Optimal. Leaf size=88 \[ \frac{a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac{a^2 (3 A+4 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+a^2 B x \]
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Rubi [A] time = 0.219088, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2975, 2968, 3021, 2735, 3770} \[ \frac{a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac{a^2 (3 A+4 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+a^2 B x \]
Antiderivative was successfully verified.
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Rule 2975
Rule 2968
Rule 3021
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+a \cos (c+d x)) (a (3 A+2 B)+2 a B \cos (c+d x)) \sec ^2(c+d x) \, dx\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int \left (a^2 (3 A+2 B)+\left (2 a^2 B+a^2 (3 A+2 B)\right ) \cos (c+d x)+2 a^2 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int \left (a^2 (3 A+4 B)+2 a^2 B \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=a^2 B x+\frac{a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^2 (3 A+4 B)\right ) \int \sec (c+d x) \, dx\\ &=a^2 B x+\frac{a^2 (3 A+4 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [B] time = 1.1834, size = 277, normalized size = 3.15 \[ \frac{1}{16} a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (\frac{4 (2 A+B) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (2 A+B) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 (3 A+4 B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (3 A+4 B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{A}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+4 B x\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.106, size = 113, normalized size = 1.3 \begin{align*}{\frac{3\,{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{a}^{2}Bx+{\frac{B{a}^{2}c}{d}}+2\,{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.980099, size = 192, normalized size = 2.18 \begin{align*} \frac{4 \,{\left (d x + c\right )} B 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 )} + 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 = 1.39228, size = 297, normalized size = 3.38 \begin{align*} \frac{4 \, B a^{2} d x \cos \left (d x + c\right )^{2} +{\left (3 \, A + 4 \, B\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\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (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 = 1.32333, size = 208, normalized size = 2.36 \begin{align*} \frac{2 \,{\left (d x + c\right )} B a^{2} +{\left (3 \, A a^{2} + 4 \, B 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}\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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