Optimal. Leaf size=59 \[ \frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]
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Rubi [A] time = 0.0536586, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3788, 3767, 8, 4046, 3770} \[ \frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 3788
Rule 3767
Rule 8
Rule 4046
Rule 3770
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \sec (c+d x))^2 \, dx &=(2 a b) \int \sec ^2(c+d x) \, dx+\int \sec (c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b^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-\frac{(2 a b) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.104026, size = 45, normalized size = 0.76 \[ \frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))+b \tan (c+d x) (4 a+b \sec (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.025, size = 78, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{ab\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}\sec \left ( dx+c \right ) \tan \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 = 1.16169, size = 108, normalized size = 1.83 \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 ) - 8 \, 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 = 1.68205, size = 236, normalized size = 4. \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 ) + 2 \,{\left (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 + b \sec{\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 = 1.2196, size = 174, normalized size = 2.95 \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 (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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