Optimal. Leaf size=83 \[ \frac{2 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{2 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rubi [A] time = 0.080896, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3796, 3794} \[ \frac{2 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{2 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 3796
Rule 3794
Rubi steps
\begin{align*} \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\frac{\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{2 \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a}\\ &=\frac{\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{2 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{2 \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac{\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{2 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{2 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.2189, size = 86, normalized size = 1.04 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-30 \sin \left (c+\frac{d x}{2}\right )+20 \sin \left (c+\frac{3 d x}{2}\right )-15 \sin \left (2 c+\frac{3 d x}{2}\right )+7 \sin \left (2 c+\frac{5 d x}{2}\right )+40 \sin \left (\frac{d x}{2}\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{240 a^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.039, size = 45, normalized size = 0.5 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+\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 = 1.11988, size = 90, normalized size = 1.08 \begin{align*} \frac{\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}}}{60 \, a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.60977, size = 186, normalized size = 2.24 \begin{align*} \frac{{\left (7 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2\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{\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}{a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.34133, size = 62, normalized size = 0.75 \begin{align*} \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, \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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