Optimal. Leaf size=209 \[ -\frac{b \left (13 a^2 A b+4 a^3 B-8 a b^2 B-2 A b^3\right ) \tan (c+d x)}{2 d}+\frac{b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{1}{2} a^2 x \left (a^2 A+8 a b B+12 A b^2\right )+\frac{a (2 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]
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Rubi [A] time = 0.462845, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4025, 4094, 4048, 3770, 3767, 8} \[ -\frac{b \left (13 a^2 A b+4 a^3 B-8 a b^2 B-2 A b^3\right ) \tan (c+d x)}{2 d}+\frac{b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{1}{2} a^2 x \left (a^2 A+8 a b B+12 A b^2\right )+\frac{a (2 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]
Antiderivative was successfully verified.
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Rule 4025
Rule 4094
Rule 4048
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (-a (5 A b+2 a B)-\left (a^2 A+2 A b^2+4 a b B\right ) \sec (c+d x)+2 b (a A-b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{1}{2} \int (a+b \sec (c+d x)) \left (-a \left (a^2 A+12 A b^2+8 a b B\right )+b \left (a^2 A-2 A b^2-6 a b B\right ) \sec (c+d x)+2 b \left (6 a A b+2 a^2 B-b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{4} \int \left (-2 a^2 \left (a^2 A+12 A b^2+8 a b B\right )-2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \sec (c+d x)+2 b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (b^2 \left (8 a A b+12 a^2 B+b^2 B\right )\right ) \int \sec (c+d x) \, dx-\frac{1}{2} \left (b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac{b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\left (b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=\frac{1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac{b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \tan (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 1.89767, size = 310, normalized size = 1.48 \[ \frac{2 a^2 (c+d x) \left (a^2 A+8 a b B+12 A b^2\right )-2 b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a^3 (a B+4 A b) \sin (c+d x)+a^4 A \sin (2 (c+d x))+\frac{4 b^3 (4 a B+A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b^3 (4 a B+A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{b^4 B}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^4 B}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.064, size = 236, normalized size = 1.1 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}Ax}{2}}+{\frac{A{a}^{4}c}{2\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{A{a}^{3}b\sin \left ( dx+c \right ) }{d}}+4\,B{a}^{3}bx+4\,{\frac{B{a}^{3}bc}{d}}+6\,A{a}^{2}{b}^{2}x+6\,{\frac{A{a}^{2}{b}^{2}c}{d}}+6\,{\frac{B{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Ba{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{b}^{4}\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 = 0.981407, size = 282, normalized size = 1.35 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 16 \,{\left (d x + c\right )} B a^{3} b + 24 \,{\left (d x + c\right )} A a^{2} b^{2} - B b^{4}{\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 \, B a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{4} \sin \left (d x + c\right ) + 16 \, A a^{3} b \sin \left (d x + c\right ) + 16 \, B a b^{3} \tan \left (d x + c\right ) + 4 \, A b^{4} \tan \left (d x + c\right )}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.60597, size = 479, normalized size = 2.29 \begin{align*} \frac{2 \,{\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} d x \cos \left (d x + c\right )^{2} +{\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{4} \cos \left (d x + c\right )^{3} + B b^{4} + 2 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \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 = 1.30352, size = 713, normalized size = 3.41 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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