Optimal. Leaf size=167 \[ -\frac{b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}+\frac{1}{2} a x \left (2 a^2 C+6 A b^2+3 b^2 C\right )+\frac{3 a^2 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{6 d}-\frac{b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]
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Rubi [A] time = 0.503616, antiderivative size = 167, normalized size of antiderivative = 1., 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 = {3048, 3049, 3033, 3023, 2735, 3770} \[ -\frac{b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}+\frac{1}{2} a x \left (2 a^2 C+6 A b^2+3 b^2 C\right )+\frac{3 a^2 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{6 d}-\frac{b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]
Antiderivative was successfully verified.
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Rule 3048
Rule 3049
Rule 3033
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^2 \left (3 A b+a C \cos (c+d x)-b (3 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{1}{3} \int (a+b \cos (c+d x)) \left (9 a A b+\left (3 A b^2+3 a^2 C+2 b^2 C\right ) \cos (c+d x)-a b (6 A-5 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{1}{6} \int \left (18 a^2 A b+3 a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \cos (c+d x)-2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac{a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{1}{6} \int \left (18 a^2 A b+3 a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) x-\frac{b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac{a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\left (3 a^2 A b\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) x+\frac{3 a^2 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac{a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end{align*}
Mathematica [A] time = 0.866054, size = 185, normalized size = 1.11 \[ \frac{3 b \left (3 C \left (4 a^2+b^2\right )+4 A b^2\right ) \sin (c+d x)-36 a^2 A b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+36 a^2 A b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 a^3 A \tan (c+d x)+12 a^3 c C+12 a^3 C d x+36 a A b^2 c+36 a A b^2 d x+9 a b^2 C \sin (2 (c+d x))+18 a b^2 c C+18 a b^2 C d x+b^3 C \sin (3 (c+d x))}{12 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.055, size = 183, normalized size = 1.1 \begin{align*}{\frac{A{b}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{3\,d}}+{\frac{2\,C{b}^{3}\sin \left ( dx+c \right ) }{3\,d}}+3\,aA{b}^{2}x+3\,{\frac{Aa{b}^{2}c}{d}}+{\frac{3\,Ca{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,a{b}^{2}Cx}{2}}+{\frac{3\,Ca{b}^{2}c}{2\,d}}+3\,{\frac{A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{2}bC\sin \left ( dx+c \right ) }{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{a}^{3}Cx+{\frac{C{a}^{3}c}{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.02213, size = 190, normalized size = 1.14 \begin{align*} \frac{12 \,{\left (d x + c\right )} C a^{3} + 36 \,{\left (d x + c\right )} A a b^{2} + 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 18 \, A a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, C a^{2} b \sin \left (d x + c\right ) + 12 \, A b^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.59462, size = 394, normalized size = 2.36 \begin{align*} \frac{9 \, A a^{2} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, A a^{2} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (2 \, C a^{3} + 3 \,{\left (2 \, A + C\right )} a b^{2}\right )} d x \cos \left (d x + c\right ) +{\left (2 \, C b^{3} \cos \left (d x + c\right )^{3} + 9 \, C a b^{2} \cos \left (d x + c\right )^{2} + 6 \, A a^{3} + 2 \,{\left (9 \, C a^{2} b +{\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.7086, size = 413, normalized size = 2.47 \begin{align*} \frac{18 \, A a^{2} b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, A a^{2} b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{12 \, A a^{3} \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} + 3 \,{\left (2 \, C a^{3} + 6 \, A a b^{2} + 3 \, C a b^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b^{3} \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 )}^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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