Optimal. Leaf size=216 \[ \frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {a \left (4 a^3 B+16 a^2 A b+34 a b^2 B+19 A b^3\right ) \tan (c+d x)}{6 d}+\frac {\left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 a B+7 A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^4 B x \]
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Rubi [A] time = 0.60, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2989, 3047, 3031, 3021, 2735, 3770} \[ \frac {a \left (16 a^2 A b+4 a^3 B+34 a b^2 B+19 A b^3\right ) \tan (c+d x)}{6 d}+\frac {\left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {a (4 a B+7 A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^4 B x \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2989
Rule 3021
Rule 3031
Rule 3047
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx &=\frac {a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x))^2 \left (a (7 A b+4 a B)+\left (3 a^2 A+4 A b^2+8 a b B\right ) \cos (c+d x)+4 b^2 B \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+b \cos (c+d x)) \left (a \left (9 a^2 A+26 A b^2+32 a b B\right )+\left (23 a^2 A b+12 A b^3+8 a^3 B+36 a b^2 B\right ) \cos (c+d x)+12 b^3 B \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (-4 a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right )-3 \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \cos (c+d x)-24 b^4 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (-3 \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right )-24 b^4 B \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^4 B x+\frac {a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{8} \left (-3 a^4 A-24 a^2 A b^2-8 A b^4-16 a^3 b B-32 a b^3 B\right ) \int \sec (c+d x) \, dx\\ &=b^4 B x+\frac {\left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {a (7 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {a A (a+b \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 = 1.08, size = 160, normalized size = 0.74 \[ \frac {8 a^3 (a B+4 A b) \tan ^3(c+d x)+3 a \tan (c+d x) \left (2 a^3 A \sec ^3(c+d x)+a \left (3 a^2 A+16 a b B+24 A b^2\right ) \sec (c+d x)+8 \left (a^3 B+4 a^2 A b+6 a b^2 B+4 A b^3\right )\right )+3 \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))+24 b^4 B d x}{24 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.84, size = 250, normalized size = 1.16 \[ \frac {48 \, B b^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{4} + 16 \, {\left (B a^{4} + 4 \, A a^{3} b + 9 \, B a^{2} b^{2} + 6 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.69, size = 635, normalized size = 2.94 \[ \frac {24 \, {\left (d x + c\right )} B b^{4} + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\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^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 144 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 432 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 432 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 144 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, A a b^{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}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 338, normalized size = 1.56 \[ \frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a^{4} B \tan \left (d x +c \right )}{3 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {8 A \,a^{3} b \tan \left (d x +c \right )}{3 d}+\frac {4 A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 B \,a^{3} b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {2 B \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 A \,a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {3 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {6 B \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {4 A a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {4 B a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+b^{4} B x +\frac {B \,b^{4} c}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 317, normalized size = 1.47 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} b + 48 \, {\left (d x + c\right )} B b^{4} - 3 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{3} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 288 \, B a^{2} b^{2} \tan \left (d x + c\right ) + 192 \, A a b^{3} \tan \left (d x + c\right )}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.98, size = 1969, normalized size = 9.12 \[ \text {result too large to display} \]
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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