Optimal. Leaf size=74 \[ \frac{3 c^2 \tan (e+f x)}{a f}-\frac{3 c^2 \tanh ^{-1}(\sin (e+f x))}{a f}+\frac{2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)} \]
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Rubi [A] time = 0.102935, antiderivative size = 74, 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 = {3957, 3787, 3770, 3767, 8} \[ \frac{3 c^2 \tan (e+f x)}{a f}-\frac{3 c^2 \tanh ^{-1}(\sin (e+f x))}{a f}+\frac{2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)} \]
Antiderivative was successfully verified.
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Rule 3957
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (c-c \sec (e+f x))^2}{a+a \sec (e+f x)} \, dx &=\frac{2 \left (c^2-c^2 \sec (e+f x)\right ) \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(3 c) \int \sec (e+f x) (c-c \sec (e+f x)) \, dx}{a}\\ &=\frac{2 \left (c^2-c^2 \sec (e+f x)\right ) \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (3 c^2\right ) \int \sec (e+f x) \, dx}{a}+\frac{\left (3 c^2\right ) \int \sec ^2(e+f x) \, dx}{a}\\ &=-\frac{3 c^2 \tanh ^{-1}(\sin (e+f x))}{a f}+\frac{2 \left (c^2-c^2 \sec (e+f x)\right ) \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (3 c^2\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a f}\\ &=-\frac{3 c^2 \tanh ^{-1}(\sin (e+f x))}{a f}+\frac{3 c^2 \tan (e+f x)}{a f}+\frac{2 \left (c^2-c^2 \sec (e+f x)\right ) \tan (e+f x)}{f (a+a \sec (e+f x))}\\ \end{align*}
Mathematica [B] time = 1.57997, size = 220, normalized size = 2.97 \[ \frac{2 c^2 \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \left (4 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc \left (\frac{1}{2} (e+f x)\right )+\cot \left (\frac{1}{2} (e+f x)\right ) \left (\frac{\sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{a f (\sec (e+f x)+1)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.059, size = 116, normalized size = 1.6 \begin{align*} 4\,{\frac{{c}^{2}\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}-{\frac{{c}^{2}}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{{c}^{2}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{fa}}-{\frac{{c}^{2}}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{{c}^{2}\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 = 0.961418, size = 302, normalized size = 4.08 \begin{align*} -\frac{c^{2}{\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 )} + 2 \, c^{2}{\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 )} - \frac{c^{2} \sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.478713, size = 301, normalized size = 4.07 \begin{align*} -\frac{3 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (5 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left (a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )\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^{2} \left (\int \frac{\sec{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{2 \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 = 1.32756, size = 138, normalized size = 1.86 \begin{align*} -\frac{\frac{3 \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{3 \, c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{4 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{2 \, c^{2} \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}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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