Optimal. Leaf size=145 \[ \frac{a \left (2 a^2 A+9 a b B+8 A b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (3 a^2 A b+a^3 B+6 a b^2 B+2 A b^3\right )+\frac{a^2 (3 a B+5 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{b^3 B \tanh ^{-1}(\sin (c+d x))}{d} \]
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Rubi [A] time = 0.347466, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4025, 4074, 4047, 8, 4045, 3770} \[ \frac{a \left (2 a^2 A+9 a b B+8 A b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (3 a^2 A b+a^3 B+6 a b^2 B+2 A b^3\right )+\frac{a^2 (3 a B+5 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{b^3 B \tanh ^{-1}(\sin (c+d x))}{d} \]
Antiderivative was successfully verified.
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Rule 4025
Rule 4074
Rule 4047
Rule 8
Rule 4045
Rule 3770
Rubi steps
\begin{align*} \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}-\frac{1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (-a (5 A b+3 a B)-\left (2 a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x)-3 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \cos (c+d x) \left (2 a \left (2 a^2 A+8 A b^2+9 a b B\right )+3 \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) \sec (c+d x)+6 b^3 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \cos (c+d x) \left (2 a \left (2 a^2 A+8 A b^2+9 a b B\right )+6 b^3 B \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) \int 1 \, dx\\ &=\frac{1}{2} \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) x+\frac{a \left (2 a^2 A+8 A b^2+9 a b B\right ) \sin (c+d x)}{3 d}+\frac{a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\left (b^3 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) x+\frac{b^3 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a \left (2 a^2 A+8 A b^2+9 a b B\right ) \sin (c+d x)}{3 d}+\frac{a^2 (5 A b+3 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.355049, size = 159, normalized size = 1.1 \[ \frac{6 (c+d x) \left (3 a^2 A b+a^3 B+6 a b^2 B+2 A b^3\right )+9 a \left (a^2 A+4 a b B+4 A b^2\right ) \sin (c+d x)+3 a^2 (a B+3 A b) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))-12 b^3 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^3 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{12 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.06, size = 207, normalized size = 1.4 \begin{align*}{\frac{A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{3\,d}}+{\frac{2\,A{a}^{3}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{B{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{3}x}{2}}+{\frac{B{a}^{3}c}{2\,d}}+{\frac{3\,A{a}^{2}b\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{2}bx}{2}}+{\frac{3\,A{a}^{2}bc}{2\,d}}+3\,{\frac{B{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{\frac{Aa{b}^{2}\sin \left ( dx+c \right ) }{d}}+3\,Ba{b}^{2}x+3\,{\frac{Ba{b}^{2}c}{d}}+A{b}^{3}x+{\frac{A{b}^{3}c}{d}}+{\frac{B{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.963334, size = 205, normalized size = 1.41 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 36 \,{\left (d x + c\right )} B a b^{2} - 12 \,{\left (d x + c\right )} A b^{3} - 6 \, B b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{2} b \sin \left (d x + c\right ) - 36 \, A a b^{2} \sin \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 = 0.541763, size = 317, normalized size = 2.19 \begin{align*} \frac{3 \, B b^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B b^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} d x +{\left (2 \, A a^{3} \cos \left (d x + c\right )^{2} + 4 \, A a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 3 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \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 [B] time = 1.2674, size = 424, normalized size = 2.92 \begin{align*} \frac{6 \, B b^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, B b^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, A a b^{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 )}^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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