Optimal. Leaf size=112 \[ -\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {a^2 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+2 a^2 C x+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]
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Rubi [A] time = 0.36, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3044, 2975, 2968, 3023, 2735, 3770} \[ -\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {a^2 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+2 a^2 C x+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2968
Rule 2975
Rule 3023
Rule 3044
Rule 3770
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^2 (2 a A-a (A-2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx}{2 a}\\ &=\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x)) \left (a^2 (3 A+2 C)-a^2 (3 A-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int \left (a^3 (3 A+2 C)+\left (-a^3 (3 A-2 C)+a^3 (3 A+2 C)\right ) \cos (c+d x)-a^3 (3 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int \left (a^3 (3 A+2 C)+4 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=2 a^2 C x-\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^2 (3 A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=2 a^2 C x+\frac {a^2 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
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Mathematica [B] time = 2.20, size = 293, normalized size = 2.62 \[ \frac {1}{16} a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {2 (3 A+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (3 A+2 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {8 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {8 A \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {A}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 C \sin (c) \cos (d x)}{d}+\frac {4 C \cos (c) \sin (d x)}{d}+8 C x\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.78, size = 129, normalized size = 1.15 \[ \frac {8 \, C a^{2} d x \cos \left (d x + c\right )^{2} + {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.78, size = 152, normalized size = 1.36 \[ \frac {4 \, {\left (d x + c\right )} C a^{2} + \frac {4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + {\left (3 \, A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.31, size = 114, normalized size = 1.02 \[ \frac {3 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{2} C \sin \left (d x +c \right )}{d}+\frac {2 a^{2} A \tan \left (d x +c \right )}{d}+2 a^{2} C x +\frac {2 a^{2} C c}{d}+\frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.59, size = 142, normalized size = 1.27 \[ \frac {8 \, {\left (d x + c\right )} C a^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} \sin \left (d x + c\right ) + 8 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.95, size = 154, normalized size = 1.38 \[ \frac {C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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