Optimal. Leaf size=198 \[ \frac{\left (4 a^2 A+6 a b B+3 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan ^3(c+d x)}{15 d}+\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan (c+d x)}{5 d}+\frac{b (6 a B+5 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{b B \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))}{5 d} \]
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Rubi [A] time = 0.290915, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4026, 4047, 3767, 4046, 3768, 3770} \[ \frac{\left (4 a^2 A+6 a b B+3 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan ^3(c+d x)}{15 d}+\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan (c+d x)}{5 d}+\frac{b (6 a B+5 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{b B \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))}{5 d} \]
Antiderivative was successfully verified.
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Rule 4026
Rule 4047
Rule 3767
Rule 4046
Rule 3768
Rule 3770
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac{b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^3(c+d x) \left (a (5 a A+3 b B)+\left (4 b^2 B+5 a (2 A b+a B)\right ) \sec (c+d x)+b (5 A b+6 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^3(c+d x) \left (a (5 a A+3 b B)+b (5 A b+6 a B) \sec ^2(c+d x)\right ) \, dx+\frac{1}{5} \left (4 b^2 B+5 a (2 A b+a B)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{b (5 A b+6 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{1}{4} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int \sec ^3(c+d x) \, dx-\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan (c+d x)}{5 d}+\frac{\left (4 a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b (5 A b+6 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan ^3(c+d x)}{15 d}+\frac{1}{8} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (4 a^2 A+3 A b^2+6 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan (c+d x)}{5 d}+\frac{\left (4 a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b (5 A b+6 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan ^3(c+d x)}{15 d}\\ \end{align*}
Mathematica [A] time = 1.51414, size = 150, normalized size = 0.76 \[ \frac{15 \left (4 a^2 A+6 a b B+3 A b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \left (a^2 B+2 a A b+2 b^2 B\right ) \tan ^2(c+d x)+15 \left (a^2 B+2 a A b+b^2 B\right )+3 b^2 B \tan ^4(c+d x)\right )+15 \left (4 a^2 A+6 a b B+3 A b^2\right ) \sec (c+d x)+30 b (2 a B+A b) \sec ^3(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.042, size = 312, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,B{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{4\,Aab\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,Aab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{Bab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,Bab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{A{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,B{b}^{2}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{B{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,B{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.979049, size = 373, normalized size = 1.88 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b + 16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B b^{2} - 30 \, B a b{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, A b^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.526381, size = 521, normalized size = 2.63 \begin{align*} \frac{15 \,{\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (5 \, B a^{2} + 10 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, B b^{2} + 8 \,{\left (5 \, B a^{2} + 10 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \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*} \int \left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.25681, size = 713, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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