Optimal. Leaf size=183 \[ \frac{5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac{a^3 (15 A+20 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(15 A+20 B+12 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+\frac{(3 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+a^3 C x+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.553734, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3043, 2975, 2968, 3021, 2735, 3770} \[ \frac{5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac{a^3 (15 A+20 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(15 A+20 B+12 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+\frac{(3 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+a^3 C x+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 3043
Rule 2975
Rule 2968
Rule 3021
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^3 (a (3 A+4 B)+4 a C \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac{(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^2 \left (a^2 (15 A+20 B+12 C)+12 a^2 C \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac{(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x)) \left (15 a^3 (3 A+4 (B+C))+24 a^3 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac{(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (15 a^4 (3 A+4 (B+C))+\left (24 a^4 C+15 a^4 (3 A+4 (B+C))\right ) \cos (c+d x)+24 a^4 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac{5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac{(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (3 a^4 (15 A+20 B+28 C)+24 a^4 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=a^3 C x+\frac{5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac{(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} \left (a^3 (15 A+20 B+28 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 C x+\frac{a^3 (15 A+20 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac{(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [B] time = 6.18326, size = 793, normalized size = 4.33 \[ \frac{\sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (9 A \sin \left (\frac{1}{2} (c+d x)\right )+11 B \sin \left (\frac{1}{2} (c+d x)\right )+9 C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{24 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (9 A \sin \left (\frac{1}{2} (c+d x)\right )+11 B \sin \left (\frac{1}{2} (c+d x)\right )+9 C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{24 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{(57 A+40 B+12 C) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3}{384 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(-57 A-40 B-12 C) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3}{384 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(-15 A-20 B-28 C) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{(15 A+20 B+28 C) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{\sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (3 A \sin \left (\frac{1}{2} (c+d x)\right )+B \sin \left (\frac{1}{2} (c+d x)\right )\right )}{48 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{\sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (3 A \sin \left (\frac{1}{2} (c+d x)\right )+B \sin \left (\frac{1}{2} (c+d x)\right )\right )}{48 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{A \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3}{128 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{A \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3}{128 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{C (c+d x) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^3}{8 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.098, size = 262, normalized size = 1.4 \begin{align*}{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,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}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{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{5\,{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}}+{a}^{3}Cx+{\frac{C{a}^{3}c}{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.0297, size = 494, normalized size = 2.7 \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} + 48 \,{\left (d x + c\right )} C 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 )} - 12 \, C 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 )} + 72 \, 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 )} + 48 \, A a^{3} \tan \left (d x + c\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 144 \, C 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 = 2.16669, size = 447, normalized size = 2.44 \begin{align*} \frac{48 \, C a^{3} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (15 \, A + 20 \, B + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (15 \, A + 20 \, B + 28 \, C\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 + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 3 \,{\left (15 \, A + 12 \, B + 4 \, C\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 = 1.26833, size = 406, normalized size = 2.22 \begin{align*} \frac{24 \,{\left (d x + c\right )} C a^{3} + 3 \,{\left (15 \, A a^{3} + 20 \, B a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (15 \, A a^{3} + 20 \, B a^{3} + 28 \, C 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} + 60 \, C 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} - 204 \, C 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} + 228 \, C 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 ) - 84 \, 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}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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