Optimal. Leaf size=184 \[ \frac{\tan (c+d x) \left (2 a^2 B+4 a A b+6 a b C+3 b^2 B\right )}{3 d}+\frac{\left (a^2 (3 A+4 C)+8 a b B+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)+8 a b B+2 A b^2\right )}{8 d}+\frac{a (2 a B+A b) \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]
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Rubi [A] time = 0.465376, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac{\tan (c+d x) \left (2 a^2 B+4 a A b+6 a b C+3 b^2 B\right )}{3 d}+\frac{\left (a^2 (3 A+4 C)+8 a b B+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)+8 a b B+2 A b^2\right )}{8 d}+\frac{a (2 a B+A b) \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]
Antiderivative was successfully verified.
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Rule 3047
Rule 3031
Rule 3021
Rule 2748
Rule 3767
Rule 8
Rule 3770
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 ^5(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x)) \left (2 (A b+2 a B)+(3 a A+4 b B+4 a C) \cos (c+d x)+b (A+4 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a (A b+2 a B) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{12} \int \left (-3 \left (2 A b^2+8 a b B+a^2 (3 A+4 C)\right )-4 \left (4 a A b+2 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)-3 b^2 (A+4 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (2 A b^2+8 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (A b+2 a B) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-8 \left (4 a A b+2 a^2 B+3 b^2 B+6 a b C\right )-3 \left (8 a b B+4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (2 A b^2+8 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (A b+2 a B) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{3} \left (-4 a A b-2 a^2 B-3 b^2 B-6 a b C\right ) \int \sec ^2(c+d x) \, dx-\frac{1}{8} \left (-8 a b B-4 b^2 (A+2 C)-a^2 (3 A+4 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (8 a b B+4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (2 A b^2+8 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (A b+2 a B) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{\left (4 a A b+2 a^2 B+3 b^2 B+6 a b C\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{\left (8 a b B+4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a A b+2 a^2 B+3 b^2 B+6 a b C\right ) \tan (c+d x)}{3 d}+\frac{\left (2 A b^2+8 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (A b+2 a B) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.07744, size = 137, normalized size = 0.74 \[ \frac{3 \left (a^2 (3 A+4 C)+8 a b B+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (3 a^2 B+a (a B+2 A b) \tan ^2(c+d x)+6 a b (A+C)+3 b^2 B\right )+3 \sec (c+d x) \left (a^2 (3 A+4 C)+8 a b B+4 A b^2\right )+6 a^2 A \sec ^3(c+d x)\right )}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.069, size = 321, normalized size = 1.7 \begin{align*}{\frac{A{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{2}B\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,aAb\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,aAb \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{abB\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{abB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{abC\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,{a}^{2}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}B \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.98514, size = 423, normalized size = 2.3 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b - 3 \, A 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 )} - 12 \, C 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 )} - 24 \, B 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 )} - 12 \, 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 )} + 24 \, C b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, C a b \tan \left (d x + c\right ) + 48 \, B b^{2} \tan \left (d x + c\right )}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.91735, size = 510, normalized size = 2.77 \begin{align*} \frac{3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 8 \, B a b + 4 \,{\left (A + 2 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 8 \, B a b + 4 \,{\left (A + 2 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (2 \, B a^{2} + 2 \,{\left (2 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, A a^{2} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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 = 1.28139, size = 851, normalized size = 4.62 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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