Optimal. Leaf size=86 \[ -\frac {a^2 c \tan ^3(e+f x)}{3 f}+\frac {a^2 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a^2 c \tan (e+f x) \sec ^3(e+f x)}{4 f}+\frac {a^2 c \tan (e+f x) \sec (e+f x)}{8 f} \]
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Rubi [A] time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3962, 2607, 30, 2611, 3768, 3770} \[ -\frac {a^2 c \tan ^3(e+f x)}{3 f}+\frac {a^2 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a^2 c \tan (e+f x) \sec ^3(e+f x)}{4 f}+\frac {a^2 c \tan (e+f x) \sec (e+f x)}{8 f} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2607
Rule 2611
Rule 3768
Rule 3770
Rule 3962
Rubi steps
\begin {align*} \int \sec ^2(e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx &=-\left ((a c) \int \left (a \sec ^2(e+f x) \tan ^2(e+f x)+a \sec ^3(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^2 c\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx\right )-\left (a^2 c\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {a^2 c \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {1}{4} \left (a^2 c\right ) \int \sec ^3(e+f x) \, dx-\frac {\left (a^2 c\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^2 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^2 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {a^2 c \tan ^3(e+f x)}{3 f}+\frac {1}{8} \left (a^2 c\right ) \int \sec (e+f x) \, dx\\ &=\frac {a^2 c \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac {a^2 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^2 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {a^2 c \tan ^3(e+f x)}{3 f}\\ \end {align*}
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Mathematica [A] time = 0.18, size = 57, normalized size = 0.66 \[ \frac {a^2 c \left (3 \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \left (-8 \tan ^2(e+f x)-6 \sec ^3(e+f x)+3 \sec (e+f x)\right )\right )}{24 f} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 117, normalized size = 1.36 \[ \frac {3 \, a^{2} c \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a^{2} c \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (8 \, a^{2} c \cos \left (f x + e\right )^{3} + 3 \, a^{2} c \cos \left (f x + e\right )^{2} - 8 \, a^{2} c \cos \left (f x + e\right ) - 6 \, a^{2} c\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \]
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 = 1.37, size = 107, normalized size = 1.24 \[ \frac {a^{2} c \sec \left (f x +e \right ) \tan \left (f x +e \right )}{8 f}+\frac {a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8 f}+\frac {a^{2} c \tan \left (f x +e \right )}{3 f}-\frac {a^{2} c \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{3 f}-\frac {a^{2} c \left (\sec ^{3}\left (f x +e \right )\right ) \tan \left (f x +e \right )}{4 f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 160, normalized size = 1.86 \[ -\frac {16 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c - 3 \, a^{2} c {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 12 \, a^{2} c {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 48 \, a^{2} c \tan \left (f x + e\right )}{48 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.92, size = 146, normalized size = 1.70 \[ \frac {a^2\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f}-\frac {\frac {c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {11\,c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{12}+\frac {53\,c\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12}+\frac {c\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]
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 \[ - a^{2} c \left (\int \left (- \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int \sec ^{4}{\left (e + f x \right )}\, dx + \int \sec ^{5}{\left (e + f x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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