Optimal. Leaf size=79 \[ -\frac {(4 A-B) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.18, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2978, 12, 3770} \[ -\frac {(4 A-B) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2978
Rule 3770
Rubi steps
\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {(3 a A-a (A-B) \cos (c+d x)) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(4 A-B) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int 3 a^2 A \sec (c+d x) \, dx}{3 a^4}\\ &=-\frac {(4 A-B) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {A \int \sec (c+d x) \, dx}{a^2}\\ &=\frac {A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(4 A-B) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end {align*}
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Mathematica [B] time = 0.54, size = 170, normalized size = 2.15 \[ -\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left ((A-B) \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+(A-B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (4 A-B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+6 A \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.80, size = 131, normalized size = 1.66 \[ \frac {3 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right ) + 5 \, A - 2 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.44, size = 113, normalized size = 1.43 \[ \frac {\frac {6 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 119, normalized size = 1.51 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 145, normalized size = 1.84 \[ -\frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.23, size = 74, normalized size = 0.94 \[ \frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A}{a^2}+\frac {A-B}{2\,a^2}\right )}{d} \]
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 \[ \frac {\int \frac {A \sec {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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