Optimal. Leaf size=180 \[ -\frac{\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin (c+d x)}{5 d}+\frac{\left (3 a^2 B+6 a A b+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2 B+6 a A b+4 b^2 B\right )+\frac{a^2 A \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac{a (a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d} \]
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Rubi [A] time = 0.268378, antiderivative size = 180, 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 = {4024, 4047, 2633, 4045, 2635, 8} \[ -\frac{\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin (c+d x)}{5 d}+\frac{\left (3 a^2 B+6 a A b+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2 B+6 a A b+4 b^2 B\right )+\frac{a^2 A \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac{a (a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 4024
Rule 4047
Rule 2633
Rule 4045
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac{a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} \int \cos ^4(c+d x) \left (-5 a (2 A b+a B)+\left (A \left (-4 a^2-5 b^2\right )-10 a b B\right ) \sec (c+d x)-5 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} \int \cos ^4(c+d x) \left (-5 a (2 A b+a B)-5 b^2 B \sec ^2(c+d x)\right ) \, dx-\frac{1}{5} \left (-4 a^2 A-5 A b^2-10 a b B\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac{a (2 A b+a B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{1}{4} \left (-6 a A b-3 a^2 B-4 b^2 B\right ) \int \cos ^2(c+d x) \, dx-\frac{\left (4 a^2 A+5 A b^2+10 a b B\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin (c+d x)}{5 d}+\frac{\left (6 a A b+3 a^2 B+4 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a (2 A b+a B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin ^3(c+d x)}{15 d}-\frac{1}{8} \left (-6 a A b-3 a^2 B-4 b^2 B\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (6 a A b+3 a^2 B+4 b^2 B\right ) x+\frac{\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin (c+d x)}{5 d}+\frac{\left (6 a A b+3 a^2 B+4 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a (2 A b+a B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}
Mathematica [A] time = 0.545049, size = 146, normalized size = 0.81 \[ \frac{60 (c+d x) \left (3 a^2 B+6 a A b+4 b^2 B\right )+60 \left (5 a^2 A+12 a b B+6 A b^2\right ) \sin (c+d x)+120 \left (a^2 B+2 a A b+b^2 B\right ) \sin (2 (c+d x))+10 \left (5 a^2 A+8 a b B+4 A b^2\right ) \sin (3 (c+d x))+6 a^2 A \sin (5 (c+d x))+15 a (a B+2 A b) \sin (4 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 184, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}A\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+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 ) +2\,Aab \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{\frac{2\,Bab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{A{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+B{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.963242, size = 238, normalized size = 1.32 \begin{align*} \frac{32 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} + 15 \,{\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} + 30 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2}}{480 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.50198, size = 350, normalized size = 1.94 \begin{align*} \frac{15 \,{\left (3 \, B a^{2} + 6 \, A a b + 4 \, B b^{2}\right )} d x +{\left (24 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \,{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{3} + 64 \, A a^{2} + 160 \, B a b + 80 \, A b^{2} + 8 \,{\left (4 \, A a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (3 \, B a^{2} + 6 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, 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 = 1.19317, size = 657, normalized size = 3.65 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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