Optimal. Leaf size=232 \[ \frac{\tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{15 d}+\frac{\left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac{a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
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Rubi [A] time = 0.507059, antiderivative size = 232, normalized size of antiderivative = 1., 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 = {3047, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{15 d}+\frac{\left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac{a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 3047
Rule 3031
Rule 3021
Rule 2748
Rule 3768
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x)) \left (2 A b+5 a B+(4 a A+5 b B+5 a C) \cos (c+d x)+b (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{20} \int \left (-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )-5 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x)-4 b^2 (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{60} \int \left (-15 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right )-4 \left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{4} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{15} \left (-20 a b B-5 b^2 (2 A+3 C)-2 a^2 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{8} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \sec (c+d x) \, dx-\frac{\left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 2.47562, size = 167, normalized size = 0.72 \[ \frac{15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \tan ^2(c+d x) \left (a^2 (2 A+C)+2 a b B+A b^2\right )+15 \left (a^2 (A+C)+2 a b B+b^2 (A+C)\right )+3 a^2 A \tan ^4(c+d x)\right )+15 \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+30 a (a B+2 A b) \sec ^3(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.069, size = 404, normalized size = 1.7 \begin{align*}{\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}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{aAb \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,aAb\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,aAb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{4\,abB\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,abB\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{abC\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+{\frac{abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,A{a}^{2}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{2}B \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.033, size = 482, normalized size = 2.08 \begin{align*} \frac{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 )} A a^{2} + 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} - 15 \, B a^{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 )} - 30 \, A 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 )} - 120 \, C 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 )} - 60 \, B 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 \, C b^{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 = 1.87832, size = 598, normalized size = 2.58 \begin{align*} \frac{15 \,{\left (3 \, B a^{2} + 2 \,{\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, B a^{2} + 2 \,{\left (3 \, A + 4 \, C\right )} a b + 4 \, 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 (2 \,{\left (4 \, A + 5 \, C\right )} a^{2} + 20 \, B a b + 5 \,{\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (3 \, B a^{2} + 2 \,{\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, A a^{2} + 8 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (B a^{2} + 2 \, A a b\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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.20769, size = 1034, normalized size = 4.46 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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