Optimal. Leaf size=90 \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 \sin (c+d x)}{2 d}+\frac{a b \tan (c+d x)}{d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^2}{2 d}-2 a b x \]
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Rubi [A] time = 0.42642, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {4397, 2889, 3048, 3031, 3023, 2735, 3770} \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 \sin (c+d x)}{2 d}+\frac{a b \tan (c+d x)}{d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^2}{2 d}-2 a b x \]
Antiderivative was successfully verified.
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Rule 4397
Rule 2889
Rule 3048
Rule 3031
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx &=\int (b+a \cos (c+d x))^2 \sec (c+d x) \tan ^2(c+d x) \, dx\\ &=\int (b+a \cos (c+d x))^2 \left (1-\cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (b+a \cos (c+d x)) \left (2 a-b \cos (c+d x)-3 a \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a b \tan (c+d x)}{d}+\frac{(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-2 a^2+b^2+4 a b \cos (c+d x)+3 a^2 \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{3 a^2 \sin (c+d x)}{2 d}+\frac{a b \tan (c+d x)}{d}+\frac{(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-2 a^2+b^2+4 a b \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=-2 a b x-\frac{3 a^2 \sin (c+d x)}{2 d}+\frac{a b \tan (c+d x)}{d}+\frac{(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \left (-2 a^2+b^2\right ) \int \sec (c+d x) \, dx\\ &=-2 a b x+\frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{3 a^2 \sin (c+d x)}{2 d}+\frac{a b \tan (c+d x)}{d}+\frac{(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.192681, size = 75, normalized size = 0.83 \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))-2 a^2 \sin (c+d x)-4 a b \tan ^{-1}(\tan (c+d x))+4 a b \tan (c+d x)+b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.063, size = 123, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-2\,abx+2\,{\frac{ab\tan \left ( dx+c \right ) }{d}}-2\,{\frac{abc}{d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.71679, size = 138, normalized size = 1.53 \begin{align*} -\frac{8 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a b + b^{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^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.545411, size = 305, normalized size = 3.39 \begin{align*} -\frac{8 \, a b d x \cos \left (d x + c\right )^{2} -{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) - b^{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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}\right )^{2} \sec{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 2.75911, size = 231, normalized size = 2.57 \begin{align*} -\frac{4 \,{\left (d x + c\right )} a b -{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) +{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{4 \, 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} + \frac{2 \,{\left (4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{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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