Optimal. Leaf size=116 \[ -\frac{a^3 (4 B+3 C) \sin ^3(c+d x)}{12 d}+\frac{a^3 (4 B+3 C) \sin (c+d x)}{d}+\frac{3 a^3 (4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (4 B+3 C)+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.166009, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {3029, 2751, 2645, 2637, 2635, 8, 2633} \[ -\frac{a^3 (4 B+3 C) \sin ^3(c+d x)}{12 d}+\frac{a^3 (4 B+3 C) \sin (c+d x)}{d}+\frac{3 a^3 (4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (4 B+3 C)+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 3029
Rule 2751
Rule 2645
Rule 2637
Rule 2635
Rule 8
Rule 2633
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\int (a+a \cos (c+d x))^3 (B+C \cos (c+d x)) \, dx\\ &=\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} (4 B+3 C) \int (a+a \cos (c+d x))^3 \, dx\\ &=\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} (4 B+3 C) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac{1}{4} a^3 (4 B+3 C) x+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \left (a^3 (4 B+3 C)\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \cos (c+d x) \, dx+\frac{1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{1}{4} a^3 (4 B+3 C) x+\frac{3 a^3 (4 B+3 C) \sin (c+d x)}{4 d}+\frac{3 a^3 (4 B+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{8} \left (3 a^3 (4 B+3 C)\right ) \int 1 \, dx-\frac{\left (a^3 (4 B+3 C)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac{5}{8} a^3 (4 B+3 C) x+\frac{a^3 (4 B+3 C) \sin (c+d x)}{d}+\frac{3 a^3 (4 B+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{a^3 (4 B+3 C) \sin ^3(c+d x)}{12 d}\\ \end{align*}
Mathematica [A] time = 0.301944, size = 86, normalized size = 0.74 \[ \frac{a^3 (24 (15 B+13 C) \sin (c+d x)+24 (3 B+4 C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+240 B d x+24 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+180 C d x)}{96 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.055, size = 176, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{3}C \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{{a}^{3}B \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{3}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,{a}^{3}B \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{3}C \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{3}B\sin \left ( dx+c \right ) +{a}^{3}C\sin \left ( dx+c \right ) +{a}^{3}B \left ( dx+c \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.06128, size = 225, normalized size = 1.94 \begin{align*} -\frac{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} - 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 = 1.68229, size = 216, normalized size = 1.86 \begin{align*} \frac{15 \,{\left (4 \, B + 3 \, C\right )} a^{3} d x +{\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} + 9 \,{\left (4 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \,{\left (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 = 1.49682, size = 238, normalized size = 2.05 \begin{align*} \frac{15 \,{\left (4 \, B a^{3} + 3 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (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} + 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} + 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} + 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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