Optimal. Leaf size=173 \[ \frac {5 a^4 (7 A+8 B) \tan (c+d x)}{8 d}+\frac {a^4 (35 A+48 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+a^4 B x+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.52, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2975, 2968, 3021, 2735, 3770} \[ \frac {5 a^4 (7 A+8 B) \tan (c+d x)}{8 d}+\frac {a^4 (35 A+48 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(7 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}+\frac {(35 A+32 B) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+a^4 B x+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2968
Rule 2975
Rule 3021
Rule 3770
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx &=\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+a \cos (c+d x))^3 (a (7 A+4 B)+4 a B \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac {(7 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+a \cos (c+d x))^2 \left (a^2 (35 A+32 B)+12 a^2 B \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {(35 A+32 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(7 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{24} \int (a+a \cos (c+d x)) \left (15 a^3 (7 A+8 B)+24 a^3 B \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {(35 A+32 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(7 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (15 a^4 (7 A+8 B)+\left (24 a^4 B+15 a^4 (7 A+8 B)\right ) \cos (c+d x)+24 a^4 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{8 d}+\frac {(35 A+32 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(7 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (3 a^4 (35 A+48 B)+24 a^4 B \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=a^4 B x+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{8 d}+\frac {(35 A+32 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(7 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (a^4 (35 A+48 B)\right ) \int \sec (c+d x) \, dx\\ &=a^4 B x+\frac {a^4 (35 A+48 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^4 (7 A+8 B) \tan (c+d x)}{8 d}+\frac {(35 A+32 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(7 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a 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 = 2.03, size = 326, normalized size = 1.88 \[ \frac {a^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^4 \left (\sec (c) (105 A \sin (2 c+d x)+544 A \sin (c+2 d x)-96 A \sin (3 c+2 d x)+81 A \sin (2 c+3 d x)+81 A \sin (4 c+3 d x)+160 A \sin (3 c+4 d x)-480 A \sin (c)+105 A \sin (d x)+48 B \sin (2 c+d x)+496 B \sin (c+2 d x)-144 B \sin (3 c+2 d x)+48 B \sin (2 c+3 d x)+48 B \sin (4 c+3 d x)+160 B \sin (3 c+4 d x)+72 B d x \cos (c)+48 B d x \cos (c+2 d x)+48 B d x \cos (3 c+2 d x)+12 B d x \cos (3 c+4 d x)+12 B d x \cos (5 c+4 d x)-480 B \sin (c)+48 B \sin (d x))-24 (35 A+48 B) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3072 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 157, normalized size = 0.91 \[ \frac {48 \, B a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (35 \, A + 48 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (35 \, A + 48 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (160 \, {\left (A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (27 \, A + 16 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.68, size = 223, normalized size = 1.29 \[ \frac {24 \, {\left (d x + c\right )} B a^{4} + 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 385 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 279 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 204, normalized size = 1.18 \[ \frac {35 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+a^{4} B x +\frac {a^{4} B c}{d}+\frac {20 A \,a^{4} \tan \left (d x +c \right )}{3 d}+\frac {6 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {27 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {20 a^{4} B \tan \left (d x +c \right )}{3 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 307, normalized size = 1.77 \[ \frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 48 \, {\left (d x + c\right )} B a^{4} - 3 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, A a^{4} \tan \left (d x + c\right ) + 288 \, B a^{4} \tan \left (d x + c\right )}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.38, size = 255, normalized size = 1.47 \[ \frac {35\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {2\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {27\,A\,a^4\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4}+\frac {20\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \]
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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