Optimal. Leaf size=147 \[ -\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}-\frac{(A-3 B) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (A-3 B)}{a^3}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(4 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.457031, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2977, 2968, 3023, 12, 2735, 2648} \[ -\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}-\frac{(A-3 B) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (A-3 B)}{a^3}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(4 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2977
Rule 2968
Rule 3023
Rule 12
Rule 2735
Rule 2648
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos ^2(c+d x) (3 a (A-B)-a (A-6 B) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos (c+d x) \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{2 a^2 (4 A-9 B) \cos (c+d x)-a^2 (7 A-27 B) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{15 a^3 (A-3 B) \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5}\\ &=-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(A-3 B) \int \frac{\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(A-3 B) x}{a^3}-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(A-3 B) \int \frac{1}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(A-3 B) x}{a^3}-\frac{(7 A-27 B) \sin (c+d x)}{15 a^3 d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(A-3 B) \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}
Mathematica [B] time = 0.832577, size = 361, normalized size = 2.46 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (300 d x (A-3 B) \cos \left (c+\frac{d x}{2}\right )+300 d x (A-3 B) \cos \left (\frac{d x}{2}\right )+540 A \sin \left (c+\frac{d x}{2}\right )-460 A \sin \left (c+\frac{3 d x}{2}\right )+180 A \sin \left (2 c+\frac{3 d x}{2}\right )-128 A \sin \left (2 c+\frac{5 d x}{2}\right )+150 A d x \cos \left (c+\frac{3 d x}{2}\right )+150 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+30 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+30 A d x \cos \left (3 c+\frac{5 d x}{2}\right )-740 A \sin \left (\frac{d x}{2}\right )-1125 B \sin \left (c+\frac{d x}{2}\right )+1215 B \sin \left (c+\frac{3 d x}{2}\right )-225 B \sin \left (2 c+\frac{3 d x}{2}\right )+363 B \sin \left (2 c+\frac{5 d x}{2}\right )+75 B \sin \left (3 c+\frac{5 d x}{2}\right )+15 B \sin \left (3 c+\frac{7 d x}{2}\right )+15 B \sin \left (4 c+\frac{7 d x}{2}\right )-450 B d x \cos \left (c+\frac{3 d x}{2}\right )-450 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-90 B d x \cos \left (2 c+\frac{5 d x}{2}\right )-90 B d x \cos \left (3 c+\frac{5 d x}{2}\right )+1755 B \sin \left (\frac{d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.065, size = 189, normalized size = 1.3 \begin{align*} -{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{d{a}^{3}}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.56786, size = 312, normalized size = 2.12 \begin{align*} \frac{3 \, B{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - A{\left (\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.36359, size = 429, normalized size = 2.92 \begin{align*} \frac{15 \,{\left (A - 3 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (A - 3 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (A - 3 \, B\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (A - 3 \, B\right )} d x +{\left (15 \, B \cos \left (d x + c\right )^{3} -{\left (32 \, A - 117 \, B\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (17 \, A - 57 \, B\right )} \cos \left (d x + c\right ) - 22 \, A + 72 \, B\right )} \sin \left (d x + c\right )}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 16.5648, size = 496, normalized size = 3.37 \begin{align*} \begin{cases} \frac{60 A d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{60 A d x}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{3 A \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{17 A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{85 A \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{105 A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{180 B d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{180 B d x}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{3 B \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} - \frac{27 B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{225 B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{375 B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \left (A + B \cos{\left (c \right )}\right ) \cos ^{3}{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\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 = 1.2116, size = 209, normalized size = 1.42 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )}{\left (A - 3 \, B\right )}}{a^{3}} + \frac{120 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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