Optimal. Leaf size=86 \[ \frac{\left (2 a^2 C+4 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 B x+\frac{b (3 a C+2 b B) \tan (c+d x)}{2 d}+\frac{b C \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]
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Rubi [A] time = 0.141975, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {4072, 3918, 3770, 3767, 8} \[ \frac{\left (2 a^2 C+4 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 B x+\frac{b (3 a C+2 b B) \tan (c+d x)}{2 d}+\frac{b C \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 3918
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \cos (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 (a+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac{1}{2} \int \left (2 a^2 B+\left (4 a b B+2 a^2 C+b^2 C\right ) \sec (c+d x)+b (2 b B+3 a C) \sec ^2(c+d x)\right ) \, dx\\ &=a^2 B x+\frac{b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac{1}{2} (b (2 b B+3 a C)) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (4 a b B+2 a^2 C+b^2 C\right ) \int \sec (c+d x) \, dx\\ &=a^2 B x+\frac{\left (4 a b B+2 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}-\frac{(b (2 b B+3 a C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^2 B x+\frac{\left (4 a b B+2 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b (2 b B+3 a C) \tan (c+d x)}{2 d}+\frac{b C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.26881, size = 67, normalized size = 0.78 \[ \frac{\left (2 a^2 C+4 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))+2 a^2 B d x+b \tan (c+d x) (4 a C+2 b B+b C \sec (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.056, size = 133, normalized size = 1.6 \begin{align*}{a}^{2}Bx+{\frac{B{a}^{2}c}{d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{abC\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}C\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 = 0.963041, size = 189, normalized size = 2.2 \begin{align*} \frac{4 \,{\left (d x + c\right )} B a^{2} - C 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 )} + 2 \, C a^{2}{\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 b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \tan \left (d x + c\right ) + 4 \, B b^{2} \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 = 0.524719, size = 335, normalized size = 3.9 \begin{align*} \frac{4 \, B a^{2} d x \cos \left (d x + c\right )^{2} +{\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (C b^{2} + 2 \,{\left (2 \, C a b + B b^{2}\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 = 1.23939, size = 259, normalized size = 3.01 \begin{align*} \frac{2 \,{\left (d x + c\right )} B a^{2} +{\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (4 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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