Optimal. Leaf size=179 \[ \frac{\left (4 a^2 b B+a^3 (-C)+8 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{6 b d}+\frac{\left (4 a^2 C+8 a b B+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (-2 a^2 C+8 a b B+9 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(4 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d} \]
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Rubi [A] time = 0.347556, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {4072, 4010, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{\left (4 a^2 b B+a^3 (-C)+8 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{6 b d}+\frac{\left (4 a^2 C+8 a b B+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (-2 a^2 C+8 a b B+9 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(4 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4010
Rule 4002
Rule 3997
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 (3 b C+(4 b B-a C) \sec (c+d x)) \, dx}{4 b}\\ &=\frac{(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (b (8 b B+7 a C)+\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x)\right ) \, dx}{12 b}\\ &=\frac{\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{\int \sec (c+d x) \left (3 b \left (8 a b B+4 a^2 C+3 b^2 C\right )+4 \left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \sec (c+d x)\right ) \, dx}{24 b}\\ &=\frac{\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{1}{8} \left (8 a b B+4 a^2 C+3 b^2 C\right ) \int \sec (c+d x) \, dx+\frac{\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \int \sec ^2(c+d x) \, dx}{6 b}\\ &=\frac{\left (8 a b B+4 a^2 C+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}-\frac{\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 b d}\\ &=\frac{\left (8 a b B+4 a^2 C+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \tan (c+d x)}{6 b d}+\frac{\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}\\ \end{align*}
Mathematica [A] time = 0.761261, size = 120, normalized size = 0.67 \[ \frac{3 \left (4 a^2 C+8 a b B+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 a^2 C+8 a b B+3 b^2 C\right ) \sec (c+d x)+24 \left (a^2 B+2 a b C+b^2 B\right )+8 b (2 a C+b B) \tan ^2(c+d x)+6 b^2 C \sec ^3(c+d x)\right )}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.037, size = 241, normalized size = 1.4 \begin{align*}{\frac{B{a}^{2}\tan \left ( dx+c \right ) }{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}}+{\frac{Bab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,abC\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,abC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{2\,B{b}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}C\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 = 0.977326, size = 308, normalized size = 1.72 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{2} - 3 \, C b^{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 )} + 48 \, B a^{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 = 0.534846, size = 443, normalized size = 2.47 \begin{align*} \frac{3 \,{\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (3 \, B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, C b^{2} + 3 \,{\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (2 \, C a b + B b^{2}\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 (B + C \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.42798, size = 645, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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