Optimal. Leaf size=65 \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a b \sec (c+d x)}{2 d}+\frac{b \sec (c+d x) (a+b \tan (c+d x))}{2 d} \]
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Rubi [A] time = 0.051964, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3508, 3486, 3770} \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a b \sec (c+d x)}{2 d}+\frac{b \sec (c+d x) (a+b \tan (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Rule 3508
Rule 3486
Rule 3770
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac{b \sec (c+d x) (a+b \tan (c+d x))}{2 d}+\frac{1}{2} \int \sec (c+d x) \left (2 a^2-b^2+3 a b \tan (c+d x)\right ) \, dx\\ &=\frac{3 a b \sec (c+d x)}{2 d}+\frac{b \sec (c+d x) (a+b \tan (c+d x))}{2 d}+\frac{1}{2} \left (2 a^2-b^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a b \sec (c+d x)}{2 d}+\frac{b \sec (c+d x) (a+b \tan (c+d x))}{2 d}\\ \end{align*}
Mathematica [A] time = 0.038608, size = 67, normalized size = 1.03 \[ \frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a b \sec (c+d x)}{d}-\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.021, size = 98, normalized size = 1.5 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{\sin \left ( dx+c \right ){b}^{2}}{2\,d}}-{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{ab}{d\cos \left ( dx+c \right ) }}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.36742, size = 111, normalized size = 1.71 \begin{align*} -\frac{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 )} - 4 \, a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - \frac{8 \, a b}{\cos \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 = 1.79038, size = 234, normalized size = 3.6 \begin{align*} \frac{{\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 ) + 8 \, a b \cos \left (d x + c\right ) + 2 \, b^{2} \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 + 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 = 1.48975, size = 165, normalized size = 2.54 \begin{align*} \frac{{\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{2 \,{\left (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 )^{2} + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a b\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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