Optimal. Leaf size=98 \[ \frac{2 (5 A-2 B) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-B) \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{x (2 A-B)}{a^2}-\frac{(A-B) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.23049, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {4020, 3787, 2637, 8} \[ \frac{2 (5 A-2 B) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-B) \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{x (2 A-B)}{a^2}-\frac{(A-B) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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))^2} \, dx &=-\frac{(A-B) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\cos (c+d x) (a (4 A-B)-2 a (A-B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(2 A-B) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \cos (c+d x) \left (2 a^2 (5 A-2 B)-3 a^2 (2 A-B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(2 A-B) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(2 (5 A-2 B)) \int \cos (c+d x) \, dx}{3 a^2}-\frac{(2 A-B) \int 1 \, dx}{a^2}\\ &=-\frac{(2 A-B) x}{a^2}+\frac{2 (5 A-2 B) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-B) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}
Mathematica [B] time = 0.586479, size = 245, normalized size = 2.5 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-18 d x (2 A-B) \cos \left (c+\frac{d x}{2}\right )-18 d x (2 A-B) \cos \left (\frac{d x}{2}\right )-30 A \sin \left (c+\frac{d x}{2}\right )+41 A \sin \left (c+\frac{3 d x}{2}\right )+9 A \sin \left (2 c+\frac{3 d x}{2}\right )+3 A \sin \left (2 c+\frac{5 d x}{2}\right )+3 A \sin \left (3 c+\frac{5 d x}{2}\right )-12 A d x \cos \left (c+\frac{3 d x}{2}\right )-12 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+66 A \sin \left (\frac{d x}{2}\right )+24 B \sin \left (c+\frac{d x}{2}\right )-20 B \sin \left (c+\frac{3 d x}{2}\right )+6 B d x \cos \left (c+\frac{3 d x}{2}\right )+6 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-36 B \sin \left (\frac{d x}{2}\right )\right )}{12 a^2 d (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.082, size = 149, normalized size = 1.5 \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{5\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.49117, size = 258, normalized size = 2.63 \begin{align*} \frac{A{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac{a^{2} \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 )}}\right )} - B{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.464818, size = 296, normalized size = 3.02 \begin{align*} -\frac{3 \,{\left (2 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (2 \, A - B\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (2 \, A - B\right )} d x -{\left (3 \, A \cos \left (d x + c\right )^{2} +{\left (14 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 10 \, A - 4 \, B\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \cos{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{B \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.20842, size = 163, normalized size = 1.66 \begin{align*} -\frac{\frac{6 \,{\left (d x + c\right )}{\left (2 \, A - B\right )}}{a^{2}} - \frac{12 \, 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^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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