Optimal. Leaf size=124 \[ -\frac {(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
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Rubi [A] time = 0.35, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3041, 2978, 12, 3770} \[ -\frac {(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2978
Rule 3041
Rule 3770
Rubi steps
\begin {align*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(5 a A-a (2 A-2 B-3 C) \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (15 a^2 A-a^2 (7 A-2 B-3 C) \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int 15 a^3 A \sec (c+d x) \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {A \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 1.69, size = 276, normalized size = 2.23 \[ -\frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (5 (29 A-4 B-3 C) \sin \left (\frac {d x}{2}\right )+15 (C-5 A) \sin \left (c+\frac {d x}{2}\right )+95 A \sin \left (c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {3 d x}{2}\right )+22 A \sin \left (2 c+\frac {5 d x}{2}\right )-10 B \sin \left (c+\frac {3 d x}{2}\right )-2 B \sin \left (2 c+\frac {5 d x}{2}\right )-15 C \sin \left (c+\frac {3 d x}{2}\right )-3 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )+240 A \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{15 a^3 d (\cos (c+d x)+1)^3 (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 193, normalized size = 1.56 \[ \frac {15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (22 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 32 \, A - 7 \, B - 3 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.89, size = 180, normalized size = 1.45 \[ \frac {\frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.21, size = 197, normalized size = 1.59 \[ \frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{3}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 232, normalized size = 1.87 \[ -\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.23, size = 134, normalized size = 1.08 \[ \frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{12\,a^3}-\frac {B-3\,A+C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A+B-C}{4\,a^3}+\frac {A-B+C}{4\,a^3}-\frac {B-3\,A+C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \]
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 \[ \frac {\int \frac {A \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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