Optimal. Leaf size=145 \[ -\frac{(6 A-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}-\frac{(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac{3 a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^3 x (6 A+5 C)+\frac{5 a^3 C \sin (c+d x)}{2 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]
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Rubi [A] time = 0.452336, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3044, 2976, 2968, 3023, 2735, 3770} \[ -\frac{(6 A-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}-\frac{(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac{3 a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^3 x (6 A+5 C)+\frac{5 a^3 C \sin (c+d x)}{2 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]
Antiderivative was successfully verified.
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Rule 3044
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 ^2(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x))^3 (3 a A-a (3 A-C) \cos (c+d x)) \sec (c+d x) \, dx}{a}\\ &=-\frac{(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}+\frac{A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x))^2 \left (9 a^2 A-a^2 (6 A-5 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{3 a}\\ &=-\frac{(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac{(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x)) \left (18 a^3 A+15 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac{(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac{(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{\int \left (18 a^4 A+\left (18 a^4 A+15 a^4 C\right ) \cos (c+d x)+15 a^4 C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac{5 a^3 C \sin (c+d x)}{2 d}-\frac{(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac{(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{\int \left (18 a^4 A+3 a^4 (6 A+5 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac{1}{2} a^3 (6 A+5 C) x+\frac{5 a^3 C \sin (c+d x)}{2 d}-\frac{(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac{(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\left (3 a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^3 (6 A+5 C) x+\frac{3 a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 C \sin (c+d x)}{2 d}-\frac{(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac{(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end{align*}
Mathematica [B] time = 1.9216, size = 298, normalized size = 2.06 \[ \frac{1}{96} a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (\frac{3 (4 A+15 C) \sin (c) \cos (d x)}{d}+\frac{3 (4 A+15 C) \cos (c) \sin (d x)}{d}+\frac{12 A \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{12 A \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{36 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{36 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+6 x (6 A+5 C)+\frac{9 C \sin (2 c) \cos (2 d x)}{d}+\frac{C \sin (3 c) \cos (3 d x)}{d}+\frac{9 C \cos (2 c) \sin (2 d x)}{d}+\frac{C \cos (3 c) \sin (3 d x)}{d}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.07, size = 146, normalized size = 1. \begin{align*}{\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}}{3\,d}}+{\frac{11\,{a}^{3}C\sin \left ( dx+c \right ) }{3\,d}}+3\,A{a}^{3}x+3\,{\frac{A{a}^{3}c}{d}}+{\frac{3\,{a}^{3}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{5\,{a}^{3}Cx}{2}}+{\frac{5\,{a}^{3}Cc}{2\,d}}+3\,{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.04451, size = 185, normalized size = 1.28 \begin{align*} \frac{36 \,{\left (d x + c\right )} A a^{3} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \,{\left (d x + c\right )} C a^{3} + 18 \, A a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.56, size = 350, normalized size = 2.41 \begin{align*} \frac{3 \,{\left (6 \, A + 5 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 9 \, A a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, A a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 9 \, C a^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, A + 11 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \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 = 1.25744, size = 284, normalized size = 1.96 \begin{align*} \frac{18 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{12 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + 3 \,{\left (6 \, A a^{3} + 5 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 33 \, 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 )}^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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