Optimal. Leaf size=286 \[ \frac{\tan (c+d x) \left (4 a^2 b^2 (20 A+13 C)+15 a^3 b B-3 a^4 C+60 a b^3 B+4 b^4 (5 A+4 C)\right )}{30 b d}+\frac{\left (4 a^3 (2 A+C)+12 a^2 b B+3 a b^2 (4 A+3 C)+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) \left (30 a^2 b B-6 a^3 C+a b^2 (100 A+71 C)+45 b^3 B\right )}{120 d}+\frac{\tan (c+d x) \left (3 a (5 b B-a C)+4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2}{60 b d}+\frac{(5 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]
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Rubi [A] time = 0.587281, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.18, Rules used = {4082, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{\tan (c+d x) \left (4 a^2 b^2 (20 A+13 C)+15 a^3 b B-3 a^4 C+60 a b^3 B+4 b^4 (5 A+4 C)\right )}{30 b d}+\frac{\left (4 a^3 (2 A+C)+12 a^2 b B+3 a b^2 (4 A+3 C)+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) \left (30 a^2 b B-6 a^3 C+a b^2 (100 A+71 C)+45 b^3 B\right )}{120 d}+\frac{\tan (c+d x) \left (3 a (5 b B-a C)+4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2}{60 b d}+\frac{(5 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]
Antiderivative was successfully verified.
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Rule 4082
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 (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, 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 (b (5 A+4 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 (20 a A+15 b B+13 a C)+\left (4 b^2 (5 A+4 C)+3 a (5 b B-a C)\right ) \sec (c+d x)\right ) \, dx}{20 b}\\ &=\frac{\left (4 b^2 (5 A+4 C)+3 a (5 b B-a 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+8 b^2 (5 A+4 C)+a^2 (60 A+33 C)\right )+\left (30 a^2 b B+45 b^3 B-6 a^3 C+a b^2 (100 A+71 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+a b^2 (100 A+71 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (4 b^2 (5 A+4 C)+3 a (5 b B-a 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 (2 A+C)+3 a b^2 (4 A+3 C)\right )+4 \left (15 a^3 b B+60 a b^3 B-3 a^4 C+4 b^4 (5 A+4 C)+4 a^2 b^2 (20 A+13 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+a b^2 (100 A+71 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (4 b^2 (5 A+4 C)+3 a (5 b B-a 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 (2 A+C)+3 a b^2 (4 A+3 C)\right ) \int \sec (c+d x) \, dx+\frac{\left (15 a^3 b B+60 a b^3 B-3 a^4 C+4 b^4 (5 A+4 C)+4 a^2 b^2 (20 A+13 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 (2 A+C)+3 a b^2 (4 A+3 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+a b^2 (100 A+71 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (4 b^2 (5 A+4 C)+3 a (5 b B-a 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+4 b^4 (5 A+4 C)+4 a^2 b^2 (20 A+13 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 (2 A+C)+3 a b^2 (4 A+3 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+4 b^4 (5 A+4 C)+4 a^2 b^2 (20 A+13 C)\right ) \tan (c+d x)}{30 b d}+\frac{\left (30 a^2 b B+45 b^3 B-6 a^3 C+a b^2 (100 A+71 C)\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (4 b^2 (5 A+4 C)+3 a (5 b B-a 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 = 2.93432, size = 451, normalized size = 1.58 \[ -\frac{\sec ^5(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (120 \cos ^5(c+d x) \left (4 a^3 (2 A+C)+12 a^2 b B+3 a b^2 (4 A+3 C)+3 b^3 B\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-2 \sin (c+d x) \left (15 \cos (c+d x) \left (36 a^2 b B+12 a^3 C+3 a b^2 (12 A+17 C)+17 b^3 B\right )+48 \cos (2 (c+d x)) \left (15 a^2 b (A+C)+5 a^3 B+15 a b^2 B+b^3 (5 A+4 C)\right )+180 a^2 A b \cos (4 (c+d x))+540 a^2 A b+180 a^2 b B \cos (3 (c+d x))+120 a^2 b C \cos (4 (c+d x))+600 a^2 b C+60 a^3 B \cos (4 (c+d x))+180 a^3 B+60 a^3 C \cos (3 (c+d x))+180 a A b^2 \cos (3 (c+d x))+120 a b^2 B \cos (4 (c+d x))+600 a b^2 B+135 a b^2 C \cos (3 (c+d x))+40 A b^3 \cos (4 (c+d x))+200 A b^3+45 b^3 B \cos (3 (c+d x))+32 b^3 C \cos (4 (c+d x))+256 b^3 C\right )\right )}{480 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.061, size = 504, normalized size = 1.8 \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}}+{\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}}+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}}+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}}+{\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{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{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}}+{\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 = 1.05533, size = 601, normalized size = 2.1 \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} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} 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 )} 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 )} - 180 \, 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 )} + 240 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 240 \, B a^{3} \tan \left (d x + c\right ) + 720 \, A a^{2} 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 = 0.589616, size = 711, normalized size = 2.49 \begin{align*} \frac{15 \,{\left (4 \,{\left (2 \, A + C\right )} a^{3} + 12 \, B a^{2} b + 3 \,{\left (4 \, A + 3 \, C\right )} 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 \,{\left (2 \, A + C\right )} a^{3} + 12 \, B a^{2} b + 3 \,{\left (4 \, A + 3 \, C\right )} 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} + 15 \,{\left (3 \, A + 2 \, C\right )} a^{2} b + 30 \, B a b^{2} + 2 \,{\left (5 \, A + 4 \, C\right )} 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 + 3 \,{\left (4 \, A + 3 \, C\right )} 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} +{\left (5 \, A + 4 \, C\right )} 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 (a + b \sec{\left (c + d x \right )}\right )^{3} \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \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 = 1.26946, size = 1335, normalized size = 4.67 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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