Optimal. Leaf size=134 \[ \frac{a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac{a^2 (2 A+3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+a^2 C x+\frac{A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]
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Rubi [A] time = 0.38518, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3043, 2975, 2968, 3021, 2735, 3770} \[ \frac{a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac{a^2 (2 A+3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+a^2 C x+\frac{A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 3043
Rule 2975
Rule 2968
Rule 3021
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (c+d x))^2 (a (2 A+3 B)+3 a C \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac{(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (c+d x)) \left (3 a^2 (2 A+3 B+2 C)+6 a^2 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac{(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int \left (3 a^3 (2 A+3 B+2 C)+\left (6 a^3 C+3 a^3 (2 A+3 B+2 C)\right ) \cos (c+d x)+6 a^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac{a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac{(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int \left (3 a^3 (2 A+3 B+4 C)+6 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=a^2 C x+\frac{a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac{(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} \left (a^2 (2 A+3 B+4 C)\right ) \int \sec (c+d x) \, dx\\ &=a^2 C x+\frac{a^2 (2 A+3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac{(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}
Mathematica [B] time = 5.1629, size = 315, normalized size = 2.35 \[ \frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (-6 (2 A+3 B+4 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \tan (c+d x) \sec ^2(c+d x) ((5 A+6 B+3 C) \cos (2 (c+d x))+A (-\cos (c+d x))+7 A+6 B+3 C)+\frac{7 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{7 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+12 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{3 B}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{3 B}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+18 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+24 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 c C+12 C d x\right )}{48 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.078, size = 193, normalized size = 1.4 \begin{align*}{\frac{5\,A{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{{a}^{2}B\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{a}^{2}Cx+{\frac{C{a}^{2}c}{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.0195, size = 302, normalized size = 2.25 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 12 \,{\left (d x + c\right )} C a^{2} - 6 \, A 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 )} - 3 \, B 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 )} + 6 \, 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 )} + 12 \, 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 )} + 12 \, A a^{2} \tan \left (d x + c\right ) + 24 \, B a^{2} \tan \left (d x + c\right ) + 12 \, C a^{2} \tan \left (d x + c\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.16061, size = 379, normalized size = 2.83 \begin{align*} \frac{12 \, C a^{2} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, A + 3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, A + 3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (5 \, A + 6 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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 = 1.27702, size = 338, normalized size = 2.52 \begin{align*} \frac{6 \,{\left (d x + c\right )} C a^{2} + 3 \,{\left (2 \, A a^{2} + 3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, A a^{2} + 3 \, B a^{2} + 4 \, 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 (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 16 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, 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 )}^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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