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 = 0.17, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 8
Rule 2648
Rule 2952
Rule 3767
Rule 3770
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*}
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Mathematica [B] time = 0.91, 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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fricas [A] time = 1.14, size = 108, normalized size = 1.66 \[ -\frac {3 \, a^{2} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 3 \, a^{2} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 10 \, a^{2} \cos \left (f x + e\right )^{2} - 8 \, a^{2} \cos \left (f x + e\right ) + 2 \, a^{2}}{2 \, c f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.21, size = 116, normalized size = 1.78 \[ \frac {4 a^{2}}{f c \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}+\frac {a^{2}}{f c \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}+\frac {3 a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f c}+\frac {a^{2}}{f c \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}-\frac {3 a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.58, size = 225, normalized size = 3.46 \[ -\frac {a^{2} {\left (\frac {\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 2 \, a^{2} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac {a^{2} {\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.62, size = 77, normalized size = 1.18 \[ \frac {6\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-4\,a^2}{c\,f\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}-\frac {6\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c\,f} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{2} \left (\int \frac {\sec ^{2}{\left (e + f x \right )}}{\cos {\left (e + f x \right )} - 1}\, dx + \int \frac {2 \cos {\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{\cos {\left (e + f x \right )} - 1}\, dx + \int \frac {\cos ^{2}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{\cos {\left (e + f x \right )} - 1}\, dx\right )}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
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