Optimal. Leaf size=80 \[ \frac{\left (a^2 A+4 a b B+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 A \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a (a B+2 A b) \tan (c+d x)}{d}+b^2 B x \]
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Rubi [A] time = 0.19951, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2988, 3021, 2735, 3770} \[ \frac{\left (a^2 A+4 a b B+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 A \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a (a B+2 A b) \tan (c+d x)}{d}+b^2 B x \]
Antiderivative was successfully verified.
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Rule 2988
Rule 3021
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx &=\frac{a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-2 a (2 A b+a B)-\left (a^2 A+2 A b^2+4 a b B\right ) \cos (c+d x)-2 b^2 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a (2 A b+a B) \tan (c+d x)}{d}+\frac{a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-a^2 A-2 A b^2-4 a b B-2 b^2 B \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^2 B x+\frac{a (2 A b+a B) \tan (c+d x)}{d}+\frac{a^2 A \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \left (-a^2 A-2 A b^2-4 a b B\right ) \int \sec (c+d x) \, dx\\ &=b^2 B x+\frac{\left (a^2 A+2 A b^2+4 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (2 A b+a B) \tan (c+d x)}{d}+\frac{a^2 A \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.265642, size = 67, normalized size = 0.84 \[ \frac{\left (a^2 A+4 a b B+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))+a \tan (c+d x) (a A \sec (c+d x)+2 a B+4 A b)+2 b^2 B d x}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.088, size = 133, normalized size = 1.7 \begin{align*}{\frac{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{Aab\tan \left ( dx+c \right ) }{d}}+2\,{\frac{Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{2}Bx+{\frac{B{b}^{2}c}{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.03715, size = 189, normalized size = 2.36 \begin{align*} \frac{4 \,{\left (d x + c\right )} B b^{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 )} + 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 \, B a^{2} \tan \left (d x + c\right ) + 8 \, A a b \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 = 1.41339, size = 335, normalized size = 4.19 \begin{align*} \frac{4 \, B b^{2} d x \cos \left (d x + c\right )^{2} +{\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{2} + 2 \,{\left (B a^{2} + 2 \, A a b\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 [B] time = 1.5992, size = 257, normalized size = 3.21 \begin{align*} \frac{2 \,{\left (d x + c\right )} B b^{2} +{\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (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} - 4 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 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 ) + 4 \, A a b \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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