Optimal. Leaf size=156 \[ -\frac {(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac {(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.43, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4019, 4008, 3787, 3770, 3767, 8} \[ -\frac {(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac {(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3787
Rule 4008
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))^3} \, dx &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^3(c+d x) (3 a (A-B)-a (A-6 B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec (c+d x) \left (-15 a^3 (A-3 B)+a^3 (7 A-27 B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(7 A-27 B) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {(A-3 B) \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(7 A-27 B) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 4.28, size = 480, normalized size = 3.08 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (5 (32 A-51 B) \sin \left (\frac {d x}{2}\right )+(567 B-167 A) \sin \left (\frac {3 d x}{2}\right )+170 A \sin \left (c-\frac {d x}{2}\right )-170 A \sin \left (c+\frac {d x}{2}\right )+160 A \sin \left (2 c+\frac {d x}{2}\right )+75 A \sin \left (c+\frac {3 d x}{2}\right )-167 A \sin \left (2 c+\frac {3 d x}{2}\right )+75 A \sin \left (3 c+\frac {3 d x}{2}\right )-95 A \sin \left (c+\frac {5 d x}{2}\right )+15 A \sin \left (2 c+\frac {5 d x}{2}\right )-95 A \sin \left (3 c+\frac {5 d x}{2}\right )+15 A \sin \left (4 c+\frac {5 d x}{2}\right )-22 A \sin \left (2 c+\frac {7 d x}{2}\right )-22 A \sin \left (4 c+\frac {7 d x}{2}\right )-600 B \sin \left (c-\frac {d x}{2}\right )+375 B \sin \left (c+\frac {d x}{2}\right )-480 B \sin \left (2 c+\frac {d x}{2}\right )-60 B \sin \left (c+\frac {3 d x}{2}\right )+402 B \sin \left (2 c+\frac {3 d x}{2}\right )-225 B \sin \left (3 c+\frac {3 d x}{2}\right )+315 B \sin \left (c+\frac {5 d x}{2}\right )+30 B \sin \left (2 c+\frac {5 d x}{2}\right )+240 B \sin \left (3 c+\frac {5 d x}{2}\right )-45 B \sin \left (4 c+\frac {5 d x}{2}\right )+72 B \sin \left (2 c+\frac {7 d x}{2}\right )+15 B \sin \left (3 c+\frac {7 d x}{2}\right )+57 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )-960 (A-3 B) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 256, normalized size = 1.64 \[ \frac {15 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (11 \, A - 36 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (17 \, A - 57 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A - 117 \, B\right )} \cos \left (d x + c\right ) - 15 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.31, size = 186, normalized size = 1.19 \[ \frac {\frac {60 \, {\left (A - 3 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (A - 3 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {120 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.60, size = 245, normalized size = 1.57 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A}{d \,a^{3}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{3}}-\frac {B}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A}{d \,a^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{3}}-\frac {B}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 286, normalized size = 1.83 \[ \frac {3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.04, size = 168, normalized size = 1.08 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-3\,B\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{4\,a^3}-\frac {3\,B}{2\,a^3}+\frac {2\,A-4\,B}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {2\,A-4\,B}{12\,a^3}\right )}{d}-\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \]
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 \[ \frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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