Optimal. Leaf size=225 \[ -\frac{\left (-4 a^2 b^2 (20 A+13 C)+3 a^4 C-4 b^4 (5 A+4 C)\right ) \sin (c+d x)}{30 b d}-\frac{\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}+\frac{a \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac{1}{8} a x \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right )+\frac{C \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}-\frac{a C \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d} \]
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Rubi [A] time = 0.340941, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3024, 2753, 2734} \[ -\frac{\left (-4 a^2 b^2 (20 A+13 C)+3 a^4 C-4 b^4 (5 A+4 C)\right ) \sin (c+d x)}{30 b d}-\frac{\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}+\frac{a \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac{1}{8} a x \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right )+\frac{C \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}-\frac{a C \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d} \]
Antiderivative was successfully verified.
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Rule 3024
Rule 2753
Rule 2734
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac{\int (a+b \cos (c+d x))^3 (b (5 A+4 C)-a C \cos (c+d x)) \, dx}{5 b}\\ &=-\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac{\int (a+b \cos (c+d x))^2 \left (a b (20 A+13 C)-\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \cos (c+d x)\right ) \, dx}{20 b}\\ &=-\frac{\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}-\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac{\int (a+b \cos (c+d x)) \left (b \left (8 b^2 (5 A+4 C)+a^2 (60 A+33 C)\right )+a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \cos (c+d x)\right ) \, dx}{60 b}\\ &=\frac{1}{8} a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) x-\frac{\left (3 a^4 C-4 b^4 (5 A+4 C)-4 a^2 b^2 (20 A+13 C)\right ) \sin (c+d x)}{30 b d}+\frac{a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{120 d}-\frac{\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}-\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}\\ \end{align*}
Mathematica [A] time = 0.682024, size = 160, normalized size = 0.71 \[ \frac{60 a (c+d x) \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right )+60 b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \sin (c+d x)+10 b \left (12 a^2 C+4 A b^2+5 b^2 C\right ) \sin (3 (c+d x))+120 a \left (C \left (a^2+3 b^2\right )+3 A b^2\right ) \sin (2 (c+d x))+45 a b^2 C \sin (4 (c+d x))+6 b^3 C \sin (5 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.018, size = 201, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{C{b}^{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\,Ca{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 ) +{\frac{A{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{2}bC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,aA{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{3}C \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +3\,A{a}^{2}b\sin \left ( dx+c \right ) +A{a}^{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 = 1.01009, size = 262, normalized size = 1.16 \begin{align*} \frac{480 \,{\left (d x + c\right )} A a^{3} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 480 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b + 360 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} + 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 )} C a b^{2} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{3} + 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 )} C b^{3} + 1440 \, A a^{2} b \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 = 1.43832, size = 367, normalized size = 1.63 \begin{align*} \frac{15 \,{\left (4 \,{\left (2 \, A + C\right )} a^{3} + 3 \,{\left (4 \, A + 3 \, C\right )} a b^{2}\right )} d x +{\left (24 \, C b^{3} \cos \left (d x + c\right )^{4} + 90 \, C a b^{2} \cos \left (d x + c\right )^{3} + 120 \,{\left (3 \, A + 2 \, C\right )} a^{2} b + 16 \,{\left (5 \, A + 4 \, C\right )} b^{3} + 8 \,{\left (15 \, C a^{2} b +{\left (5 \, A + 4 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (4 \, C a^{3} + 3 \,{\left (4 \, A + 3 \, C\right )} a 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 [A] time = 3.88849, size = 440, normalized size = 1.96 \begin{align*} \begin{cases} A a^{3} x + \frac{3 A a^{2} b \sin{\left (c + d x \right )}}{d} + \frac{3 A a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 A a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 A a b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 A b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A b^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{C a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{C a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{C a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 C a^{2} b \sin ^{3}{\left (c + d x \right )}}{d} + \frac{3 C a^{2} b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{9 C a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{9 C a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{9 C a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{9 C a b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{15 C a b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{8 C b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 C b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{C b^{3} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19548, size = 235, normalized size = 1.04 \begin{align*} \frac{C b^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{3 \, C a b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{1}{8} \,{\left (8 \, A a^{3} + 4 \, C a^{3} + 12 \, A a b^{2} + 9 \, C a b^{2}\right )} x + \frac{{\left (12 \, C a^{2} b + 4 \, A b^{3} + 5 \, C b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (C a^{3} + 3 \, A a b^{2} + 3 \, C a b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (24 \, A a^{2} b + 18 \, C a^{2} b + 6 \, A b^{3} + 5 \, C b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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