Optimal. Leaf size=163 \[ \frac{a \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} b x \left (3 a^2 (A+2 C)+2 A b^2\right )+\frac{A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac{A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{3 a b^2 C \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^3 (5 A-6 C) \tan (c+d x)}{6 d} \]
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Rubi [A] time = 0.523604, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4094, 4076, 4047, 8, 4045, 3770} \[ \frac{a \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} b x \left (3 a^2 (A+2 C)+2 A b^2\right )+\frac{A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac{A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{3 a b^2 C \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^3 (5 A-6 C) \tan (c+d x)}{6 d} \]
Antiderivative was successfully verified.
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Rule 4095
Rule 4094
Rule 4076
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 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (2 A+3 C) \sec (c+d x)-b (A-3 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{1}{6} \int \cos (c+d x) (a+b \sec (c+d x)) \left (2 \left (3 A b^2+\frac{1}{2} a^2 (4 A+6 C)\right )+a b (5 A+12 C) \sec (c+d x)-b^2 (5 A-6 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^3 (5 A-6 C) \tan (c+d x)}{6 d}+\frac{1}{6} \int \cos (c+d x) \left (2 a \left (3 A b^2+\frac{1}{2} a^2 (4 A+6 C)\right )+3 b \left (2 A b^2+3 a^2 (A+2 C)\right ) \sec (c+d x)+18 a b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^3 (5 A-6 C) \tan (c+d x)}{6 d}+\frac{1}{6} \int \cos (c+d x) \left (2 a \left (3 A b^2+\frac{1}{2} a^2 (4 A+6 C)\right )+18 a b^2 C \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (b \left (2 A b^2+3 a^2 (A+2 C)\right )\right ) \int 1 \, dx\\ &=\frac{1}{2} b \left (2 A b^2+3 a^2 (A+2 C)\right ) x+\frac{a \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}+\frac{A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^3 (5 A-6 C) \tan (c+d x)}{6 d}+\left (3 a b^2 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} b \left (2 A b^2+3 a^2 (A+2 C)\right ) x+\frac{3 a b^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}+\frac{A b \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^3 (5 A-6 C) \tan (c+d x)}{6 d}\\ \end{align*}
Mathematica [A] time = 0.911944, size = 184, normalized size = 1.13 \[ \frac{3 a \left (a^2 (3 A+4 C)+12 A b^2\right ) \sin (c+d x)+9 a^2 A b \sin (2 (c+d x))+18 a^2 A b c+18 a^2 A b d x+a^3 A \sin (3 (c+d x))+36 a^2 b c C+36 a^2 b C d x-36 a b^2 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+36 a b^2 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 A b^3 c+12 A b^3 d x+12 b^3 C \tan (c+d x)}{12 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.069, size = 183, normalized size = 1.1 \begin{align*}{\frac{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{3}}{3\,d}}+{\frac{2\,A{a}^{3}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+{\frac{3\,A{a}^{2}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}Abx}{2}}+{\frac{3\,A{a}^{2}bc}{2\,d}}+3\,{a}^{2}bCx+3\,{\frac{C{a}^{2}bc}{d}}+3\,{\frac{Aa{b}^{2}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+A{b}^{3}x+{\frac{A{b}^{3}c}{d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.977301, size = 190, normalized size = 1.17 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A 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 )} C a^{2} b - 12 \,{\left (d x + c\right )} A b^{3} - 18 \, C 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 )} - 12 \, C a^{3} \sin \left (d x + c\right ) - 36 \, A a b^{2} \sin \left (d x + c\right ) - 12 \, C b^{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 = 0.550193, size = 394, normalized size = 2.42 \begin{align*} \frac{9 \, C a b^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, C a b^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (3 \,{\left (A + 2 \, C\right )} a^{2} b + 2 \, A b^{3}\right )} d x \cos \left (d x + c\right ) +{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, A a^{2} b \cos \left (d x + c\right )^{2} + 6 \, C b^{3} + 2 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{3} + 9 \, A a b^{2}\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 = 1.23739, size = 413, normalized size = 2.53 \begin{align*} \frac{18 \, C a b^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, C a b^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{12 \, C b^{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 (3 \, A a^{2} b + 6 \, C a^{2} b + 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} + 6 \, C 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 \, 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} + 12 \, C a^{3} \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 ) + 6 \, C 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 \, 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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