Optimal. Leaf size=131 \[ -\frac {(3 A-4 B) \tan ^3(c+d x)}{3 a d}-\frac {(3 A-4 B) \tan (c+d x)}{a d}+\frac {3 (A-B) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 (A-B) \tan (c+d x) \sec (c+d x)}{2 a d} \]
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Rubi [A] time = 0.17, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4019, 3787, 3768, 3770, 3767} \[ -\frac {(3 A-4 B) \tan ^3(c+d x)}{3 a d}-\frac {(3 A-4 B) \tan (c+d x)}{a d}+\frac {3 (A-B) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 (A-B) \tan (c+d x) \sec (c+d x)}{2 a d} \]
Antiderivative was successfully verified.
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Rule 3767
Rule 3768
Rule 3770
Rule 3787
Rule 4019
Rubi steps
\begin {align*} \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \sec ^3(c+d x) (3 a (A-B)-a (3 A-4 B) \sec (c+d x)) \, dx}{a^2}\\ &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A-4 B) \int \sec ^4(c+d x) \, dx}{a}+\frac {(3 (A-B)) \int \sec ^3(c+d x) \, dx}{a}\\ &=\frac {3 (A-B) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 (A-B)) \int \sec (c+d x) \, dx}{2 a}+\frac {(3 A-4 B) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=\frac {3 (A-B) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {(3 A-4 B) \tan (c+d x)}{a d}+\frac {3 (A-B) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A-4 B) \tan ^3(c+d x)}{3 a d}\\ \end {align*}
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Mathematica [B] time = 6.30, size = 635, normalized size = 4.85 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^3(c+d x) \left (12 A \sin \left (c-\frac {d x}{2}\right )+6 A \sin \left (c+\frac {d x}{2}\right )+24 A \sin \left (2 c+\frac {d x}{2}\right )-9 A \sin \left (c+\frac {3 d x}{2}\right )-9 A \sin \left (2 c+\frac {3 d x}{2}\right )+9 A \sin \left (3 c+\frac {3 d x}{2}\right )-3 A \sin \left (c+\frac {5 d x}{2}\right )+3 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 A \sin \left (3 c+\frac {5 d x}{2}\right )+9 A \sin \left (4 c+\frac {5 d x}{2}\right )-12 A \sin \left (2 c+\frac {7 d x}{2}\right )-6 A \sin \left (3 c+\frac {7 d x}{2}\right )-6 A \sin \left (4 c+\frac {7 d x}{2}\right )+6 A \sin \left (\frac {d x}{2}\right )-27 A \sin \left (\frac {3 d x}{2}\right )-24 B \sin \left (c-\frac {d x}{2}\right )-6 B \sin \left (c+\frac {d x}{2}\right )-24 B \sin \left (2 c+\frac {d x}{2}\right )+21 B \sin \left (c+\frac {3 d x}{2}\right )+9 B \sin \left (2 c+\frac {3 d x}{2}\right )-9 B \sin \left (3 c+\frac {3 d x}{2}\right )+7 B \sin \left (c+\frac {5 d x}{2}\right )+B \sin \left (2 c+\frac {5 d x}{2}\right )-3 B \sin \left (3 c+\frac {5 d x}{2}\right )-9 B \sin \left (4 c+\frac {5 d x}{2}\right )+16 B \sin \left (2 c+\frac {7 d x}{2}\right )+10 B \sin \left (3 c+\frac {7 d x}{2}\right )+6 B \sin \left (4 c+\frac {7 d x}{2}\right )+6 B \sin \left (\frac {d x}{2}\right )+39 B \sin \left (\frac {3 d x}{2}\right )\right ) (A+B \sec (c+d x))}{48 d (a \sec (c+d x)+a) (A \cos (c+d x)+B)}+\frac {3 (B-A) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) (A+B \sec (c+d x)) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \sec (c+d x)+a) (A \cos (c+d x)+B)}-\frac {3 (B-A) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) (A+B \sec (c+d x)) \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \sec (c+d x)+a) (A \cos (c+d x)+B)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 170, normalized size = 1.30 \[ \frac {9 \, {\left ({\left (A - B\right )} \cos \left (d x + c\right )^{4} + {\left (A - B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left ({\left (A - B\right )} \cos \left (d x + c\right )^{4} + {\left (A - B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (3 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (3 \, A - B\right )} \cos \left (d x + c\right ) - 2 \, B\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.80, size = 182, normalized size = 1.39 \[ \frac {\frac {9 \, {\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {9 \, {\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.57, size = 340, normalized size = 2.60 \[ -\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {B}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {B}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 a d}-\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {5 B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {B}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {5 B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 A}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 a d}+\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.34, size = 368, normalized size = 2.81 \[ \frac {B {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, A {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.44, size = 152, normalized size = 1.16 \[ \frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-B\right )}{a\,d}-\frac {\left (3\,A-5\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {16\,B}{3}-4\,A\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A-3\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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