Optimal. Leaf size=125 \[ \frac{(4 A-29 B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{B \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{(A-B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(2 A-7 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.315356, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4019, 4008, 3998, 3770, 3794} \[ \frac{(4 A-29 B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{B \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{(A-B) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(2 A-7 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4019
Rule 4008
Rule 3998
Rule 3770
Rule 3794
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^2(c+d x) (2 a (A-B)+5 a B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(2 A-7 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) \left (-2 a^2 (2 A-7 B)-15 a^2 B \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(2 A-7 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(4 A-29 B) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}+\frac{B \int \sec (c+d x) \, dx}{a^3}\\ &=\frac{B \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{(A-B) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(2 A-7 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(4 A-29 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.893983, size = 197, normalized size = 1.58 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (5 (4 A-29 B) \sin \left (\frac{d x}{2}\right )+10 A \sin \left (c+\frac{3 d x}{2}\right )+2 A \sin \left (2 c+\frac{5 d x}{2}\right )+75 B \sin \left (c+\frac{d x}{2}\right )-95 B \sin \left (c+\frac{3 d x}{2}\right )+15 B \sin \left (2 c+\frac{3 d x}{2}\right )-22 B \sin \left (2 c+\frac{5 d x}{2}\right )\right )-240 B \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.054, size = 159, normalized size = 1.3 \begin{align*} -{\frac{B}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{6\,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{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}{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 ) }+{\frac{B}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.03031, size = 252, normalized size = 2.02 \begin{align*} -\frac{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{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \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}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.486405, size = 481, normalized size = 3.85 \begin{align*} \frac{15 \,{\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 32 \, B\right )} \sin \left (d x + c\right )}{30 \,{\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 )}}{\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 ^{4}{\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 = 1.38604, size = 198, normalized size = 1.58 \begin{align*} \frac{\frac{60 \, B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\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} + 10 \, 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} + 15 \, 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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