Optimal. Leaf size=147 \[ -\frac{(7 B-27 C) \sin (c+d x)}{15 a^3 d}-\frac{(B-3 C) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (B-3 C)}{a^3}+\frac{(B-C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.51487, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {3029, 2977, 2968, 3023, 12, 2735, 2648} \[ -\frac{(7 B-27 C) \sin (c+d x)}{15 a^3 d}-\frac{(B-3 C) \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{x (B-3 C)}{a^3}+\frac{(B-C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(4 B-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 3029
Rule 2977
Rule 2968
Rule 3023
Rule 12
Rule 2735
Rule 2648
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx &=\int \frac{\cos ^3(c+d x) (B+C \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\\ &=\frac{(B-C) \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 (B-C)-a (B-6 C) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 B-9 C) \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 B-9 C)-a^2 (7 B-27 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 B-9 C) \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 B-9 C) \cos (c+d x)-a^2 (7 B-27 C) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(7 B-27 C) \sin (c+d x)}{15 a^3 d}+\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 B-9 C) \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 (B-3 C) \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5}\\ &=-\frac{(7 B-27 C) \sin (c+d x)}{15 a^3 d}+\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 B-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(B-3 C) \int \frac{\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(B-3 C) x}{a^3}-\frac{(7 B-27 C) \sin (c+d x)}{15 a^3 d}+\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 B-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(B-3 C) \int \frac{1}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=\frac{(B-3 C) x}{a^3}-\frac{(7 B-27 C) \sin (c+d x)}{15 a^3 d}+\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(4 B-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(B-3 C) \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}
Mathematica [B] time = 0.861192, size = 361, normalized size = 2.46 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (300 d x (B-3 C) \cos \left (c+\frac{d x}{2}\right )+540 B \sin \left (c+\frac{d x}{2}\right )-460 B \sin \left (c+\frac{3 d x}{2}\right )+180 B \sin \left (2 c+\frac{3 d x}{2}\right )-128 B \sin \left (2 c+\frac{5 d x}{2}\right )+150 B d x \cos \left (c+\frac{3 d x}{2}\right )+150 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+30 B d x \cos \left (2 c+\frac{5 d x}{2}\right )+30 B d x \cos \left (3 c+\frac{5 d x}{2}\right )+300 d x (B-3 C) \cos \left (\frac{d x}{2}\right )-740 B \sin \left (\frac{d x}{2}\right )-1125 C \sin \left (c+\frac{d x}{2}\right )+1215 C \sin \left (c+\frac{3 d x}{2}\right )-225 C \sin \left (2 c+\frac{3 d x}{2}\right )+363 C \sin \left (2 c+\frac{5 d x}{2}\right )+75 C \sin \left (3 c+\frac{5 d x}{2}\right )+15 C \sin \left (3 c+\frac{7 d x}{2}\right )+15 C \sin \left (4 c+\frac{7 d x}{2}\right )-450 C d x \cos \left (c+\frac{3 d x}{2}\right )-450 C d x \cos \left (2 c+\frac{3 d x}{2}\right )-90 C d x \cos \left (2 c+\frac{5 d x}{2}\right )-90 C d x \cos \left (3 c+\frac{5 d x}{2}\right )+1755 C \sin \left (\frac{d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.031, size = 189, normalized size = 1.3 \begin{align*} -{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{3}}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.96908, size = 312, normalized size = 2.12 \begin{align*} \frac{3 \, C{\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 )} - B{\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 = 1.67054, size = 429, normalized size = 2.92 \begin{align*} \frac{15 \,{\left (B - 3 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (B - 3 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (B - 3 \, C\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (B - 3 \, C\right )} d x +{\left (15 \, C \cos \left (d x + c\right )^{3} -{\left (32 \, B - 117 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (17 \, B - 57 \, C\right )} \cos \left (d x + c\right ) - 22 \, B + 72 \, C\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 = 25.2218, size = 502, normalized size = 3.41 \begin{align*} \begin{cases} \frac{60 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{60 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{17 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{85 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{105 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} - \frac{180 C 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 C d x}{60 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 60 a^{3} d} + \frac{3 C \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 C \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 C \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 C \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 (B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\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 = 1.52184, size = 209, normalized size = 1.42 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )}{\left (B - 3 \, C\right )}}{a^{3}} + \frac{120 \, C \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 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, C 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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