Optimal. Leaf size=94 \[ \frac{a^3 c^2 \tan ^5(e+f x)}{5 f}+\frac{3 a^3 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^3 c^2 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{3 a^3 c^2 \tan (e+f x) \sec (e+f x)}{8 f} \]
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Rubi [A] time = 0.147801, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3958, 2611, 3770, 2607, 30} \[ \frac{a^3 c^2 \tan ^5(e+f x)}{5 f}+\frac{3 a^3 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^3 c^2 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{3 a^3 c^2 \tan (e+f x) \sec (e+f x)}{8 f} \]
Antiderivative was successfully verified.
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Rule 3958
Rule 2611
Rule 3770
Rule 2607
Rule 30
Rubi steps
\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \left (a \sec (e+f x) \tan ^4(e+f x)+a \sec ^2(e+f x) \tan ^4(e+f x)\right ) \, dx\\ &=\left (a^3 c^2\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx+\left (a^3 c^2\right ) \int \sec ^2(e+f x) \tan ^4(e+f x) \, dx\\ &=\frac{a^3 c^2 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac{1}{4} \left (3 a^3 c^2\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx+\frac{\left (a^3 c^2\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{3 a^3 c^2 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^3 c^2 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac{a^3 c^2 \tan ^5(e+f x)}{5 f}+\frac{1}{8} \left (3 a^3 c^2\right ) \int \sec (e+f x) \, dx\\ &=\frac{3 a^3 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{3 a^3 c^2 \sec (e+f x) \tan (e+f x)}{8 f}+\frac{a^3 c^2 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac{a^3 c^2 \tan ^5(e+f x)}{5 f}\\ \end{align*}
Mathematica [A] time = 0.848097, size = 81, normalized size = 0.86 \[ \frac{a^3 c^2 \left (120 \tanh ^{-1}(\sin (e+f x))+(40 \sin (e+f x)-10 \sin (2 (e+f x))-20 \sin (3 (e+f x))-25 \sin (4 (e+f x))+4 \sin (5 (e+f x))) \sec ^5(e+f x)\right )}{320 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.027, size = 142, normalized size = 1.5 \begin{align*} -{\frac{5\,{a}^{3}{c}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{3\,{a}^{3}{c}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{{a}^{3}{c}^{2}\tan \left ( fx+e \right ) }{5\,f}}-{\frac{2\,{a}^{3}{c}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{5\,f}}+{\frac{{a}^{3}{c}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{{a}^{3}{c}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5\,f}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 0.987646, size = 306, normalized size = 3.26 \begin{align*} \frac{16 \,{\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{3} c^{2} - 160 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{2} - 15 \, a^{3} c^{2}{\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 )} + 120 \, a^{3} c^{2}{\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 )} + 240 \, a^{3} c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 240 \, a^{3} c^{2} \tan \left (f x + e\right )}{240 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.493915, size = 356, normalized size = 3.79 \begin{align*} \frac{15 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (8 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} - 25 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} - 16 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} + 10 \, a^{3} c^{2} \cos \left (f x + e\right ) + 8 \, a^{3} c^{2}\right )} \sin \left (f x + e\right )}{80 \, f \cos \left (f x + e\right )^{5}} \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*} a^{3} c^{2} \left (\int \sec{\left (e + f x \right )}\, dx + \int \sec ^{2}{\left (e + f x \right )}\, dx + \int - 2 \sec ^{3}{\left (e + f x \right )}\, dx + \int - 2 \sec ^{4}{\left (e + f x \right )}\, dx + \int \sec ^{5}{\left (e + f x \right )}\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.35731, size = 225, normalized size = 2.39 \begin{align*} \frac{15 \, a^{3} c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, a^{3} c^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 70 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 128 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 70 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 15 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{5}}}{40 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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