Optimal. Leaf size=179 \[ \frac{\left (6 a^2 b B+2 a^3 C+9 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{3 d}+\frac{a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (12 a^2 b C+3 a^3 B+12 a b^2 B+8 b^3 C\right )+\frac{a^2 (2 a C+3 b B) \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac{a B \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]
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Rubi [A] time = 0.491216, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4025, 4074, 4047, 2637, 4045, 8} \[ \frac{\left (6 a^2 b B+2 a^3 C+9 a b^2 C+3 b^3 B\right ) \sin (c+d x)}{3 d}+\frac{a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (12 a^2 b C+3 a^3 B+12 a b^2 B+8 b^3 C\right )+\frac{a^2 (2 a C+3 b B) \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac{a B \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4025
Rule 4074
Rule 4047
Rule 2637
Rule 4045
Rule 8
Rubi steps
\begin{align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^4(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (-2 a (3 b B+2 a C)-\left (3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x)-b (a B+4 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (3 b B+2 a C) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac{a B \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{1}{12} \int \cos ^2(c+d x) \left (3 a \left (3 a^2 B+10 b^2 B+12 a b C\right )+4 \left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \sec (c+d x)+3 b^2 (a B+4 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (3 b B+2 a C) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac{a B \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{1}{12} \int \cos ^2(c+d x) \left (3 a \left (3 a^2 B+10 b^2 B+12 a b C\right )+3 b^2 (a B+4 b C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} \left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \int \cos (c+d x) \, dx\\ &=\frac{\left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \sin (c+d x)}{3 d}+\frac{a \left (3 a^2 B+10 b^2 B+12 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (3 b B+2 a C) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac{a B \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{1}{8} \left (3 a^3 B+12 a b^2 B+12 a^2 b C+8 b^3 C\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (3 a^3 B+12 a b^2 B+12 a^2 b C+8 b^3 C\right ) x+\frac{\left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \sin (c+d x)}{3 d}+\frac{a \left (3 a^2 B+10 b^2 B+12 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (3 b B+2 a C) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac{a B \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.415275, size = 140, normalized size = 0.78 \[ \frac{12 (c+d x) \left (12 a^2 b C+3 a^3 B+12 a b^2 B+8 b^3 C\right )+24 a \left (a^2 B+3 a b C+3 b^2 B\right ) \sin (2 (c+d x))+24 \left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \sin (c+d x)+8 a^2 (a C+3 b B) \sin (3 (c+d x))+3 a^3 B \sin (4 (c+d x))}{96 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.067, size = 180, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( B{a}^{3} \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 ) +B{a}^{2}b \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{{a}^{3}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+3\,Ba{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{2}bC \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +B{b}^{3}\sin \left ( dx+c \right ) +3\,Ca{b}^{2}\sin \left ( dx+c \right ) +C{b}^{3} \left ( dx+c \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.988208, size = 231, normalized size = 1.29 \begin{align*} \frac{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 )} B a^{3} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} - 96 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 96 \,{\left (d x + c\right )} C b^{3} + 288 \, C a b^{2} \sin \left (d x + c\right ) + 96 \, B b^{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 = 0.520797, size = 321, normalized size = 1.79 \begin{align*} \frac{3 \,{\left (3 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 8 \, C b^{3}\right )} d x +{\left (6 \, B a^{3} \cos \left (d x + c\right )^{3} + 16 \, C a^{3} + 48 \, B a^{2} b + 72 \, C a b^{2} + 24 \, B b^{3} + 8 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )^{2} + 9 \,{\left (B a^{3} + 4 \, C a^{2} b + 4 \, B a b^{2}\right )} \cos \left (d x + c\right )\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 [B] time = 1.24079, size = 724, normalized size = 4.04 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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