Optimal. Leaf size=252 \[ \frac{\left (52 a^2 b^2 C+15 a^3 b B-3 a^4 C+60 a b^3 B+16 b^4 C\right ) \tan (c+d x)}{30 b d}+\frac{\left (12 a^2 b B+4 a^3 C+9 a b^2 C+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (-3 a^2 C+15 a b B+16 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac{\left (30 a^2 b B-6 a^3 C+71 a b^2 C+45 b^3 B\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(5 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]
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Rubi [A] time = 0.498703, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {4072, 4010, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{\left (52 a^2 b^2 C+15 a^3 b B-3 a^4 C+60 a b^3 B+16 b^4 C\right ) \tan (c+d x)}{30 b d}+\frac{\left (12 a^2 b B+4 a^3 C+9 a b^2 C+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (-3 a^2 C+15 a b B+16 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac{\left (30 a^2 b B-6 a^3 C+71 a b^2 C+45 b^3 B\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(5 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4010
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 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 (4 b C+(5 b B-a C) \sec (c+d x)) \, dx}{5 b}\\ &=\frac{(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (b (15 b B+13 a C)+\left (15 a b B-3 a^2 C+16 b^2 C\right ) \sec (c+d x)\right ) \, dx}{20 b}\\ &=\frac{\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (75 a b B+33 a^2 C+32 b^2 C\right )+\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x)\right ) \, dx}{60 b}\\ &=\frac{\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) \left (15 b \left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right )+4 \left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=\frac{\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{1}{8} \left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right ) \int \sec (c+d x) \, dx+\frac{\left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \int \sec ^2(c+d x) \, dx}{30 b}\\ &=\frac{\left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}-\frac{\left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b d}\\ &=\frac{\left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \tan (c+d x)}{30 b d}+\frac{\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}\\ \end{align*}
Mathematica [A] time = 3.29128, size = 181, normalized size = 0.72 \[ \frac{15 \left (12 a^2 b B+4 a^3 C+9 a b^2 C+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 b \left (3 a^2 C+3 a b B+2 b^2 C\right ) \tan ^2(c+d x)+15 \left (3 a^2 b C+a^3 B+3 a b^2 B+b^3 C\right )+3 b^3 C \tan ^4(c+d x)\right )+15 \left (12 a^2 b B+4 a^3 C+9 a b^2 C+3 b^3 B\right ) \sec (c+d x)+30 b^2 (3 a C+b B) \sec ^3(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.043, size = 382, normalized size = 1.5 \begin{align*}{\frac{B{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,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}}+2\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}bC\tan \left ( dx+c \right ) \left ( \sec \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{3\,Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,Ca{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,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}}+{\frac{8\,C{b}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,C{b}^{3}\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 = 0.984851, size = 460, normalized size = 1.83 \begin{align*} \frac{240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{2} + 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^{3} - 45 \, C 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 )} - 15 \, 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 )} - 60 \, C a^{3}{\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 )} - 180 \, 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 )} + 240 \, B a^{3} \tan \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.561328, size = 612, normalized size = 2.43 \begin{align*} \frac{15 \,{\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (15 \, B a^{3} + 30 \, C a^{2} b + 30 \, B a b^{2} + 8 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, C b^{3} + 15 \,{\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (15 \, C a^{2} b + 15 \, B a b^{2} + 4 \, C b^{3}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (3 \, C a b^{2} + B b^{3}\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 (B + C \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{3} \sec ^{2}{\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 = 1.28937, size = 975, 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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