cos \left (c+d\,x\right )} \]
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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