Optimal. Leaf size=80 \[ \frac {\left (a^2 A+4 a b B+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a (a B+2 A b) \tan (c+d x)}{d}+b^2 B x \]
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Rubi [A] time = 0.20, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2988, 3021, 2735, 3770} \[ \frac {\left (a^2 A+4 a b B+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a (a B+2 A b) \tan (c+d x)}{d}+b^2 B x \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2988
Rule 3021
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx &=\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 a (2 A b+a B)-\left (a^2 A+2 A b^2+4 a b B\right ) \cos (c+d x)-2 b^2 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a (2 A b+a B) \tan (c+d x)}{d}+\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-a^2 A-2 A b^2-4 a b B-2 b^2 B \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^2 B x+\frac {a (2 A b+a B) \tan (c+d x)}{d}+\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \left (-a^2 A-2 A b^2-4 a b B\right ) \int \sec (c+d x) \, dx\\ &=b^2 B x+\frac {\left (a^2 A+2 A b^2+4 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (2 A b+a B) \tan (c+d x)}{d}+\frac {a^2 A \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.27, size = 67, normalized size = 0.84 \[ \frac {\left (a^2 A+4 a b B+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))+a \tan (c+d x) (a A \sec (c+d x)+2 a B+4 A b)+2 b^2 B d x}{2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.61, size = 136, normalized size = 1.70 \[ \frac {4 \, B b^{2} d x \cos \left (d x + c\right )^{2} + {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.52, size = 190, normalized size = 2.38 \[ \frac {2 \, {\left (d x + c\right )} B b^{2} + {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 133, normalized size = 1.66 \[ \frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {B \,a^{2} \tan \left (d x +c \right )}{d}+\frac {2 A a b \tan \left (d x +c \right )}{d}+\frac {2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+b^{2} B x +\frac {b^{2} B c}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 140, normalized size = 1.75 \[ \frac {4 \, {\left (d x + c\right )} B b^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \tan \left (d x + c\right ) + 8 \, A a b \tan \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.98, size = 176, normalized size = 2.20 \[ \frac {2\,\left (\frac {A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2}+A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]
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 \[ \int \left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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