Optimal. Leaf size=116 \[ \frac{(4 A-29 B) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{B x}{a^3}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(2 A-7 B) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.321075, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2977, 2968, 3019, 2735, 2648} \[ \frac{(4 A-29 B) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{B x}{a^3}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(2 A-7 B) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2977
Rule 2968
Rule 3019
Rule 2735
Rule 2648
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos (c+d x) (2 a (A-B)+5 a B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{2 a (A-B) \cos (c+d x)+5 a B \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(2 A-7 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{\int \frac{-2 a^2 (2 A-7 B)-15 a^2 B \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{B x}{a^3}+\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(2 A-7 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(4 A-29 B) \int \frac{1}{a+a \cos (c+d x)} \, dx}{15 a^2}\\ &=\frac{B x}{a^3}+\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(2 A-7 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(4 A-29 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}
Mathematica [B] time = 0.5373, size = 241, normalized size = 2.08 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (-60 A \sin \left (c+\frac{d x}{2}\right )+40 A \sin \left (c+\frac{3 d x}{2}\right )-30 A \sin \left (2 c+\frac{3 d x}{2}\right )+14 A \sin \left (2 c+\frac{5 d x}{2}\right )+80 A \sin \left (\frac{d x}{2}\right )+270 B \sin \left (c+\frac{d x}{2}\right )-230 B \sin \left (c+\frac{3 d x}{2}\right )+90 B \sin \left (2 c+\frac{3 d x}{2}\right )-64 B \sin \left (2 c+\frac{5 d x}{2}\right )+150 B d x \cos \left (c+\frac{d x}{2}\right )+75 B d x \cos \left (c+\frac{3 d x}{2}\right )+75 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+15 B d x \cos \left (2 c+\frac{5 d x}{2}\right )+15 B d x \cos \left (3 c+\frac{5 d x}{2}\right )-370 B \sin \left (\frac{d x}{2}\right )+150 B d x \cos \left (\frac{d x}{2}\right )\right )}{480 a^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.056, size = 137, normalized size = 1.2 \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}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\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 = 1.48969, size = 216, normalized size = 1.86 \begin{align*} -\frac{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 )} - \frac{A{\left (\frac{15 \, \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{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.37787, size = 351, normalized size = 3.03 \begin{align*} \frac{15 \, B d x \cos \left (d x + c\right )^{3} + 45 \, B d x \cos \left (d x + c\right )^{2} + 45 \, B d x \cos \left (d x + c\right ) + 15 \, B d x +{\left ({\left (7 \, A - 32 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 22 \, 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 = 9.22754, size = 148, normalized size = 1.28 \begin{align*} \begin{cases} \frac{A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} - \frac{A \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{3} d} + \frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} + \frac{B x}{a^{3}} - \frac{B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} + \frac{B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{3} d} - \frac{7 B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \left (A + B \cos{\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 = 1.23817, size = 162, normalized size = 1.4 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )} B}{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} - 10 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, 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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