Optimal. Leaf size=179 \[ -\frac {4 (2 A-3 B) \tan ^3(c+d x)}{3 a^2 d}-\frac {4 (2 A-3 B) \tan (c+d x)}{a^2 d}+\frac {(7 A-10 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(7 A-10 B) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {(7 A-10 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.32, antiderivative size = 179, 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 = {4019, 3787, 3768, 3770, 3767} \[ -\frac {4 (2 A-3 B) \tan ^3(c+d x)}{3 a^2 d}-\frac {4 (2 A-3 B) \tan (c+d x)}{a^2 d}+\frac {(7 A-10 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(7 A-10 B) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {(7 A-10 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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 ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^4(c+d x) (4 a (A-B)-3 a (A-2 B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \sec ^3(c+d x) \left (3 a^2 (7 A-10 B)-12 a^2 (2 A-3 B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(7 A-10 B) \int \sec ^3(c+d x) \, dx}{a^2}-\frac {(4 (2 A-3 B)) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=\frac {(7 A-10 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(7 A-10 B) \int \sec (c+d x) \, dx}{2 a^2}+\frac {(4 (2 A-3 B)) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=\frac {(7 A-10 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac {4 (2 A-3 B) \tan (c+d x)}{a^2 d}+\frac {(7 A-10 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 (2 A-3 B) \tan ^3(c+d x)}{3 a^2 d}\\ \end {align*}
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Mathematica [B] time = 6.43, size = 764, normalized size = 4.27 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4(c+d x) \left (195 A \sin \left (c-\frac {d x}{2}\right )-51 A \sin \left (c+\frac {d x}{2}\right )+189 A \sin \left (2 c+\frac {d x}{2}\right )-A \sin \left (c+\frac {3 d x}{2}\right )-81 A \sin \left (2 c+\frac {3 d x}{2}\right )+119 A \sin \left (3 c+\frac {3 d x}{2}\right )-129 A \sin \left (c+\frac {5 d x}{2}\right )-9 A \sin \left (2 c+\frac {5 d x}{2}\right )-57 A \sin \left (3 c+\frac {5 d x}{2}\right )+63 A \sin \left (4 c+\frac {5 d x}{2}\right )-75 A \sin \left (2 c+\frac {7 d x}{2}\right )-15 A \sin \left (3 c+\frac {7 d x}{2}\right )-39 A \sin \left (4 c+\frac {7 d x}{2}\right )+21 A \sin \left (5 c+\frac {7 d x}{2}\right )-32 A \sin \left (3 c+\frac {9 d x}{2}\right )-12 A \sin \left (4 c+\frac {9 d x}{2}\right )-20 A \sin \left (5 c+\frac {9 d x}{2}\right )+45 A \sin \left (\frac {d x}{2}\right )-201 A \sin \left (\frac {3 d x}{2}\right )-306 B \sin \left (c-\frac {d x}{2}\right )+42 B \sin \left (c+\frac {d x}{2}\right )-270 B \sin \left (2 c+\frac {d x}{2}\right )+50 B \sin \left (c+\frac {3 d x}{2}\right )+90 B \sin \left (2 c+\frac {3 d x}{2}\right )-170 B \sin \left (3 c+\frac {3 d x}{2}\right )+198 B \sin \left (c+\frac {5 d x}{2}\right )+42 B \sin \left (2 c+\frac {5 d x}{2}\right )+66 B \sin \left (3 c+\frac {5 d x}{2}\right )-90 B \sin \left (4 c+\frac {5 d x}{2}\right )+114 B \sin \left (2 c+\frac {7 d x}{2}\right )+36 B \sin \left (3 c+\frac {7 d x}{2}\right )+48 B \sin \left (4 c+\frac {7 d x}{2}\right )-30 B \sin \left (5 c+\frac {7 d x}{2}\right )+48 B \sin \left (3 c+\frac {9 d x}{2}\right )+22 B \sin \left (4 c+\frac {9 d x}{2}\right )+26 B \sin \left (5 c+\frac {9 d x}{2}\right )-6 B \sin \left (\frac {d x}{2}\right )+310 B \sin \left (\frac {3 d x}{2}\right )\right ) (A+B \sec (c+d x))}{96 d (a \sec (c+d x)+a)^2 (A \cos (c+d x)+B)}+\frac {2 (10 B-7 A) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (c+d x) (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)^2 (A \cos (c+d x)+B)}-\frac {2 (10 B-7 A) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (c+d x) (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)^2 (A \cos (c+d x)+B)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 245, normalized size = 1.37 \[ \frac {3 \, {\left ({\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (43 \, A - 66 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 2 \, B\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.57, size = 226, normalized size = 1.26 \[ \frac {\frac {3 \, {\left (7 \, A - 10 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (7 \, A - 10 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.05, size = 382, normalized size = 2.13 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {9 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {7 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{2}}-\frac {5 B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {7 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{2}}+\frac {3 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {5 B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {5 A}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B}{3 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.36, size = 425, normalized size = 2.37 \[ \frac {B {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.98, size = 202, normalized size = 1.13 \[ \frac {\left (5\,A-10\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {40\,B}{3}-8\,A\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-6\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A-B\right )}{a^2}+\frac {3\,A-5\,B}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,A-10\,B\right )}{a^2\,d} \]
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 \[ \frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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