Optimal. Leaf size=169 \[ \frac {5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac {a^3 (15 A+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{8 d}+a^3 C x+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 a d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.50, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3044, 2975, 2968, 3021, 2735, 3770} \[ \frac {5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac {a^3 (15 A+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{8 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 a d}+a^3 C x+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2968
Rule 2975
Rule 3021
Rule 3044
Rule 3770
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^3 (3 a A+4 a C \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^2 \left (3 a^2 (5 A+4 C)+12 a^2 C \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac {(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x)) \left (15 a^3 (3 A+4 C)+24 a^3 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac {(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (15 a^4 (3 A+4 C)+\left (24 a^4 C+15 a^4 (3 A+4 C)\right ) \cos (c+d x)+24 a^4 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac {5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac {(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (3 a^4 (15 A+28 C)+24 a^4 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=a^3 C x+\frac {5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac {(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (a^3 (15 A+28 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 C x+\frac {a^3 (15 A+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac {(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 1.46, size = 334, normalized size = 1.98 \[ \frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (\sec (c) (23 A \sin (2 c+d x)+88 A \sin (c+2 d x)-8 A \sin (3 c+2 d x)+15 A \sin (2 c+3 d x)+15 A \sin (4 c+3 d x)+24 A \sin (3 c+4 d x)-72 A \sin (c)+23 A \sin (d x)+4 C \sin (2 c+d x)+72 C \sin (c+2 d x)-24 C \sin (3 c+2 d x)+4 C \sin (2 c+3 d x)+4 C \sin (4 c+3 d x)+24 C \sin (3 c+4 d x)+24 C d x \cos (c)+16 C d x \cos (c+2 d x)+16 C d x \cos (3 c+2 d x)+4 C d x \cos (3 c+4 d x)+4 C d x \cos (5 c+4 d x)-72 C \sin (c)+4 C \sin (d x))-8 (15 A+28 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{512 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 151, normalized size = 0.89 \[ \frac {16 \, C a^{3} d x \cos \left (d x + c\right )^{4} + {\left (15 \, A + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (15 \, A + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, {\left (A + C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (15 \, A + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, A a^{3} \cos \left (d x + c\right ) + 2 \, A a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.32, size = 222, normalized size = 1.31 \[ \frac {8 \, {\left (d x + c\right )} C a^{3} + {\left (15 \, A a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (15 \, A a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 20 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 55 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 68 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 73 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 76 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 49 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 28 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.43, size = 180, normalized size = 1.07 \[ \frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+a^{3} C x +\frac {C \,a^{3} c}{d}+\frac {15 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {7 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C \,a^{3} \tan \left (d x +c \right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.68, size = 257, normalized size = 1.52 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \, {\left (d x + c\right )} C a^{3} - A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, A a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.04, size = 231, normalized size = 1.37 \[ \frac {15\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {15\,A\,a^3\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]
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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