Optimal. Leaf size=214 \[ \frac{\left (2 a^2 C+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right )-\frac{2 a b (5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac{2 a b (5 A+4 C) \sin (c+d x)}{5 d}+\frac{a b C \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac{C \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))^2}{6 d} \]
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Rubi [A] time = 0.489375, antiderivative size = 214, 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 = {3050, 3033, 3023, 2748, 2635, 8, 2633} \[ \frac{\left (2 a^2 C+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right )-\frac{2 a b (5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac{2 a b (5 A+4 C) \sin (c+d x)}{5 d}+\frac{a b C \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac{C \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))^2}{6 d} \]
Antiderivative was successfully verified.
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Rule 3050
Rule 3033
Rule 3023
Rule 2748
Rule 2635
Rule 8
Rule 2633
Rubi steps
\begin{align*} \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (3 a (2 A+C)+b (6 A+5 C) \cos (c+d x)+2 a C \cos ^2(c+d x)\right ) \, dx\\ &=\frac{a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{30} \int \cos ^2(c+d x) \left (15 a^2 (2 A+C)+12 a b (5 A+4 C) \cos (c+d x)+5 \left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{120} \int \cos ^2(c+d x) \left (15 \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right )+48 a b (5 A+4 C) \cos (c+d x)\right ) \, dx\\ &=\frac{\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{5} (2 a b (5 A+4 C)) \int \cos ^3(c+d x) \, dx+\frac{1}{8} \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{\left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{16} \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \int 1 \, dx-\frac{(2 a b (5 A+4 C)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{1}{16} \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) x+\frac{2 a b (5 A+4 C) \sin (c+d x)}{5 d}+\frac{\left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (2 a^2 C+b^2 (6 A+5 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a b C \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{C \cos ^3(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}-\frac{2 a b (5 A+4 C) \sin ^3(c+d x)}{15 d}\\ \end{align*}
Mathematica [A] time = 0.7054, size = 160, normalized size = 0.75 \[ \frac{60 (c+d x) \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right )+15 \left (16 a^2 (A+C)+b^2 (16 A+15 C)\right ) \sin (2 (c+d x))+15 \left (2 a^2 C+2 A b^2+3 b^2 C\right ) \sin (4 (c+d x))+240 a b (6 A+5 C) \sin (c+d x)+40 a b (4 A+5 C) \sin (3 (c+d x))+24 a b C \sin (5 (c+d x))+5 b^2 C \sin (6 (c+d x))}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.02, size = 209, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( A{b}^{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 ) +{b}^{2}C \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{2\,aAb \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{2\,abC\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 ) }+A{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +{a}^{2}C \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 ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.01576, size = 273, normalized size = 1.28 \begin{align*} \frac{240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A 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 )} C a^{2} - 640 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 128 \,{\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 )} C a b + 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 b^{2} - 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 )} C b^{2}}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.49532, size = 383, normalized size = 1.79 \begin{align*} \frac{15 \,{\left (2 \,{\left (4 \, A + 3 \, C\right )} a^{2} +{\left (6 \, A + 5 \, C\right )} b^{2}\right )} d x +{\left (40 \, C b^{2} \cos \left (d x + c\right )^{5} + 96 \, C a b \cos \left (d x + c\right )^{4} + 32 \,{\left (5 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{2} + 10 \,{\left (6 \, C a^{2} +{\left (6 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} + 64 \,{\left (5 \, A + 4 \, C\right )} a b + 15 \,{\left (2 \,{\left (4 \, A + 3 \, C\right )} a^{2} +{\left (6 \, A + 5 \, C\right )} b^{2}\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 [A] time = 6.52721, size = 592, normalized size = 2.77 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.41274, size = 247, normalized size = 1.15 \begin{align*} \frac{C b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{C a b \sin \left (5 \, d x + 5 \, c\right )}{40 \, d} + \frac{1}{16} \,{\left (8 \, A a^{2} + 6 \, C a^{2} + 6 \, A b^{2} + 5 \, C b^{2}\right )} x + \frac{{\left (2 \, C a^{2} + 2 \, A b^{2} + 3 \, C b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (4 \, A a b + 5 \, C a b\right )} \sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac{{\left (16 \, A a^{2} + 16 \, C a^{2} + 16 \, A b^{2} + 15 \, C b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{{\left (6 \, A a b + 5 \, C a b\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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