Optimal. Leaf size=218 \[ \frac{a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \sin (c+d x)}{15 d}+\frac{a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{30 d}+\frac{3 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{40 d}+\frac{1}{8} b x \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac{3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d} \]
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Rubi [A] time = 0.650911, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4094, 4074, 4047, 2637, 4045, 8} \[ \frac{a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \sin (c+d x)}{15 d}+\frac{a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{30 d}+\frac{3 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{40 d}+\frac{1}{8} b x \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac{3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d} \]
Antiderivative was successfully verified.
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Rule 4095
Rule 4094
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 (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (4 A+5 C) \sec (c+d x)+b (A+5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (2 \left (3 A b^2+2 a^2 (4 A+5 C)\right )+a b (29 A+40 C) \sec (c+d x)+b^2 (7 A+20 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac{1}{60} \int \cos ^2(c+d x) \left (-9 b \left (2 A b^2+5 a^2 (3 A+4 C)\right )-4 a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sec (c+d x)-3 b^3 (7 A+20 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac{1}{60} \int \cos ^2(c+d x) \left (-9 b \left (2 A b^2+5 a^2 (3 A+4 C)\right )-3 b^3 (7 A+20 C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{15} \left (a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \int \cos (c+d x) \, dx\\ &=\frac{a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac{3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{8} \left (b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right )\right ) \int 1 \, dx\\ &=\frac{1}{8} b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) x+\frac{a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac{3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 0.644841, size = 155, normalized size = 0.71 \[ \frac{60 b (c+d x) \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+60 a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \sin (c+d x)+10 a \left (a^2 (5 A+4 C)+12 A b^2\right ) \sin (3 (c+d x))+120 b \left (3 a^2 (A+C)+A b^2\right ) \sin (2 (c+d x))+45 a^2 A b \sin (4 (c+d x))+6 a^3 A \sin (5 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.073, size = 201, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{A{a}^{3}\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 ) }+3\,A{a}^{2}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 ) +Aa{b}^{2} \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}}+A{b}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{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 ) +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 = 1.00878, size = 262, normalized size = 1.2 \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^{3} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 45 \,{\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 + 360 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 480 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \,{\left (d x + c\right )} C b^{3} + 1440 \, C a b^{2} \sin \left (d x + c\right )}{480 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.527668, size = 367, normalized size = 1.68 \begin{align*} \frac{15 \,{\left (3 \,{\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \,{\left (A + 2 \, C\right )} b^{3}\right )} d x +{\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 90 \, A a^{2} b \cos \left (d x + c\right )^{3} + 16 \,{\left (4 \, A + 5 \, C\right )} a^{3} + 120 \,{\left (2 \, A + 3 \, C\right )} a b^{2} + 8 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (3 \,{\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, A b^{3}\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 = 1.23341, size = 818, normalized size = 3.75 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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