Optimal. Leaf size=56 \[ -\frac {a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac {4 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))}+\frac {a^2 x}{c} \]
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Rubi [A] time = 0.16, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3903, 3777, 8, 3794, 3789, 3770} \[ -\frac {a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac {4 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))}+\frac {a^2 x}{c} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3770
Rule 3777
Rule 3789
Rule 3794
Rule 3903
Rubi steps
\begin {align*} \int \frac {(a+a \sec (e+f x))^2}{c-c \sec (e+f x)} \, dx &=\frac {\int \left (\frac {a^2}{1-\sec (e+f x)}+\frac {2 a^2 \sec (e+f x)}{1-\sec (e+f x)}+\frac {a^2 \sec ^2(e+f x)}{1-\sec (e+f x)}\right ) \, dx}{c}\\ &=\frac {a^2 \int \frac {1}{1-\sec (e+f x)} \, dx}{c}+\frac {a^2 \int \frac {\sec ^2(e+f x)}{1-\sec (e+f x)} \, dx}{c}+\frac {\left (2 a^2\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{c}\\ &=-\frac {3 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))}-\frac {a^2 \int -1 \, dx}{c}-\frac {a^2 \int \sec (e+f x) \, dx}{c}+\frac {a^2 \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{c}\\ &=\frac {a^2 x}{c}-\frac {a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac {4 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))}\\ \end {align*}
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Mathematica [B] time = 0.29, size = 169, normalized size = 3.02 \[ \frac {a^2 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (-\cos \left (\frac {f x}{2}\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+f x\right )+\cos \left (e+\frac {f x}{2}\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+f x\right )+8 \sin \left (\frac {f x}{2}\right )\right )}{c f (\cos (e+f x)-1)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 87, normalized size = 1.55 \[ \frac {2 \, a^{2} f x \sin \left (f x + e\right ) - a^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + a^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 8 \, a^{2} \cos \left (f x + e\right ) + 8 \, a^{2}}{2 \, c f \sin \left (f x + e\right )} \]
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 = 0.78, size = 90, normalized size = 1.61 \[ \frac {a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f c}-\frac {a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f c}+\frac {4 a^{2}}{f c \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}+\frac {2 a^{2} \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 153, normalized size = 2.73 \[ \frac {a^{2} {\left (\frac {2 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} + \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - a^{2} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} + \frac {2 \, a^{2} {\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.48, size = 46, normalized size = 0.82 \[ \frac {a^2\,x}{c}-\frac {a^2\,\left (2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {4}{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{c\,f} \]
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 \[ - \frac {a^{2} \left (\int \frac {2 \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {1}{\sec {\left (e + f x \right )} - 1}\, dx\right )}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
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