Optimal. Leaf size=185 \[ \frac{a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac{a^3 (13 A+15 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (43 A+45 B) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{a^3 (13 A+15 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{(7 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{20 d}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
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Rubi [A] time = 0.44741, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac{a^3 (13 A+15 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (43 A+45 B) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{a^3 (13 A+15 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{(7 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{20 d}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 2975
Rule 2968
Rule 3021
Rule 2748
Rule 3768
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx &=\frac{a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+a \cos (c+d x))^2 (a (7 A+5 B)+a (2 A+5 B) \cos (c+d x)) \sec ^5(c+d x) \, dx\\ &=\frac{(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int (a+a \cos (c+d x)) \left (a^2 (43 A+45 B)+2 a^2 (11 A+15 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int \left (a^3 (43 A+45 B)+\left (2 a^3 (11 A+15 B)+a^3 (43 A+45 B)\right ) \cos (c+d x)+2 a^3 (11 A+15 B) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{60} \int \left (15 a^3 (13 A+15 B)+4 a^3 (38 A+45 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} \left (a^3 (13 A+15 B)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{15} \left (a^3 (38 A+45 B)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (13 A+15 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{8} \left (a^3 (13 A+15 B)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^3 (38 A+45 B)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a^3 (13 A+15 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac{a^3 (13 A+15 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 1.41489, size = 294, normalized size = 1.59 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (13 A+15 B) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (3 A+5 B) \sin (2 c+d x)+80 (29 A+30 B) \sin (d x)+750 A \sin (c+2 d x)+750 A \sin (3 c+2 d x)+1520 A \sin (2 c+3 d x)+195 A \sin (3 c+4 d x)+195 A \sin (5 c+4 d x)+304 A \sin (4 c+5 d x)+570 B \sin (c+2 d x)+570 B \sin (3 c+2 d x)+1680 B \sin (2 c+3 d x)-120 B \sin (4 c+3 d x)+225 B \sin (3 c+4 d x)+225 B \sin (5 c+4 d x)+360 B \sin (4 c+5 d x))\right )}{15360 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.114, size = 234, normalized size = 1.3 \begin{align*}{\frac{13\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{13\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+3\,{\frac{{a}^{3}B\tan \left ( dx+c \right ) }{d}}+{\frac{38\,A{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{19\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{15\,{a}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.02777, size = 455, normalized size = 2.46 \begin{align*} \frac{16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 45 \, A a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3}{\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 )} - 180 \, B a^{3}{\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 )} + 240 \, B a^{3} \tan \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.47847, size = 431, normalized size = 2.33 \begin{align*} \frac{15 \,{\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (38 \, A + 45 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \,{\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (19 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 24 \, A a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \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 = 1.27076, size = 332, normalized size = 1.79 \begin{align*} \frac{15 \,{\left (13 \, A a^{3} + 15 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (13 \, A a^{3} + 15 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (195 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 225 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 910 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1050 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1920 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1330 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1830 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 735 \, B 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 )}^{5}}}{120 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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