Optimal. Leaf size=136 \[ \frac{2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac{(3 A-B) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{x (3 A-B)}{a^3}-\frac{(9 A-4 B) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rubi [A] time = 0.366747, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {4020, 3787, 2637, 8} \[ \frac{2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac{(3 A-B) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{x (3 A-B)}{a^3}-\frac{(9 A-4 B) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 4020
Rule 3787
Rule 2637
Rule 8
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=-\frac{(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\cos (c+d x) (a (6 A-B)-3 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\cos (c+d x) \left (a^2 (27 A-7 B)-2 a^2 (9 A-4 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \cos (c+d x) \left (2 a^3 (36 A-11 B)-15 a^3 (3 A-B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(2 (36 A-11 B)) \int \cos (c+d x) \, dx}{15 a^3}-\frac{(3 A-B) \int 1 \, dx}{a^3}\\ &=-\frac{(3 A-B) x}{a^3}+\frac{2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac{(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [B] time = 1.0293, size = 365, normalized size = 2.68 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-300 d x (3 A-B) \cos \left (c+\frac{d x}{2}\right )-300 d x (3 A-B) \cos \left (\frac{d x}{2}\right )-1125 A \sin \left (c+\frac{d x}{2}\right )+1215 A \sin \left (c+\frac{3 d x}{2}\right )-225 A \sin \left (2 c+\frac{3 d x}{2}\right )+363 A \sin \left (2 c+\frac{5 d x}{2}\right )+75 A \sin \left (3 c+\frac{5 d x}{2}\right )+15 A \sin \left (3 c+\frac{7 d x}{2}\right )+15 A \sin \left (4 c+\frac{7 d x}{2}\right )-450 A d x \cos \left (c+\frac{3 d x}{2}\right )-450 A d x \cos \left (2 c+\frac{3 d x}{2}\right )-90 A d x \cos \left (2 c+\frac{5 d x}{2}\right )-90 A d x \cos \left (3 c+\frac{5 d x}{2}\right )+1755 A \sin \left (\frac{d x}{2}\right )+540 B \sin \left (c+\frac{d x}{2}\right )-460 B \sin \left (c+\frac{3 d x}{2}\right )+180 B \sin \left (2 c+\frac{3 d x}{2}\right )-128 B \sin \left (2 c+\frac{5 d x}{2}\right )+150 B d x \cos \left (c+\frac{3 d x}{2}\right )+150 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+30 B d x \cos \left (2 c+\frac{5 d x}{2}\right )+30 B d x \cos \left (3 c+\frac{5 d x}{2}\right )-740 B \sin \left (\frac{d x}{2}\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.093, size = 189, normalized size = 1.4 \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}{2\,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{17\,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{A\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 ) }}-6\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}+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 = 1.48971, size = 312, normalized size = 2.29 \begin{align*} \frac{3 \, A{\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 = 0.475016, size = 431, normalized size = 3.17 \begin{align*} -\frac{15 \,{\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (3 \, A - B\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (3 \, A - B\right )} d x -{\left (15 \, A \cos \left (d x + c\right )^{3} +{\left (117 \, A - 32 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (57 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 72 \, 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 [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.42345, size = 212, normalized size = 1.56 \begin{align*} -\frac{\frac{60 \,{\left (d x + c\right )}{\left (3 \, A - B\right )}}{a^{3}} - \frac{120 \, A \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} - 30 \, 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} + 255 \, 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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