Optimal. Leaf size=200 \[ \frac{b \left (34 a^2 A b+19 a^3 B+16 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 d}+\frac{\left (32 a^3 A b+24 a^2 b^2 B+8 a^4 B+16 a A b^3+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \left (26 a^2 B+32 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+a^4 A x+\frac{b (7 a B+4 A b) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{b B \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
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Rubi [A] time = 0.327352, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3918, 4056, 4048, 3770, 3767, 8} \[ \frac{b \left (34 a^2 A b+19 a^3 B+16 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 d}+\frac{\left (32 a^3 A b+24 a^2 b^2 B+8 a^4 B+16 a A b^3+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \left (26 a^2 B+32 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+a^4 A x+\frac{b (7 a B+4 A b) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{b B \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 3918
Rule 4056
Rule 4048
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \sec (c+d x))^2 \left (4 a^2 A+\left (8 a A b+4 a^2 B+3 b^2 B\right ) \sec (c+d x)+b (4 A b+7 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \sec (c+d x)) \left (12 a^3 A+\left (36 a^2 A b+8 A b^3+12 a^3 B+23 a b^2 B\right ) \sec (c+d x)+b \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{24} \int \left (24 a^4 A+3 \left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \sec (c+d x)+4 b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^4 A x+\frac{b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{6} \left (b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \int \sec (c+d x) \, dx\\ &=a^4 A x+\frac{\left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=a^4 A x+\frac{\left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac{b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.02719, size = 160, normalized size = 0.8 \[ \frac{3 \left (32 a^3 A b+24 a^2 b^2 B+8 a^4 B+16 a A b^3+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))+3 b \tan (c+d x) \left (b \left (24 a^2 B+16 a A b+3 b^2 B\right ) \sec (c+d x)+8 \left (6 a^2 A b+4 a^3 B+4 a b^2 B+A b^3\right )+2 b^3 B \sec ^3(c+d x)\right )+24 a^4 A d x+8 b^3 (4 a B+A b) \tan ^3(c+d x)}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.049, size = 338, normalized size = 1.7 \begin{align*}{a}^{4}Ax+{\frac{A{a}^{4}c}{d}}+{\frac{B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{B{a}^{3}b\tan \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+3\,{\frac{B{a}^{2}{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+3\,{\frac{B{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Aa{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,Ba{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,Ba{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{2\,A{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{B{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,B{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,B{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.980455, size = 409, normalized size = 2.04 \begin{align*} \frac{48 \,{\left (d x + c\right )} A a^{4} + 64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{3} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{4} - 3 \, B b^{4}{\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 )} - 72 \, B a^{2} 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 )} - 48 \, A a b^{3}{\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 )} + 48 \, B a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 192 \, A a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 192 \, B a^{3} b \tan \left (d x + c\right ) + 288 \, A a^{2} 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 = 0.604623, size = 603, normalized size = 3.02 \begin{align*} \frac{48 \, A a^{4} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, B b^{4} + 16 \,{\left (6 \, B a^{3} b + 9 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (4 \, B a b^{3} + A b^{4}\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{4}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.31484, size = 857, normalized size = 4.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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