Optimal. Leaf size=123 \[ -\frac{a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac{a^2 (3 A+4 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(A+B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+a^2 x (B+2 C)+\frac{A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]
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Rubi [A] time = 0.392462, antiderivative size = 123, 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, 3023, 2735, 3770} \[ -\frac{a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac{a^2 (3 A+4 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(A+B) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+a^2 x (B+2 C)+\frac{A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]
Antiderivative was successfully verified.
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Rule 3043
Rule 2975
Rule 2968
Rule 3023
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 ^3(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x))^2 (2 a (A+B)-a (A-2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx}{2 a}\\ &=\frac{(A+B) \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x)) \left (a^2 (3 A+4 B+2 C)-a^2 (3 A+2 B-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=\frac{(A+B) \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int \left (a^3 (3 A+4 B+2 C)+\left (-a^3 (3 A+2 B-2 C)+a^3 (3 A+4 B+2 C)\right ) \cos (c+d x)-a^3 (3 A+2 B-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac{a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac{(A+B) \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int \left (a^3 (3 A+4 B+2 C)+2 a^3 (B+2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=a^2 (B+2 C) x-\frac{a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac{(A+B) \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^2 (3 A+4 B+2 C)\right ) \int \sec (c+d x) \, dx\\ &=a^2 (B+2 C) x+\frac{a^2 (3 A+4 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac{(A+B) \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [B] time = 1.44448, size = 259, normalized size = 2.11 \[ \frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (-2 (3 A+4 B+2 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 (3 A+4 B+2 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 (2 A+B) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 (2 A+B) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+4 (B+2 C) (c+d x)+4 C \sin (c+d x)\right )}{16 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.076, size = 166, normalized size = 1.4 \begin{align*}{\frac{A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{2}B\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{A{a}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{a}^{2}Cx+2\,{\frac{C{a}^{2}c}{d}}+{a}^{2}Bx+{\frac{B{a}^{2}c}{d}}+{\frac{{a}^{2}C\sin \left ( dx+c \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.01207, size = 259, normalized size = 2.11 \begin{align*} \frac{4 \,{\left (d x + c\right )} B a^{2} + 8 \,{\left (d x + c\right )} C a^{2} - 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 )} + 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 \, C a^{2} \sin \left (d x + c\right ) + 8 \, A a^{2} \tan \left (d x + c\right ) + 4 \, B 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 = 2.05396, size = 358, normalized size = 2.91 \begin{align*} \frac{4 \,{\left (B + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right )^{2} +{\left (3 \, A + 4 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A + 4 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A 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 = 1.288, size = 275, normalized size = 2.24 \begin{align*} \frac{\frac{4 \, C 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 \, C a^{2}\right )}{\left (d x + c\right )} +{\left (3 \, A a^{2} + 4 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (3 \, A a^{2} + 4 \, B a^{2} + 2 \, 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 (3 \, A 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 )^{3} - 5 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B 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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