Optimal. Leaf size=147 \[ \frac{5 a^3 (4 A+3 C) \sin (c+d x)}{8 d}+\frac{(4 A+5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{8 d}+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{8} a^3 x (28 A+15 C)+\frac{C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 a d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.438851, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3046, 2976, 2968, 3023, 2735, 3770} \[ \frac{5 a^3 (4 A+3 C) \sin (c+d x)}{8 d}+\frac{(4 A+5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{8 d}+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{8} a^3 x (28 A+15 C)+\frac{C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 a d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 3046
Rule 2976
Rule 2968
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^3 \left (A+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+3 a 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{C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{\int (a+a \cos (c+d x))^2 \left (12 a^2 A+3 a^2 (4 A+5 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{C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{8 d}+\frac{\int (a+a \cos (c+d x)) \left (24 a^3 A+15 a^3 (4 A+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{C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{8 d}+\frac{\int \left (24 a^4 A+\left (24 a^4 A+15 a^4 (4 A+3 C)\right ) \cos (c+d x)+15 a^4 (4 A+3 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{5 a^3 (4 A+3 C) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{8 d}+\frac{\int \left (24 a^4 A+3 a^4 (28 A+15 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{1}{8} a^3 (28 A+15 C) x+\frac{5 a^3 (4 A+3 C) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{8 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^3 (28 A+15 C) x+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 (4 A+3 C) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{8 d}\\ \end{align*}
Mathematica [A] time = 0.324389, size = 124, normalized size = 0.84 \[ \frac{a^3 \left (8 (12 A+13 C) \sin (c+d x)+8 (A+4 C) \sin (2 (c+d x))-32 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+32 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+112 A d x+8 C \sin (3 (c+d x))+C \sin (4 (c+d x))+60 C d x\right )}{32 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.06, size = 175, normalized size = 1.2 \begin{align*}{\frac{A{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{7\,A{a}^{3}x}{2}}+{\frac{7\,A{a}^{3}c}{2\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,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{C \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{3}}{d}}+3\,{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+{\frac{A{a}^{3}\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 = 1.03475, size = 220, normalized size = 1.5 \begin{align*} \frac{8 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 96 \,{\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 )} C 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 )} C a^{3} + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 32 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, A a^{3} \sin \left (d x + c\right ) + 32 \, 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 = 1.49299, size = 284, normalized size = 1.93 \begin{align*} \frac{{\left (28 \, A + 15 \, C\right )} a^{3} d x + 4 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 8 \, C a^{3} \cos \left (d x + c\right )^{2} +{\left (4 \, A + 15 \, 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 = 1.31912, size = 288, normalized size = 1.96 \begin{align*} \frac{8 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (28 \, A a^{3} + 15 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (20 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 68 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 55 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 76 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 73 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 28 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 49 \, 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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