Optimal. Leaf size=106 \[ -\frac{(22 A-3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{A x}{a^3}-\frac{(7 A-3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rubi [A] time = 0.180472, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4053, 3922, 3919, 3794} \[ -\frac{(22 A-3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{A x}{a^3}-\frac{(7 A-3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 4053
Rule 3922
Rule 3919
Rule 3794
Rubi steps
\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac{(A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\int \frac{-5 a A+a (2 A-3 C) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(7 A-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{15 a^2 A-a^2 (7 A-3 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{A x}{a^3}-\frac{(A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(7 A-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(22 A-3 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac{A x}{a^3}-\frac{(A+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(7 A-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(22 A-3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [B] time = 0.844389, size = 227, normalized size = 2.14 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (270 A \sin \left (c+\frac{d x}{2}\right )-230 A \sin \left (c+\frac{3 d x}{2}\right )+90 A \sin \left (2 c+\frac{3 d x}{2}\right )-64 A \sin \left (2 c+\frac{5 d x}{2}\right )+150 A d x \cos \left (c+\frac{d x}{2}\right )+75 A d x \cos \left (c+\frac{3 d x}{2}\right )+75 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+15 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+15 A d x \cos \left (3 c+\frac{5 d x}{2}\right )-370 A \sin \left (\frac{d x}{2}\right )+150 A d x \cos \left (\frac{d x}{2}\right )-30 C \sin \left (c+\frac{d x}{2}\right )+30 C \sin \left (c+\frac{3 d x}{2}\right )+6 C \sin \left (2 c+\frac{5 d x}{2}\right )+30 C \sin \left (\frac{d x}{2}\right )\right )}{480 a^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 117, normalized size = 1.1 \begin{align*} -{\frac{A}{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{A}{3\,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{C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.42385, size = 189, normalized size = 1.78 \begin{align*} -\frac{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 )} - \frac{3 \, C{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\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 = 0.476095, size = 351, normalized size = 3.31 \begin{align*} \frac{15 \, A d x \cos \left (d x + c\right )^{3} + 45 \, A d x \cos \left (d x + c\right )^{2} + 45 \, A d x \cos \left (d x + c\right ) + 15 \, A d x -{\left ({\left (32 \, A - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (17 \, A - 3 \, C\right )} \cos \left (d x + c\right ) + 22 \, A - 3 \, 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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.21603, size = 140, normalized size = 1.32 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )} A}{a^{3}} - \frac{3 \, A 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 \, A 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 ) - 15 \, 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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