Optimal. Leaf size=250 \[ \frac{\left (95 a^3 A b+112 a^2 b^2 B+12 a^4 B+80 a A b^3+16 b^4 B\right ) \tan (c+d x)}{30 d}+\frac{\left (24 a^2 A b^2+8 a^4 A+16 a^3 b B+12 a b^3 B+3 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (12 a^2 B+35 a A b+16 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 d}+\frac{b \left (130 a^2 A b+24 a^3 B+116 a b^2 B+45 A b^3\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(4 a B+5 A b) \tan (c+d x) (a+b \sec (c+d x))^3}{20 d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^4}{5 d} \]
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Rubi [A] time = 0.519875, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{\left (95 a^3 A b+112 a^2 b^2 B+12 a^4 B+80 a A b^3+16 b^4 B\right ) \tan (c+d x)}{30 d}+\frac{\left (24 a^2 A b^2+8 a^4 A+16 a^3 b B+12 a b^3 B+3 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (12 a^2 B+35 a A b+16 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 d}+\frac{b \left (130 a^2 A b+24 a^3 B+116 a b^2 B+45 A b^3\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(4 a B+5 A b) \tan (c+d x) (a+b \sec (c+d x))^3}{20 d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^4}{5 d} \]
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))^4 (A+B \sec (c+d x)) \, dx &=\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec (c+d x) (a+b \sec (c+d x))^3 (5 a A+4 b B+(5 A b+4 a B) \sec (c+d x)) \, dx\\ &=\frac{(5 A b+4 a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{20} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (20 a^2 A+15 A b^2+28 a b B+\left (35 a A b+12 a^2 B+16 b^2 B\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{\left (35 a A b+12 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 A b+4 a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{60} \int \sec (c+d x) (a+b \sec (c+d x)) \left (60 a^3 A+115 a A b^2+108 a^2 b B+32 b^3 B+\left (130 a^2 A b+45 A b^3+24 a^3 B+116 a b^2 B\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b \left (130 a^2 A b+45 A b^3+24 a^3 B+116 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (35 a A b+12 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 A b+4 a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{120} \int \sec (c+d x) \left (15 \left (8 a^4 A+24 a^2 A b^2+3 A b^4+16 a^3 b B+12 a b^3 B\right )+4 \left (95 a^3 A b+80 a A b^3+12 a^4 B+112 a^2 b^2 B+16 b^4 B\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b \left (130 a^2 A b+45 A b^3+24 a^3 B+116 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (35 a A b+12 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 A b+4 a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{8} \left (8 a^4 A+24 a^2 A b^2+3 A b^4+16 a^3 b B+12 a b^3 B\right ) \int \sec (c+d x) \, dx+\frac{1}{30} \left (95 a^3 A b+80 a A b^3+12 a^4 B+112 a^2 b^2 B+16 b^4 B\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{\left (8 a^4 A+24 a^2 A b^2+3 A b^4+16 a^3 b B+12 a b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (130 a^2 A b+45 A b^3+24 a^3 B+116 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (35 a A b+12 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 A b+4 a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}-\frac{\left (95 a^3 A b+80 a A b^3+12 a^4 B+112 a^2 b^2 B+16 b^4 B\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 d}\\ &=\frac{\left (8 a^4 A+24 a^2 A b^2+3 A b^4+16 a^3 b B+12 a b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (95 a^3 A b+80 a A b^3+12 a^4 B+112 a^2 b^2 B+16 b^4 B\right ) \tan (c+d x)}{30 d}+\frac{b \left (130 a^2 A b+45 A b^3+24 a^3 B+116 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (35 a A b+12 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{(5 A b+4 a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 3.91443, size = 198, normalized size = 0.79 \[ \frac{15 \left (24 a^2 A b^2+8 a^4 A+16 a^3 b B+12 a b^3 B+3 A b^4\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (80 b^2 \left (3 a^2 B+2 a A b+b^2 B\right ) \tan ^2(c+d x)+15 b \left (24 a^2 A b+16 a^3 B+12 a b^2 B+3 A b^3\right ) \sec (c+d x)+120 \left (4 a^3 A b+6 a^2 b^2 B+a^4 B+4 a A b^3+b^4 B\right )+30 b^3 (4 a B+A b) \sec ^3(c+d x)+24 b^4 B \tan ^4(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.049, size = 431, normalized size = 1.7 \begin{align*}{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{a}^{4}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{A{a}^{3}b\tan \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{A{a}^{2}{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+3\,{\frac{A{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{B{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{8\,Aa{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,Aa{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{Ba{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,Ba{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ba{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,B{b}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{B{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,B{b}^{4}\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 = 0.983032, size = 512, normalized size = 2.05 \begin{align*} \frac{480 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b^{2} + 320 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{3} + 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^{4} - 60 \, B a 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 )} - 15 \, A b^{4}{\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 )} - 240 \, B a^{3} 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 )} - 360 \, A a^{2} 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 )} + 240 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 240 \, B a^{4} \tan \left (d x + c\right ) + 960 \, A a^{3} b \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 = 0.593967, size = 687, normalized size = 2.75 \begin{align*} \frac{15 \,{\left (8 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 12 \, B a b^{3} + 3 \, A b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (8 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 12 \, B a b^{3} + 3 \, A b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, B b^{4} + 8 \,{\left (15 \, B a^{4} + 60 \, A a^{3} b + 60 \, B a^{2} b^{2} + 40 \, A a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 12 \, B a b^{3} + 3 \, A b^{4}\right )} \cos \left (d x + c\right )^{3} + 16 \,{\left (15 \, B a^{2} b^{2} + 10 \, A a b^{3} + 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (4 \, B a b^{3} + A b^{4}\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 )^{4} \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 = 1.27717, size = 1148, normalized size = 4.59 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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