Optimal. Leaf size=110 \[ \frac{(2 A+3 B+7 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-C) \tan (c+d x)}{3 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.207278, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {4078, 4000, 3794} \[ \frac{(2 A+3 B+7 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-C) \tan (c+d x)}{3 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4078
Rule 4000
Rule 3794
Rubi steps
\begin{align*} \int \frac{\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec (c+d x) (a (4 A+B-C)-a (A-B-4 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-C) \tan (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{(2 A+3 B+7 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-C) \tan (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{(2 A+3 B+7 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.566438, size = 156, normalized size = 1.42 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (-15 (2 A+B) \sin \left (c+\frac{d x}{2}\right )+5 (8 A+3 B+4 C) \sin \left (\frac{d x}{2}\right )+20 A \sin \left (c+\frac{3 d x}{2}\right )-15 A \sin \left (2 c+\frac{3 d x}{2}\right )+7 A \sin \left (2 c+\frac{5 d x}{2}\right )+15 B \sin \left (c+\frac{3 d x}{2}\right )+3 B \sin \left (2 c+\frac{5 d x}{2}\right )+10 C \sin \left (c+\frac{3 d x}{2}\right )+2 C \sin \left (2 c+\frac{5 d x}{2}\right )\right )}{240 a^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 113, normalized size = 1. \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{A}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2\,A}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{2\,C}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+A\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +B\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +C\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.976994, size = 242, normalized size = 2.2 \begin{align*} \frac{\frac{C{\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}} + \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}} + \frac{3 \, B{\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 = 0.459965, size = 251, normalized size = 2.28 \begin{align*} \frac{{\left ({\left (7 \, A + 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A + 3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + 2 \, A + 3 \, B + 7 \, 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{\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 + \int \frac{B \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 + \int \frac{C \sec ^{3}{\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 = 1.27846, size = 155, normalized size = 1.41 \begin{align*} \frac{3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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