Optimal. Leaf size=162 \[ \frac{5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac{(12 A+20 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+\frac{1}{8} a^3 x (28 A+20 B+15 C)+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.477943, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3045, 2976, 2968, 3023, 2735, 3770} \[ \frac{5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac{(12 A+20 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+\frac{1}{8} a^3 x (28 A+20 B+15 C)+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 3045
Rule 2976
Rule 2968
Rule 3023
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 (c+d x) \, dx &=\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^3 (4 a A+a (4 B+3 C) \cos (c+d x)) \sec (c+d x) \, dx}{4 a}\\ &=\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac{\int (a+a \cos (c+d x))^2 \left (12 a^2 A+a^2 (12 A+20 B+15 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac{(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int (a+a \cos (c+d x)) \left (24 a^3 A+15 a^3 (4 A+4 B+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac{(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int \left (24 a^4 A+\left (24 a^4 A+15 a^4 (4 A+4 B+3 C)\right ) \cos (c+d x)+15 a^4 (4 A+4 B+3 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac{(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int \left (24 a^4 A+3 a^4 (28 A+20 B+15 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{1}{8} a^3 (28 A+20 B+15 C) x+\frac{5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac{(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^3 (28 A+20 B+15 C) x+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac{(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end{align*}
Mathematica [A] time = 0.479983, size = 147, normalized size = 0.91 \[ \frac{a^3 \left (24 (12 A+15 B+13 C) \sin (c+d x)+24 (A+3 B+4 C) \sin (2 (c+d x))-96 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+96 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+336 A d x+8 B \sin (3 (c+d x))+240 B d x+24 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+180 C d x\right )}{96 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.067, size = 251, normalized size = 1.6 \begin{align*}{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{5\,{a}^{3}Bx}{2}}+{\frac{5\,{a}^{3}Bc}{2\,d}}+3\,{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+{\frac{7\,A{a}^{3}x}{2}}+{\frac{7\,A{a}^{3}c}{2\,d}}+{\frac{11\,{a}^{3}B\sin \left ( dx+c \right ) }{3\,d}}+{\frac{15\,{a}^{3}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}Cx}{8}}+{\frac{15\,{a}^{3}Cc}{8\,d}}+3\,{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{3}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{d}}+{\frac{A{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{3\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) \left ( \cos \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 = 1.01984, size = 315, normalized size = 1.94 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 288 \,{\left (d x + c\right )} A a^{3} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 96 \,{\left (d x + c\right )} B a^{3} - 96 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 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 )} C a^{3} + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 96 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 288 \, A a^{3} \sin \left (d x + c\right ) + 288 \, B a^{3} \sin \left (d x + c\right ) + 96 \, C a^{3} \sin \left (d x + c\right )}{96 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.93959, size = 336, normalized size = 2.07 \begin{align*} \frac{3 \,{\left (28 \, A + 20 \, B + 15 \, C\right )} a^{3} d x + 12 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, C a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, A + 12 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \,{\left (9 \, A + 11 \, B + 9 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, 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 = 1.26506, size = 386, normalized size = 2.38 \begin{align*} \frac{24 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (28 \, A a^{3} + 20 \, B a^{3} + 15 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (60 \, 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} + 45 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 204 \, 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} + 165 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 228 \, 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} + 219 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 84 \, 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 ) + 147 \, 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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