Optimal. Leaf size=196 \[ -\frac {8 (19 A-9 B) \tan (c+d x)}{15 a^3 d}+\frac {(13 A-6 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {(13 A-6 B) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {4 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(11 A-6 B) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
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Rubi [A] time = 0.54, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2978, 2748, 3768, 3770, 3767, 8} \[ -\frac {8 (19 A-9 B) \tan (c+d x)}{15 a^3 d}+\frac {(13 A-6 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {(13 A-6 B) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {4 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(11 A-6 B) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2978
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (7 A-2 B)-4 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A-6 B) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (43 A-18 B)-3 a^2 (11 A-6 B) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A-6 B) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 A-9 B) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (15 a^3 (13 A-6 B)-8 a^3 (19 A-9 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{15 a^6}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A-6 B) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 A-9 B) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(8 (19 A-9 B)) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {(13 A-6 B) \int \sec ^3(c+d x) \, dx}{a^3}\\ &=\frac {(13 A-6 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A-6 B) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 A-9 B) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(13 A-6 B) \int \sec (c+d x) \, dx}{2 a^3}+\frac {(8 (19 A-9 B)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {(13 A-6 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {8 (19 A-9 B) \tan (c+d x)}{15 a^3 d}+\frac {(13 A-6 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A-6 B) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 A-9 B) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 5.22, size = 610, normalized size = 3.11 \[ -\frac {1920 (13 A-6 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 )+\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left ((870 B-1235 A) \sin \left (\frac {d x}{2}\right )+5 (761 A-366 B) \sin \left (\frac {3 d x}{2}\right )-4329 A \sin \left (c-\frac {d x}{2}\right )+1989 A \sin \left (c+\frac {d x}{2}\right )-3575 A \sin \left (2 c+\frac {d x}{2}\right )-475 A \sin \left (c+\frac {3 d x}{2}\right )+2005 A \sin \left (2 c+\frac {3 d x}{2}\right )-2275 A \sin \left (3 c+\frac {3 d x}{2}\right )+2673 A \sin \left (c+\frac {5 d x}{2}\right )+105 A \sin \left (2 c+\frac {5 d x}{2}\right )+1593 A \sin \left (3 c+\frac {5 d x}{2}\right )-975 A \sin \left (4 c+\frac {5 d x}{2}\right )+1325 A \sin \left (2 c+\frac {7 d x}{2}\right )+255 A \sin \left (3 c+\frac {7 d x}{2}\right )+875 A \sin \left (4 c+\frac {7 d x}{2}\right )-195 A \sin \left (5 c+\frac {7 d x}{2}\right )+304 A \sin \left (3 c+\frac {9 d x}{2}\right )+90 A \sin \left (4 c+\frac {9 d x}{2}\right )+214 A \sin \left (5 c+\frac {9 d x}{2}\right )+2094 B \sin \left (c-\frac {d x}{2}\right )-1314 B \sin \left (c+\frac {d x}{2}\right )+1650 B \sin \left (2 c+\frac {d x}{2}\right )+450 B \sin \left (c+\frac {3 d x}{2}\right )-1230 B \sin \left (2 c+\frac {3 d x}{2}\right )+1050 B \sin \left (3 c+\frac {3 d x}{2}\right )-1278 B \sin \left (c+\frac {5 d x}{2}\right )+90 B \sin \left (2 c+\frac {5 d x}{2}\right )-918 B \sin \left (3 c+\frac {5 d x}{2}\right )+450 B \sin \left (4 c+\frac {5 d x}{2}\right )-630 B \sin \left (2 c+\frac {7 d x}{2}\right )-60 B \sin \left (3 c+\frac {7 d x}{2}\right )-480 B \sin \left (4 c+\frac {7 d x}{2}\right )+90 B \sin \left (5 c+\frac {7 d x}{2}\right )-144 B \sin \left (3 c+\frac {9 d x}{2}\right )-30 B \sin \left (4 c+\frac {9 d x}{2}\right )-114 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.82, size = 295, normalized size = 1.51 \[ \frac {15 \, {\left ({\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (19 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, A - 114 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, A - 234 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 15 \, A\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.21, size = 233, normalized size = 1.19 \[ \frac {\frac {30 \, {\left (13 \, A - 6 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, A - 6 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 334, normalized size = 1.70 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}-\frac {31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {13 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{3}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{3}}+\frac {7 A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7 A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {13 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{3}}-\frac {A}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.56, size = 377, normalized size = 1.92 \[ -\frac {A {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.28, size = 216, normalized size = 1.10 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A-2\,B\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^3}+\frac {3\,\left (5\,A-3\,B\right )}{4\,a^3}+\frac {10\,A-2\,B}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^3}+\frac {5\,A-3\,B}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,A-6\,B\right )}{a^3\,d} \]
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 \[ \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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