Optimal. Leaf size=309 \[ -\frac{a \left (16 a^2 A b+4 a^3 B+27 a b^2 B+13 A b^3\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (48 a^3 A b+87 a^2 b^2 B+12 a^4 B+53 a A b^3+15 b^4 B\right ) \sin (c+d x)}{15 d}+\frac{a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}+\frac{\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right )+\frac{a (2 a B+3 A b) \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d}+\frac{a A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d} \]
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Rubi [A] time = 0.819945, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {4025, 4094, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac{a \left (16 a^2 A b+4 a^3 B+27 a b^2 B+13 A b^3\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (48 a^3 A b+87 a^2 b^2 B+12 a^4 B+53 a A b^3+15 b^4 B\right ) \sin (c+d x)}{15 d}+\frac{a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}+\frac{\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right )+\frac{a (2 a B+3 A b) \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d}+\frac{a A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d} \]
Antiderivative was successfully verified.
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Rule 4025
Rule 4094
Rule 4074
Rule 4047
Rule 2635
Rule 8
Rule 4044
Rule 3013
Rubi steps
\begin{align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac{1}{6} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (-3 a (3 A b+2 a B)-\left (5 a^2 A+6 A b^2+12 a b B\right ) \sec (c+d x)-2 b (a A+3 b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac{1}{30} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (-a \left (25 a^2 A+48 A b^2+72 a b B\right )-\left (71 a^2 A b+30 A b^3+24 a^3 B+90 a b^2 B\right ) \sec (c+d x)-2 b \left (14 a A b+6 a^2 B+15 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{1}{120} \int \cos ^3(c+d x) \left (24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )+15 \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \sec (c+d x)+8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{1}{120} \int \cos ^3(c+d x) \left (24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )+8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{8} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{1}{120} \int \cos (c+d x) \left (8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right )+24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \cos ^2(c+d x)\right ) \, dx+\frac{1}{16} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \int 1 \, dx\\ &=\frac{1}{16} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) x+\frac{\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \left (8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right )+24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )-24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{120 d}\\ &=\frac{1}{16} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) x+\frac{\left (48 a^3 A b+53 a A b^3+12 a^4 B+87 a^2 b^2 B+15 b^4 B\right ) \sin (c+d x)}{15 d}+\frac{\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac{a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}
Mathematica [A] time = 1.22771, size = 333, normalized size = 1.08 \[ \frac{120 \left (20 a^3 A b+36 a^2 b^2 B+5 a^4 B+24 a A b^3+8 b^4 B\right ) \sin (c+d x)+15 \left (96 a^2 A b^2+15 a^4 A+64 a^3 b B+64 a b^3 B+16 A b^4\right ) \sin (2 (c+d x))+180 a^2 A b^2 \sin (4 (c+d x))+2160 a^2 A b^2 c+2160 a^2 A b^2 d x+400 a^3 A b \sin (3 (c+d x))+48 a^3 A b \sin (5 (c+d x))+45 a^4 A \sin (4 (c+d x))+5 a^4 A \sin (6 (c+d x))+300 a^4 A c+300 a^4 A d x+480 a^2 b^2 B \sin (3 (c+d x))+120 a^3 b B \sin (4 (c+d x))+1440 a^3 b B c+1440 a^3 b B d x+100 a^4 B \sin (3 (c+d x))+12 a^4 B \sin (5 (c+d x))+320 a A b^3 \sin (3 (c+d x))+1920 a b^3 B c+1920 a b^3 B d x+480 A b^4 c+480 A b^4 d x}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.079, size = 316, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( A{a}^{4} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{B{a}^{4}\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 ) }+{\frac{4\,A{a}^{3}b\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 ) }+4\,B{a}^{3}b \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 ) +6\,A{a}^{2}{b}^{2} \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 ) +2\,B{a}^{2}{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{4\,Aa{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+4\,Ba{b}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +A{b}^{4} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +B{b}^{4}\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 = 0.986661, size = 414, normalized size = 1.34 \begin{align*} -\frac{5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 64 \,{\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 )} B a^{4} - 256 \,{\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^{3} b - 120 \,{\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} b - 180 \,{\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^{2} b^{2} + 1920 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b^{2} + 1280 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{3} - 960 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} - 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} - 960 \, B b^{4} \sin \left (d x + c\right )}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.609289, size = 587, normalized size = 1.9 \begin{align*} \frac{15 \,{\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} d x +{\left (40 \, A a^{4} \cos \left (d x + c\right )^{5} + 128 \, B a^{4} + 512 \, A a^{3} b + 960 \, B a^{2} b^{2} + 640 \, A a b^{3} + 240 \, B b^{4} + 48 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{4} + 10 \,{\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 32 \,{\left (2 \, B a^{4} + 8 \, A a^{3} b + 15 \, B a^{2} b^{2} + 10 \, A a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, 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 = 1.2839, size = 1521, normalized size = 4.92 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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