Optimal. Leaf size=306 \[ \frac{a \left (a^2 b^2 (30 A+17 C)+2 a^4 C+24 b^4 (5 A+4 C)\right ) \tan (c+d x)}{60 b^2 d}+\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b^2 d}+\frac{a \left (2 a^2 C+30 A b^2+21 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b^2 d}+\frac{\left (12 a^2 b^2 (5 A+3 C)+4 a^4 C+15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{240 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^4}{15 b^2 d}+\frac{C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^4}{6 b d} \]
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Rubi [A] time = 0.719236, antiderivative size = 306, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4093, 4082, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{a \left (a^2 b^2 (30 A+17 C)+2 a^4 C+24 b^4 (5 A+4 C)\right ) \tan (c+d x)}{60 b^2 d}+\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b^2 d}+\frac{a \left (2 a^2 C+30 A b^2+21 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b^2 d}+\frac{\left (12 a^2 b^2 (5 A+3 C)+4 a^4 C+15 b^4 (6 A+5 C)\right ) \tan (c+d x) \sec (c+d x)}{240 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^4}{15 b^2 d}+\frac{C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^4}{6 b d} \]
Antiderivative was successfully verified.
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Rule 4093
Rule 4082
Rule 4002
Rule 3997
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (a C+b (6 A+5 C) \sec (c+d x)-2 a C \sec ^2(c+d x)\right ) \, dx}{6 b}\\ &=-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (-3 a b C+\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) \sec (c+d x)\right ) \, dx}{30 b^2}\\ &=\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (30 A b^2-2 a^2 C+25 b^2 C\right )+3 a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (-3 a b \left (2 a^2 C-3 b^2 (50 A+39 C)\right )+3 \left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x)\right ) \, dx}{360 b^2}\\ &=\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) \left (45 b^3 \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right )+12 a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right ) \sec (c+d x)\right ) \, dx}{720 b^2}\\ &=\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac{1}{16} \left (b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right )\right ) \int \sec (c+d x) \, dx+\frac{\left (a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right )\right ) \int \sec ^2(c+d x) \, dx}{60 b^2}\\ &=\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}-\frac{\left (a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b^2 d}\\ &=\frac{b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a \left (2 a^4 C+24 b^4 (5 A+4 C)+a^2 b^2 (30 A+17 C)\right ) \tan (c+d x)}{60 b^2 d}+\frac{\left (4 a^4 C+12 a^2 b^2 (5 A+3 C)+15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}+\frac{a \left (30 A b^2+2 a^2 C+21 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac{\left (2 a^2 C+5 b^2 (6 A+5 C)\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}-\frac{a C (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}\\ \end{align*}
Mathematica [A] time = 3.61659, size = 407, normalized size = 1.33 \[ \frac{\sec ^6(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (2 \sin (c+d x) \left (16 a \left (a^2 (75 A+80 C)+24 b^2 (10 A+11 C)\right ) \cos (c+d x)+20 b \left (18 a^2 (4 A+5 C)+5 b^2 (6 A+5 C)\right ) \cos (2 (c+d x))+360 a^2 A b \cos (4 (c+d x))+1080 a^2 A b+600 a^3 A \cos (3 (c+d x))+120 a^3 A \cos (5 (c+d x))+270 a^2 b C \cos (4 (c+d x))+1530 a^2 b C+560 a^3 C \cos (3 (c+d x))+80 a^3 C \cos (5 (c+d x))+1680 a A b^2 \cos (3 (c+d x))+240 a A b^2 \cos (5 (c+d x))+1344 a b^2 C \cos (3 (c+d x))+192 a b^2 C \cos (5 (c+d x))+90 A b^3 \cos (4 (c+d x))+510 A b^3+75 b^3 C \cos (4 (c+d x))+745 b^3 C\right )-240 b \left (6 a^2 (4 A+3 C)+b^2 (6 A+5 C)\right ) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{1920 d (A \cos (2 (c+d x))+A+2 C)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.051, size = 430, normalized size = 1.4 \begin{align*}{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{2\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{3\,A{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{3\,{a}^{2}bC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,{a}^{2}bC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+2\,{\frac{Aa{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{Aa{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{8\,Ca{b}^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{A{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,C{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,C{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.02067, size = 521, normalized size = 1.7 \begin{align*} \frac{160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 480 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} + 96 \,{\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 a b^{2} - 5 \, C b^{3}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, C a^{2} 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 )} - 30 \, 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 )} - 360 \, A 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 )} + 480 \, A a^{3} \tan \left (d x + c\right )}{480 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.576212, size = 636, normalized size = 2.08 \begin{align*} \frac{15 \,{\left (6 \,{\left (4 \, A + 3 \, C\right )} a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (6 \,{\left (4 \, A + 3 \, C\right )} a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (5 \,{\left (3 \, A + 2 \, C\right )} a^{3} + 6 \,{\left (5 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{5} + 144 \, C a b^{2} \cos \left (d x + c\right ) + 15 \,{\left (6 \,{\left (4 \, A + 3 \, C\right )} a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} + 40 \, C b^{3} + 16 \,{\left (5 \, C a^{3} + 3 \,{\left (5 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left (18 \, C a^{2} b +{\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \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 + C \sec ^{2}{\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 = 1.27429, size = 1258, normalized size = 4.11 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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