Optimal. Leaf size=154 \[ \frac{a^3 (9 A+11 B) \tan (c+d x)}{3 d}+\frac{5 a^3 (3 A+4 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (27 A+28 B) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(3 A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]
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Rubi [A] time = 0.418148, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac{a^3 (9 A+11 B) \tan (c+d x)}{3 d}+\frac{5 a^3 (3 A+4 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (27 A+28 B) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(3 A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]
Antiderivative was successfully verified.
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Rule 2975
Rule 2968
Rule 3021
Rule 2748
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx &=\frac{a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+a \cos (c+d x))^2 (2 a (3 A+2 B)+a (A+4 B) \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac{(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+a \cos (c+d x)) \left (a^2 (27 A+28 B)+a^2 (9 A+16 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int \left (a^3 (27 A+28 B)+\left (a^3 (9 A+16 B)+a^3 (27 A+28 B)\right ) \cos (c+d x)+a^3 (9 A+16 B) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{24} \int \left (8 a^3 (9 A+11 B)+15 a^3 (3 A+4 B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} \left (5 a^3 (3 A+4 B)\right ) \int \sec (c+d x) \, dx+\frac{1}{3} \left (a^3 (9 A+11 B)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{5 a^3 (3 A+4 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{\left (a^3 (9 A+11 B)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{5 a^3 (3 A+4 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (9 A+11 B) \tan (c+d x)}{3 d}+\frac{a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.23551, size = 273, normalized size = 1.77 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (120 (3 A+4 B) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-24 (9 A+11 B) \sin (c)+(69 A+36 B) \sin (d x)+69 A \sin (2 c+d x)+264 A \sin (c+2 d x)-24 A \sin (3 c+2 d x)+45 A \sin (2 c+3 d x)+45 A \sin (4 c+3 d x)+72 A \sin (3 c+4 d x)+36 B \sin (2 c+d x)+280 B \sin (c+2 d x)-72 B \sin (3 c+2 d x)+36 B \sin (2 c+3 d x)+36 B \sin (4 c+3 d x)+88 B \sin (3 c+4 d x))\right )}{1536 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.121, size = 188, normalized size = 1.2 \begin{align*} 3\,{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{5\,{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{15\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{11\,{a}^{3}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{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}}{3\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00729, size = 363, normalized size = 2.36 \begin{align*} \frac{48 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 3 \, 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 )} - 36 \, 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 )} - 36 \, 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 )} + 24 \, B a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{3} \tan \left (d x + c\right ) + 144 \, B a^{3} \tan \left (d x + c\right )}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.40401, size = 366, normalized size = 2.38 \begin{align*} \frac{15 \,{\left (3 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (9 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \,{\left (5 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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 = 1.33884, size = 286, normalized size = 1.86 \begin{align*} \frac{15 \,{\left (3 \, A a^{3} + 4 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (3 \, A a^{3} + 4 \, 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 (45 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 165 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 220 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 219 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 147 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 132 \, 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 )}^{4}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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