Optimal. Leaf size=65 \[ -\frac{a^2 \tan (e+f x)}{c f}-\frac{3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac{4 a^2 \sin (e+f x)}{c f (1-\cos (e+f x))} \]
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Rubi [A] time = 0.171388, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2952, 2648, 3770, 3767, 8} \[ -\frac{a^2 \tan (e+f x)}{c f}-\frac{3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac{4 a^2 \sin (e+f x)}{c f (1-\cos (e+f x))} \]
Antiderivative was successfully verified.
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Rule 2952
Rule 2648
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{(a+a \cos (e+f x))^2 \sec ^2(e+f x)}{-c+c \cos (e+f x)} \, dx &=\int \left (\frac{4 a^2}{c (-1+\cos (e+f x))}-\frac{3 a^2 \sec (e+f x)}{c}-\frac{a^2 \sec ^2(e+f x)}{c}\right ) \, dx\\ &=-\frac{a^2 \int \sec ^2(e+f x) \, dx}{c}-\frac{\left (3 a^2\right ) \int \sec (e+f x) \, dx}{c}+\frac{\left (4 a^2\right ) \int \frac{1}{-1+\cos (e+f x)} \, dx}{c}\\ &=-\frac{3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac{4 a^2 \sin (e+f x)}{c f (1-\cos (e+f x))}+\frac{a^2 \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{c f}\\ &=-\frac{3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac{4 a^2 \sin (e+f x)}{c f (1-\cos (e+f x))}-\frac{a^2 \tan (e+f x)}{c f}\\ \end{align*}
Mathematica [B] time = 0.755996, size = 194, normalized size = 2.98 \[ \frac{2 a^2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (4 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right )+\sin \left (\frac{1}{2} (e+f x)\right ) \left (\frac{\sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}-3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{c f (\cos (e+f x)-1)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.132, size = 116, normalized size = 1.8 \begin{align*}{\frac{{a}^{2}}{cf} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{{a}^{2}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{cf}}+{\frac{{a}^{2}}{cf} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{{a}^{2}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{cf}}+4\,{\frac{{a}^{2}}{cf\tan \left ( 1/2\,fx+e/2 \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 0.998217, size = 304, normalized size = 4.68 \begin{align*} -\frac{a^{2}{\left (\frac{\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 2 \, a^{2}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac{\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac{a^{2}{\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.96813, size = 275, normalized size = 4.23 \begin{align*} -\frac{3 \, a^{2} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 3 \, a^{2} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 10 \, a^{2} \cos \left (f x + e\right )^{2} - 8 \, a^{2} \cos \left (f x + e\right ) + 2 \, a^{2}}{2 \, c f \cos \left (f x + e\right ) \sin \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{a^{2} \left (\int \frac{\sec ^{2}{\left (e + f x \right )}}{\cos{\left (e + f x \right )} - 1}\, dx + \int \frac{2 \cos{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{\cos{\left (e + f x \right )} - 1}\, dx + \int \frac{\cos ^{2}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{\cos{\left (e + f x \right )} - 1}\, dx\right )}{c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.15664, size = 142, normalized size = 2.18 \begin{align*} -\frac{\frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} - \frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} c}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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