Optimal. Leaf size=136 \[ \frac{\left (a^2 C+2 a b B+b^2 C\right ) \sin (c+d x)}{d}+\frac{\left (3 a^2 B+8 a b C+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2 B+8 a b C+4 b^2 B\right )+\frac{a^2 B \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac{a (a C+2 b B) \sin ^3(c+d x)}{3 d} \]
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Rubi [A] time = 0.320546, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4024, 4047, 2635, 8, 4044, 3013} \[ \frac{\left (a^2 C+2 a b B+b^2 C\right ) \sin (c+d x)}{d}+\frac{\left (3 a^2 B+8 a b C+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2 B+8 a b C+4 b^2 B\right )+\frac{a^2 B \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac{a (a C+2 b B) \sin ^3(c+d x)}{3 d} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4024
Rule 4047
Rule 2635
Rule 8
Rule 4044
Rule 3013
Rubi steps
\begin{align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \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))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{a^2 B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos ^3(c+d x) \left (-4 a (2 b B+a C)+\left (\left (-3 a^2-4 b^2\right ) B-8 a b C\right ) \sec (c+d x)-4 b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos ^3(c+d x) \left (-4 a (2 b B+a C)-4 b^2 C \sec ^2(c+d x)\right ) \, dx-\frac{1}{4} \left (-3 a^2 B-4 b^2 B-8 a b C\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{\left (3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos (c+d x) \left (-4 b^2 C-4 a (2 b B+a C) \cos ^2(c+d x)\right ) \, dx-\frac{1}{8} \left (-3 a^2 B-4 b^2 B-8 a b C\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac{\left (3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \left (-4 b^2 C-4 a (2 b B+a C)+4 a (2 b B+a C) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac{1}{8} \left (3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac{\left (2 a b B+a^2 C+b^2 C\right ) \sin (c+d x)}{d}+\frac{\left (3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{a (2 b B+a C) \sin ^3(c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.51283, size = 118, normalized size = 0.87 \[ \frac{12 (c+d x) \left (3 a^2 B+8 a b C+4 b^2 B\right )+24 \left (3 a^2 C+6 a b B+4 b^2 C\right ) \sin (c+d x)+24 \left (a^2 B+2 a b C+b^2 B\right ) \sin (2 (c+d x))+3 a^2 B \sin (4 (c+d x))+8 a (a C+2 b B) \sin (3 (c+d x))}{96 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 152, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( B{a}^{2} \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}^{2}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{2\,Bab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,abC \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +B{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +{b}^{2}C\sin \left ( dx+c \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.964501, size = 192, normalized size = 1.41 \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^{2} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 48 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 96 \, C b^{2} \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 = 0.507277, size = 274, normalized size = 2.01 \begin{align*} \frac{3 \,{\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} d x +{\left (6 \, B a^{2} \cos \left (d x + c\right )^{3} + 16 \, C a^{2} + 32 \, B a b + 24 \, C b^{2} + 8 \,{\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, B a^{2} + 8 \, C a b + 4 \, B 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 = 1.22382, size = 590, normalized size = 4.34 \begin{align*} \frac{3 \,{\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )}{\left (d x + c\right )} - \frac{2 \,{\left (15 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 24 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 40 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 80 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 72 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 48 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, C b^{2} \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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