Optimal. Leaf size=126 \[ \frac{\left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b \tan (c+d x) (2 a A-2 a C-b B)}{d}+a x (a B+2 A b)+\frac{A \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac{b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d} \]
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Rubi [A] time = 0.202626, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {4094, 4048, 3770, 3767, 8} \[ \frac{\left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b \tan (c+d x) (2 a A-2 a C-b B)}{d}+a x (a B+2 A b)+\frac{A \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac{b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 4094
Rule 4048
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}+\int (a+b \sec (c+d x)) \left (2 A b+a B+(b B+a C) \sec (c+d x)-b (2 A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{b^2 (2 A-C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int \left (2 a (2 A b+a B)+\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \sec (c+d x)-2 b (2 a A-b B-2 a C) \sec ^2(c+d x)\right ) \, dx\\ &=a (2 A b+a B) x+\frac{A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{b^2 (2 A-C) \sec (c+d x) \tan (c+d x)}{2 d}-(b (2 a A-b B-2 a C)) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \int \sec (c+d x) \, dx\\ &=a (2 A b+a B) x+\frac{\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{b^2 (2 A-C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{(b (2 a A-b B-2 a C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=a (2 A b+a B) x+\frac{\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{b (2 a A-b B-2 a C) \tan (c+d x)}{d}-\frac{b^2 (2 A-C) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [B] time = 1.40957, size = 453, normalized size = 3.6 \[ \frac{\sec ^2(c+d x) \left (\cos (2 (c+d x)) \left (-\left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+\left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 a (c+d x) (a B+2 A b)\right )+\left (a^2 A+2 b^2 C\right ) \sin (c+d x)+a^2 A \sin (3 (c+d x))+2 a^2 B c+2 a^2 B d x-2 a^2 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a^2 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a A b c+4 a A b d x-4 a b B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 a b B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a b C \sin (2 (c+d x))-2 A b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 A b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 b^2 B \sin (2 (c+d x))-b^2 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+b^2 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.069, size = 184, normalized size = 1.5 \begin{align*}{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+{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\,aAbx+2\,{\frac{Aabc}{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{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \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 = 1.07345, size = 255, normalized size = 2.02 \begin{align*} \frac{4 \,{\left (d x + c\right )} B a^{2} + 8 \,{\left (d x + c\right )} A a b - 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 )} + 2 \, A b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{2} \sin \left (d x + c\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 = 0.555586, size = 406, normalized size = 3.22 \begin{align*} \frac{4 \,{\left (B a^{2} + 2 \, A a b\right )} d x \cos \left (d x + c\right )^{2} +{\left (2 \, C a^{2} + 4 \, B a b +{\left (2 \, A + C\right )} 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 +{\left (2 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 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 [A] time = 1.22505, size = 325, normalized size = 2.58 \begin{align*} \frac{\frac{4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 2 \,{\left (B a^{2} + 2 \, A a b\right )}{\left (d x + c\right )} +{\left (2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2} + 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 + 2 \, A b^{2} + 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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