Optimal. Leaf size=56 \[ -\frac{c \tan (e+f x)}{a f}+\frac{2 c \tanh ^{-1}(\sin (e+f x))}{a f}-\frac{2 c \tan (e+f x)}{f (a \sec (e+f x)+a)} \]
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Rubi [A] time = 0.108156, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {4008, 3787, 3770, 3767, 8} \[ -\frac{c \tan (e+f x)}{a f}+\frac{2 c \tanh ^{-1}(\sin (e+f x))}{a f}-\frac{2 c \tan (e+f x)}{f (a \sec (e+f x)+a)} \]
Antiderivative was successfully verified.
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Rule 4008
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{\sec ^2(e+f x) (c-c \sec (e+f x))}{a+a \sec (e+f x)} \, dx &=-\frac{2 c \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\int \sec (e+f x) (-2 a c+a c \sec (e+f x)) \, dx}{a^2}\\ &=-\frac{2 c \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{c \int \sec ^2(e+f x) \, dx}{a}+\frac{(2 c) \int \sec (e+f x) \, dx}{a}\\ &=\frac{2 c \tanh ^{-1}(\sin (e+f x))}{a f}-\frac{2 c \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac{c \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a f}\\ &=\frac{2 c \tanh ^{-1}(\sin (e+f x))}{a f}-\frac{c \tan (e+f x)}{a f}-\frac{2 c \tan (e+f x)}{f (a+a \sec (e+f x))}\\ \end{align*}
Mathematica [B] time = 0.607151, size = 154, normalized size = 2.75 \[ -\frac{c \left (\frac{2 \tan \left (\frac{1}{2} (e+f x)\right )}{f}+\frac{\sin \left (\frac{1}{2} (e+f x)\right )}{f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{\sin \left (\frac{1}{2} (e+f x)\right )}{f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}-\frac{2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}\right )}{a} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.058, size = 104, normalized size = 1.9 \begin{align*} -2\,{\frac{c\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}+2\,{\frac{c\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{fa}}+{\frac{c}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}+{\frac{c}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{c\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{fa}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.02407, size = 262, normalized size = 4.68 \begin{align*} \frac{c{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (f x + e\right )}{{\left (a - \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + c{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.476078, size = 270, normalized size = 4.82 \begin{align*} \frac{{\left (c \cos \left (f x + e\right )^{2} + c \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (c \cos \left (f x + e\right )^{2} + c \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) -{\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )} \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*} - \frac{c \left (\int - \frac{\sec ^{2}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{3}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx\right )}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.20791, size = 124, normalized size = 2.21 \begin{align*} \frac{2 \,{\left (\frac{c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} a}\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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