Optimal. Leaf size=167 \[ \frac{a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \tan (c+d x)}{2 d}+\frac{b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{8 d}+a^3 A x+\frac{a C \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
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Rubi [A] time = 0.313253, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4057, 4056, 4048, 3770, 3767, 8} \[ \frac{a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \tan (c+d x)}{2 d}+\frac{b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{8 d}+a^3 A x+\frac{a C \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 4057
Rule 4056
Rule 4048
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^3 \left (A+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+b (4 A+3 C) \sec (c+d x)+3 a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 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+3 a b (8 A+5 C) \sec (c+d x)+3 \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 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 b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \sec (c+d x)+12 a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^3 A x+\frac{b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{2} \left (a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=a^3 A x+\frac{b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^3 A x+\frac{b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \tan (c+d x)}{2 d}+\frac{b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [B] time = 6.41372, size = 1241, normalized size = 7.43 \[ \frac{\left (-4 A b^3-3 C b^3-24 a^2 A b-12 a^2 C b\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \cos ^5(c+d x)}{4 d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C)}+\frac{\left (4 A b^3+3 C b^3+24 a^2 A b+12 a^2 C b\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \cos ^5(c+d x)}{4 d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C)}+\frac{2 a^3 A (c+d x) (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \cos ^5(c+d x)}{d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C)}+\frac{a b^2 C (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5(c+d x)}{d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \left (C \sin \left (\frac{1}{2} (c+d x)\right ) a^3+3 A b^2 \sin \left (\frac{1}{2} (c+d x)\right ) a+2 b^2 C \sin \left (\frac{1}{2} (c+d x)\right ) a\right ) \cos ^5(c+d x)}{d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \left (C \sin \left (\frac{1}{2} (c+d x)\right ) a^3+3 A b^2 \sin \left (\frac{1}{2} (c+d x)\right ) a+2 b^2 C \sin \left (\frac{1}{2} (c+d x)\right ) a\right ) \cos ^5(c+d x)}{d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\left (4 A b^3+3 C b^3+4 a C b^2+12 a^2 C b\right ) (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \cos ^5(c+d x)}{8 d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\left (-4 A b^3-3 C b^3-4 a C b^2-12 a^2 C b\right ) (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \cos ^5(c+d x)}{8 d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{a b^2 C (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5(c+d x)}{d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{b^3 C (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \cos ^5(c+d x)}{8 d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{b^3 C (a+b \sec (c+d x))^3 \left (C \sec ^2(c+d x)+A\right ) \cos ^5(c+d x)}{8 d (b+a \cos (c+d x))^3 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.049, size = 267, normalized size = 1.6 \begin{align*}{a}^{3}Ax+{\frac{A{a}^{3}c}{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}}+{\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}}+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{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 = 0.963603, size = 343, normalized size = 2.05 \begin{align*} \frac{16 \,{\left (d x + c\right )} A a^{3} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{2} - 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 )} - 12 \, 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 )} - 4 \, 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 \, A a^{2} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 16 \, C a^{3} \tan \left (d x + c\right ) + 48 \, A a b^{2} \tan \left (d x + c\right )}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.557219, size = 485, normalized size = 2.9 \begin{align*} \frac{16 \, A a^{3} d x \cos \left (d x + c\right )^{4} +{\left (12 \,{\left (2 \, A + C\right )} a^{2} b +{\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 ) -{\left (12 \,{\left (2 \, A + C\right )} a^{2} b +{\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 (8 \, C a b^{2} \cos \left (d x + c\right ) + 2 \, C b^{3} + 8 \,{\left (C a^{3} +{\left (3 \, A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (12 \, C a^{2} b +{\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, 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 + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.25045, size = 710, normalized size = 4.25 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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