Optimal. Leaf size=161 \[ -\frac{\left (a^2 (4 A+5 C)+2 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (a^2+b^2\right ) (4 A+5 C) \sin (c+d x)}{5 d}+\frac{a b (3 A+4 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{a A b \sin (c+d x) \cos ^3(c+d x)}{10 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}+\frac{1}{4} a b x (3 A+4 C) \]
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Rubi [A] time = 0.399549, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac{\left (a^2 (4 A+5 C)+2 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (a^2+b^2\right ) (4 A+5 C) \sin (c+d x)}{5 d}+\frac{a b (3 A+4 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{a A b \sin (c+d x) \cos ^3(c+d x)}{10 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}+\frac{1}{4} a b x (3 A+4 C) \]
Antiderivative was successfully verified.
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Rule 4095
Rule 4074
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 (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (2 A b+a (4 A+5 C) \sec (c+d x)+b (2 A+5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a A b \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{1}{20} \int \cos ^3(c+d x) \left (-4 \left (2 A b^2+a^2 (4 A+5 C)\right )-10 a b (3 A+4 C) \sec (c+d x)-4 b^2 (2 A+5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a A b \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{1}{20} \int \cos ^3(c+d x) \left (-4 \left (2 A b^2+a^2 (4 A+5 C)\right )-4 b^2 (2 A+5 C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} (a b (3 A+4 C)) \int \cos ^2(c+d x) \, dx\\ &=\frac{a b (3 A+4 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a A b \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{1}{20} \int \cos (c+d x) \left (-4 b^2 (2 A+5 C)-4 \left (2 A b^2+a^2 (4 A+5 C)\right ) \cos ^2(c+d x)\right ) \, dx+\frac{1}{4} (a b (3 A+4 C)) \int 1 \, dx\\ &=\frac{1}{4} a b (3 A+4 C) x+\frac{a b (3 A+4 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a A b \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{\operatorname{Subst}\left (\int \left (-4 b^2 (2 A+5 C)-4 \left (2 A b^2+a^2 (4 A+5 C)\right )+4 \left (2 A b^2+a^2 (4 A+5 C)\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac{1}{4} a b (3 A+4 C) x+\frac{\left (a^2+b^2\right ) (4 A+5 C) \sin (c+d x)}{5 d}+\frac{a b (3 A+4 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a A b \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{\left (2 A b^2+a^2 (4 A+5 C)\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}
Mathematica [A] time = 0.45228, size = 126, normalized size = 0.78 \[ \frac{30 \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sin (c+d x)+5 \left (a^2 (5 A+4 C)+4 A b^2\right ) \sin (3 (c+d x))+3 a^2 A \sin (5 (c+d x))+60 a b (3 A+4 C) (c+d x)+120 a b (A+C) \sin (2 (c+d x))+15 a A b \sin (4 (c+d x))}{240 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.077, size = 158, 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 ) }+{\frac{{a}^{2}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+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 ) +2\,abC \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{A{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{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 = 0.989982, size = 208, normalized size = 1.29 \begin{align*} \frac{16 \,{\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} - 80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C 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 )} A a b + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b - 80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 240 \, C b^{2} \sin \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.513828, size = 300, normalized size = 1.86 \begin{align*} \frac{15 \,{\left (3 \, A + 4 \, C\right )} a b d x +{\left (12 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, A a b \cos \left (d x + c\right )^{3} + 15 \,{\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right ) + 8 \,{\left (4 \, A + 5 \, C\right )} a^{2} + 20 \,{\left (2 \, A + 3 \, C\right )} b^{2} + 4 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, 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 = 1.19396, size = 672, normalized size = 4.17 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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