Optimal. Leaf size=145 \[ \frac {2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
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Rubi [A] time = 0.47, antiderivative size = 145, 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 = {2978, 2748, 3767, 8, 3770} \[ \frac {2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2978
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (6 A-B)-3 a (A-B) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (27 A-7 B)-2 a^2 (9 A-4 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (2 a^3 (36 A-11 B)-15 a^3 (3 A-B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{15 a^6}\\ &=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(2 (36 A-11 B)) \int \sec ^2(c+d x) \, dx}{15 a^3}-\frac {(3 A-B) \int \sec (c+d x) \, dx}{a^3}\\ &=-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(2 (36 A-11 B)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac {(3 A-B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 3.28, size = 482, normalized size = 3.32 \[ \frac {960 (3 A-B) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (-5 (51 A-32 B) \sin \left (\frac {d x}{2}\right )+(567 A-167 B) \sin \left (\frac {3 d x}{2}\right )-600 A \sin \left (c-\frac {d x}{2}\right )+375 A \sin \left (c+\frac {d x}{2}\right )-480 A \sin \left (2 c+\frac {d x}{2}\right )-60 A \sin \left (c+\frac {3 d x}{2}\right )+402 A \sin \left (2 c+\frac {3 d x}{2}\right )-225 A \sin \left (3 c+\frac {3 d x}{2}\right )+315 A \sin \left (c+\frac {5 d x}{2}\right )+30 A \sin \left (2 c+\frac {5 d x}{2}\right )+240 A \sin \left (3 c+\frac {5 d x}{2}\right )-45 A \sin \left (4 c+\frac {5 d x}{2}\right )+72 A \sin \left (2 c+\frac {7 d x}{2}\right )+15 A \sin \left (3 c+\frac {7 d x}{2}\right )+57 A \sin \left (4 c+\frac {7 d x}{2}\right )+170 B \sin \left (c-\frac {d x}{2}\right )-170 B \sin \left (c+\frac {d x}{2}\right )+160 B \sin \left (2 c+\frac {d x}{2}\right )+75 B \sin \left (c+\frac {3 d x}{2}\right )-167 B \sin \left (2 c+\frac {3 d x}{2}\right )+75 B \sin \left (3 c+\frac {3 d x}{2}\right )-95 B \sin \left (c+\frac {5 d x}{2}\right )+15 B \sin \left (2 c+\frac {5 d x}{2}\right )-95 B \sin \left (3 c+\frac {5 d x}{2}\right )+15 B \sin \left (4 c+\frac {5 d x}{2}\right )-22 B \sin \left (2 c+\frac {7 d x}{2}\right )-22 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 272, normalized size = 1.88 \[ -\frac {15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (36 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (57 \, A - 17 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (117 \, A - 32 \, B\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.44, size = 190, normalized size = 1.31 \[ -\frac {\frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 245, normalized size = 1.69 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{2 d \,a^{3}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{3}}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{3}}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 286, normalized size = 1.97 \[ \frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.28, size = 168, normalized size = 1.16 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {4\,A-2\,B}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A}{2\,a^3}+\frac {3\,\left (A-B\right )}{4\,a^3}+\frac {4\,A-2\,B}{2\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A-B\right )}{a^3\,d} \]
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 \[ \frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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