Optimal. Leaf size=121 \[ -\frac{a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}+\frac{a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (2 A+B)-\frac{(2 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^2}{d} \]
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Rubi [A] time = 0.216887, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4086, 3917, 3914, 3767, 8, 3770} \[ -\frac{a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}+\frac{a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (2 A+B)-\frac{(2 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^2}{d} \]
Antiderivative was successfully verified.
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Rule 4086
Rule 3917
Rule 3914
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}+\frac{\int (a+a \sec (c+d x))^2 (a (2 A+B)-a (2 A-C) \sec (c+d x)) \, dx}{a}\\ &=\frac{A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac{\int (a+a \sec (c+d x)) \left (2 a^2 (2 A+B)-a^2 (2 A-2 B-3 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=a^2 (2 A+B) x+\frac{A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}-\frac{1}{2} \left (a^2 (2 A-2 B-3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (a^2 (2 A+4 B+3 C)\right ) \int \sec (c+d x) \, dx\\ &=a^2 (2 A+B) x+\frac{a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac{\left (a^2 (2 A-2 B-3 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^2 (2 A+B) x+\frac{a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac{a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [B] time = 3.5695, size = 365, normalized size = 3.02 \[ \frac{a^2 \cos ^4(c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac{2 (2 A+4 B+3 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (2 A+4 B+3 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+4 x (2 A+B)+\frac{4 A \sin (c) \cos (d x)}{d}+\frac{4 A \cos (c) \sin (d x)}{d}+\frac{4 (B+2 C) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (B+2 C) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{C}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{C}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{8 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.103, size = 166, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+{a}^{2}Bx+{\frac{B{a}^{2}c}{d}}+{\frac{3\,{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{a}^{2}Ax+2\,{\frac{A{a}^{2}c}{d}}+2\,{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{{a}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.954943, size = 259, normalized size = 2.14 \begin{align*} \frac{8 \,{\left (d x + c\right )} A a^{2} + 4 \,{\left (d x + c\right )} B a^{2} - 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 )} + 2 \, A 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^{2}{\left (\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 \, A a^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \tan \left (d x + c\right ) + 8 \, C a^{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 = 0.530149, size = 358, normalized size = 2.96 \begin{align*} \frac{4 \,{\left (2 \, A + B\right )} a^{2} d x \cos \left (d x + c\right )^{2} +{\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} \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} + 2 \,{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + C a^{2}\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 = 1.23479, size = 275, normalized size = 2.27 \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 (2 \, A a^{2} + B a^{2}\right )}{\left (d x + c\right )} +{\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, C a^{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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