Optimal. Leaf size=169 \[ \frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac{1}{8} a^3 x (15 A+28 C)+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.410138, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4087, 4017, 3996, 3770} \[ \frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac{1}{8} a^3 x (15 A+28 C)+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 4087
Rule 4017
Rule 3996
Rule 3770
Rubi steps
\begin{align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (3 a A+4 a C \sec (c+d x)) \, dx}{4 a}\\ &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (3 a^2 (5 A+4 C)+12 a^2 C \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (3 A+4 C)+24 a^3 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}-\frac{\int \left (-3 a^4 (15 A+28 C)-24 a^4 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{1}{8} a^3 (15 A+28 C) x+\frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}+\left (a^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^3 (15 A+28 C) x+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}\\ \end{align*}
Mathematica [A] time = 0.304805, size = 124, normalized size = 0.73 \[ \frac{a^3 \left (8 (13 A+12 C) \sin (c+d x)+8 (4 A+C) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+A \sin (4 (c+d x))+60 A d x-32 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+32 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+112 C d x\right )}{32 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.095, size = 175, normalized size = 1. \begin{align*}{\frac{A{a}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,A{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}Ax}{8}}+{\frac{15\,A{a}^{3}c}{8\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{3}Cx}{2}}+{\frac{7\,{a}^{3}Cc}{2\,d}}+{\frac{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{3}}{d}}+3\,{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}C\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 = 0.950342, size = 231, normalized size = 1.37 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} -{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 8 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 96 \,{\left (d x + c\right )} C a^{3} - 16 \, C a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 32 \, A a^{3} \sin \left (d x + c\right ) - 96 \, C a^{3} \sin \left (d x + c\right )}{32 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.532458, size = 284, normalized size = 1.68 \begin{align*} \frac{{\left (15 \, A + 28 \, C\right )} a^{3} d x + 4 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, A a^{3} \cos \left (d x + c\right )^{2} +{\left (15 \, A + 4 \, C\right )} a^{3} \cos \left (d x + c\right ) + 24 \,{\left (A + C\right )} a^{3}\right )} \sin \left (d x + c\right )}{8 \, 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 [A] time = 1.29159, size = 288, normalized size = 1.7 \begin{align*} \frac{8 \, C a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, C a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (15 \, A a^{3} + 28 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 20 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 55 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 68 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 73 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 76 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 49 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 28 \, C a^{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 )}^{4}}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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