Optimal. Leaf size=207 \[ \frac{\tan (c+d x) \left (16 a^2 b B+3 a^3 C+6 a b^2 (3 A+2 C)+4 b^3 B\right )}{6 d}+\frac{\left (12 a^2 b (2 A+C)+8 a^3 B+12 a b^2 B+b^3 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \tan (c+d x) \sec (c+d x) \left (6 a^2 C+20 a b B+12 A b^2+9 b^2 C\right )}{24 d}+a^3 A x+\frac{(3 a C+4 b B) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
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Rubi [A] time = 0.339917, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4056, 4048, 3770, 3767, 8} \[ \frac{\tan (c+d x) \left (16 a^2 b B+3 a^3 C+6 a b^2 (3 A+2 C)+4 b^3 B\right )}{6 d}+\frac{\left (12 a^2 b (2 A+C)+8 a^3 B+12 a b^2 B+b^3 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \tan (c+d x) \sec (c+d x) \left (6 a^2 C+20 a b B+12 A b^2+9 b^2 C\right )}{24 d}+a^3 A x+\frac{(3 a C+4 b B) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 4056
Rule 4048
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (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 A+(4 A b+4 a B+3 b C) \sec (c+d x)+(4 b B+3 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (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^2 A+\left (24 a A b+12 a^2 B+8 b^2 B+15 a b C\right ) \sec (c+d x)+\left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{24} \int \left (24 a^3 A+3 \left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) \sec (c+d x)+4 \left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^3 A x+\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{6} \left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 A x+\frac{\left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=a^3 A x+\frac{\left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [B] time = 5.51976, size = 525, normalized size = 2.54 \[ \frac{\cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-12 \cos ^4(c+d x) \left (12 a^2 b (2 A+C)+8 a^3 B+12 a b^2 B+b^3 (4 A+3 C)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 \cos ^4(c+d x) \left (12 a^2 b (2 A+C)+8 a^3 B+12 a b^2 B+b^3 (4 A+3 C)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+36 a^3 A (c+d x)+48 a^3 A (c+d x) \cos (2 (c+d x))+12 a^3 A (c+d x) \cos (4 (c+d x))+72 a^2 b B \sin (2 (c+d x))+36 a^2 b B \sin (4 (c+d x))+36 a^2 b C \sin (c+d x)+36 a^2 b C \sin (3 (c+d x))+24 a^3 C \sin (2 (c+d x))+12 a^3 C \sin (4 (c+d x))+72 a A b^2 \sin (2 (c+d x))+36 a A b^2 \sin (4 (c+d x))+36 a b^2 B \sin (c+d x)+36 a b^2 B \sin (3 (c+d x))+96 a b^2 C \sin (2 (c+d x))+24 a b^2 C \sin (4 (c+d x))+12 A b^3 \sin (c+d x)+12 A b^3 \sin (3 (c+d x))+32 b^3 B \sin (2 (c+d x))+8 b^3 B \sin (4 (c+d x))+33 b^3 C \sin (c+d x)+9 b^3 C \sin (3 (c+d x))\right )}{48 d (a \cos (c+d x)+b)^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.057, size = 389, normalized size = 1.9 \begin{align*}{a}^{3}Ax+{\frac{A{a}^{3}c}{d}}+{\frac{B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}}+3\,{\frac{A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{B{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{2}bC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{Aa{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,Ba{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ba{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{Ca{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{A{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,B{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,C{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,C{b}^{3}\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 = 1.14788, size = 483, normalized size = 2.33 \begin{align*} \frac{48 \,{\left (d x + c\right )} A a^{3} + 48 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{2} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{3} - 3 \, C b^{3}{\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 )} - 36 \, C a^{2} 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 )} - 36 \, B 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 )} - 12 \, 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^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, A a^{2} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, C a^{3} \tan \left (d x + c\right ) + 144 \, B a^{2} b \tan \left (d x + c\right ) + 144 \, A a 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 = 0.594853, size = 624, normalized size = 3.01 \begin{align*} \frac{48 \, A a^{3} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (8 \, B a^{3} + 12 \,{\left (2 \, A + C\right )} a^{2} b + 12 \, B a b^{2} +{\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, B a^{3} + 12 \,{\left (2 \, A + C\right )} a^{2} b + 12 \, B a b^{2} +{\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, C b^{3} + 8 \,{\left (3 \, C a^{3} + 9 \, B a^{2} b + 3 \,{\left (3 \, A + 2 \, C\right )} a b^{2} + 2 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (12 \, C a^{2} b + 12 \, B a b^{2} +{\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, C a b^{2} + B b^{3}\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 )^{3} \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.34382, size = 1025, normalized size = 4.95 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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