Optimal. Leaf size=180 \[ \frac{\left (16 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 d}+\frac{\left (8 a^3 A+12 a^2 b B+12 a A b^2+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (6 a^2 B+20 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(3 a B+4 A b) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
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Rubi [A] time = 0.333184, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4002, 3997, 3787, 3770, 3767, 8} \[ \frac{\left (16 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 d}+\frac{\left (8 a^3 A+12 a^2 b B+12 a A b^2+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (6 a^2 B+20 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(3 a B+4 A b) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 4002
Rule 3997
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac{B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \int \sec (c+d x) (a+b \sec (c+d x))^2 (4 a A+3 b B+(4 A b+3 a B) \sec (c+d x)) \, dx\\ &=\frac{(4 A b+3 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{12} \int \sec (c+d x) (a+b \sec (c+d x)) \left (12 a^2 A+8 A b^2+15 a b B+\left (20 a A b+6 a^2 B+9 b^2 B\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b \left (20 a A b+6 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 A b+3 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{24} \int \sec (c+d x) \left (3 \left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right )+4 \left (16 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b \left (20 a A b+6 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 A b+3 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{6} \left (16 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (20 a A b+6 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 A b+3 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (16 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{\left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (16 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac{b \left (20 a A b+6 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 A b+3 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.949985, size = 140, normalized size = 0.78 \[ \frac{3 \left (8 a^3 A+12 a^2 b B+12 a A b^2+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (9 b \left (4 a^2 B+4 a A b+b^2 B\right ) \sec (c+d x)+24 \left (3 a^2 A b+a^3 B+3 a b^2 B+A b^3\right )+8 b^2 (3 a B+A b) \tan ^2(c+d x)+6 b^3 B \sec ^3(c+d x)\right )}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.041, size = 290, normalized size = 1.6 \begin{align*}{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{a}^{3}\tan \left ( dx+c \right ) }{d}}+3\,{\frac{A{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{3\,B{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,B{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{3\,Aa{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Aa{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{Ba{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{Ba{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{2\,A{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{B{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,B{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,B{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.989862, size = 359, normalized size = 1.99 \begin{align*} \frac{48 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{2} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{3} - 3 \, B b^{3}{\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 )} - 36 \, B a^{2} b{\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 )} - 36 \, A a b^{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 )} + 48 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, B a^{3} \tan \left (d x + c\right ) + 144 \, A a^{2} b \tan \left (d x + c\right )}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.52755, size = 510, normalized size = 2.83 \begin{align*} \frac{3 \,{\left (8 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, B b^{3} + 8 \,{\left (3 \, B a^{3} + 9 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \,{\left (4 \, B a^{2} b + 4 \, A a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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 )^{3} \sec{\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 = 1.25369, size = 791, normalized size = 4.39 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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