Optimal. Leaf size=233 \[ \frac{\tan (c+d x) \left (5 a^2 (3 A+2 C)+20 a b B+2 b^2 (5 A+4 C)\right )}{15 d}+\frac{\left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^2(c+d x) \left (2 a^2 C+10 a b B+5 A b^2+4 b^2 C\right )}{15 d}+\frac{\tan (c+d x) \sec (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )}{8 d}+\frac{b (2 a C+5 b B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a+b \sec (c+d x))^2}{5 d} \]
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Rubi [A] time = 0.59159, antiderivative size = 281, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {4092, 4082, 4002, 3997, 3787, 3770, 3767, 8} \[ -\frac{\tan (c+d x) \left (-4 a^2 b^2 (5 A+3 C)+5 a^3 b B-2 a^4 C-40 a b^3 B-4 b^4 (5 A+4 C)\right )}{30 b^2 d}+\frac{\left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \left (2 a^2 C-5 a b B+20 A b^2+16 b^2 C\right ) (a+b \sec (c+d x))^2}{60 b^2 d}-\frac{\tan (c+d x) \sec (c+d x) \left (10 a^2 b B-4 a^3 C-2 a b^2 (20 A+13 C)-45 b^3 B\right )}{120 b d}+\frac{(5 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b^2 d}+\frac{C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]
Antiderivative was successfully verified.
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Rule 4092
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))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a C+b (5 A+4 C) \sec (c+d x)+(5 b B-2 a C) \sec ^2(c+d x)\right ) \, dx}{5 b}\\ &=\frac{(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \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-2 a C)+\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) \sec (c+d x)\right ) \, dx}{20 b^2}\\ &=\frac{\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac{(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (40 A b^2+35 a b B-2 a^2 C+32 b^2 C\right )-\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x)\right ) \, dx}{60 b^2}\\ &=-\frac{\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac{\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac{(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) \left (15 b^2 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right )-4 \left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=-\frac{\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac{\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac{(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{1}{8} \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \int \sec (c+d x) \, dx-\frac{\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x) \, dx}{30 b^2}\\ &=\frac{\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac{\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac{(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b^2 d}\\ &=\frac{\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \tan (c+d x)}{30 b^2 d}-\frac{\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac{\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac{(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}\\ \end{align*}
Mathematica [A] time = 2.1341, size = 371, normalized size = 1.59 \[ \frac{\sec ^5(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (2 \sin (c+d x) \left (15 \cos (c+d x) \left (12 a^2 B+24 a A b+34 a b C+17 b^2 B\right )+48 \cos (2 (c+d x)) \left (5 a^2 (A+C)+10 a b B+b^2 (5 A+4 C)\right )+60 a^2 A \cos (4 (c+d x))+180 a^2 A+60 a^2 B \cos (3 (c+d x))+40 a^2 C \cos (4 (c+d x))+200 a^2 C+120 a A b \cos (3 (c+d x))+80 a b B \cos (4 (c+d x))+400 a b B+90 a b C \cos (3 (c+d x))+40 A b^2 \cos (4 (c+d x))+200 A b^2+45 b^2 B \cos (3 (c+d x))+32 b^2 C \cos (4 (c+d x))+256 b^2 C\right )-120 \cos ^5(c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{480 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.05, size = 404, normalized size = 1.7 \begin{align*}{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{Aab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{Aab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,Bab\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,Bab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{abC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,abC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{2\,A{b}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{B{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,B{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,{b}^{2}C\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{b}^{2}C\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 = 1.06861, size = 482, normalized size = 2.07 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} 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^{2} - 30 \, C a 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 )} - 15 \, B 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 )} - 60 \, B a^{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 )} - 120 \, A a 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 \, A a^{2} \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 = 0.560782, size = 598, normalized size = 2.57 \begin{align*} \frac{15 \,{\left (4 \, B a^{2} + 2 \,{\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, B a^{2} + 2 \,{\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (5 \,{\left (3 \, A + 2 \, C\right )} a^{2} + 20 \, B a b + 2 \,{\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (4 \, B a^{2} + 2 \,{\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, C b^{2} + 8 \,{\left (5 \, C a^{2} + 10 \, B a b +{\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (2 \, C a b + B b^{2}\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 )^{2} \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \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 = 1.25892, size = 1034, normalized size = 4.44 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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