Optimal. Leaf size=165 \[ \frac{\tan ^3(c+d x) (5 a B+5 A b+4 b C)}{15 d}+\frac{\tan (c+d x) (5 a B+5 A b+4 b C)}{5 d}+\frac{(4 a A+3 a C+3 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) (4 a A+3 a C+3 b B)}{8 d}+\frac{(a C+b B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]
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Rubi [A] time = 0.232319, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4076, 4047, 3767, 4046, 3768, 3770} \[ \frac{\tan ^3(c+d x) (5 a B+5 A b+4 b C)}{15 d}+\frac{\tan (c+d x) (5 a B+5 A b+4 b C)}{5 d}+\frac{(4 a A+3 a C+3 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) (4 a A+3 a C+3 b B)}{8 d}+\frac{(a C+b B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]
Antiderivative was successfully verified.
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Rule 4076
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)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^3(c+d x) \left (5 a A+(5 A b+5 a B+4 b C) \sec (c+d x)+5 (b B+a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^3(c+d x) \left (5 a A+5 (b B+a C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{5} (5 A b+5 a B+4 b C) \int \sec ^4(c+d x) \, dx\\ &=\frac{(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} (4 a A+3 b B+3 a C) \int \sec ^3(c+d x) \, dx-\frac{(5 A b+5 a B+4 b C) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{(5 A b+5 a B+4 b C) \tan (c+d x)}{5 d}+\frac{(4 a A+3 b B+3 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{(5 A b+5 a B+4 b C) \tan ^3(c+d x)}{15 d}+\frac{1}{8} (4 a A+3 b B+3 a C) \int \sec (c+d x) \, dx\\ &=\frac{(4 a A+3 b B+3 a C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(5 A b+5 a B+4 b C) \tan (c+d x)}{5 d}+\frac{(4 a A+3 b B+3 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{(5 A b+5 a B+4 b C) \tan ^3(c+d x)}{15 d}\\ \end{align*}
Mathematica [A] time = 1.21854, size = 124, normalized size = 0.75 \[ \frac{15 (4 a A+3 a C+3 b B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \tan ^2(c+d x) (a B+A b+2 b C)+15 (a B+A b+b C)+3 b C \tan ^4(c+d x)\right )+15 \sec (c+d x) (4 a A+3 a C+3 b B)+30 (a C+b B) \sec ^3(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.043, size = 287, normalized size = 1.7 \begin{align*}{\frac{Aa\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{Aa\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,Ba\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Ba\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{aC \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,aC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,Ab\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Ab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{Bb\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,Bb\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{8\,d}}+{\frac{3\,Bb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,Cb\tan \left ( dx+c \right ) }{15\,d}}+{\frac{Cb \left ( \sec \left ( dx+c \right ) \right ) ^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{4\,Cb \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{15\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.05941, size = 359, normalized size = 2.18 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} 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 )} C b - 15 \, C a{\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 \, B 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 )} - 60 \, A a{\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 = 0.555424, size = 467, normalized size = 2.83 \begin{align*} \frac{15 \,{\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\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 +{\left (5 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, B a +{\left (5 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 24 \, C b + 30 \,{\left (C a + B b\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 )} + C \sec ^{2}{\left (c + d x \right )}\right ) \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 = 1.43949, size = 639, normalized size = 3.87 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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