Optimal. Leaf size=75 \[ \frac{b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^3 x-\frac{a b^2 \tan (c+d x)}{2 d}-\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]
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Rubi [A] time = 0.08655, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4042, 3918, 3770, 3767, 8} \[ \frac{b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^3 x-\frac{a b^2 \tan (c+d x)}{2 d}-\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Rule 4042
Rule 3918
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \sec (c+d x)) \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx &=-\int (-a+b \sec (c+d x)) (a+b \sec (c+d x))^2 \, dx\\ &=-\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-2 a^3-b \left (2 a^2-b^2\right ) \sec (c+d x)+a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=a^3 x-\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}-\frac{1}{2} \left (a b^2\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (b \left (2 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=a^3 x+\frac{b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac{\left (a b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^3 x+\frac{b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a b^2 \tan (c+d x)}{2 d}-\frac{b^2 (a+b \sec (c+d x)) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.0204045, size = 75, normalized size = 1. \[ \frac{a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+a^3 x-\frac{a b^2 \tan (c+d x)}{d}-\frac{b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^3 \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.039, size = 94, normalized size = 1.3 \begin{align*}{a}^{3}x+{\frac{{a}^{3}c}{d}}-{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}-{\frac{{b}^{3}\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 = 0.970634, size = 126, normalized size = 1.68 \begin{align*} \frac{4 \,{\left (d x + c\right )} a^{3} + b^{3}{\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} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, a b^{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 = 0.515588, size = 281, normalized size = 3.75 \begin{align*} \frac{4 \, a^{3} d x \cos \left (d x + c\right )^{2} +{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \, a b^{2} \cos \left (d x + c\right ) + b^{3}\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 ) \left (a + b \sec{\left (c + d x \right )}\right )^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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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