Optimal. Leaf size=150 \[ -\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac{7 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac{a^2 c^4 \tan ^3(e+f x) \sec ^3(e+f x)}{6 f}-\frac{a^2 c^4 \tan (e+f x) \sec ^3(e+f x)}{8 f}+\frac{a^2 c^4 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{5 a^2 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]
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Rubi [A] time = 0.24069, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3958, 2611, 3770, 2607, 30, 3768} \[ -\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac{7 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac{a^2 c^4 \tan ^3(e+f x) \sec ^3(e+f x)}{6 f}-\frac{a^2 c^4 \tan (e+f x) \sec ^3(e+f x)}{8 f}+\frac{a^2 c^4 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac{5 a^2 c^4 \tan (e+f x) \sec (e+f x)}{16 f} \]
Antiderivative was successfully verified.
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Rule 3958
Rule 2611
Rule 3770
Rule 2607
Rule 30
Rule 3768
Rubi steps
\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^4 \, dx &=\left (a^2 c^2\right ) \int \left (c^2 \sec (e+f x) \tan ^4(e+f x)-2 c^2 \sec ^2(e+f x) \tan ^4(e+f x)+c^2 \sec ^3(e+f x) \tan ^4(e+f x)\right ) \, dx\\ &=\left (a^2 c^4\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx+\left (a^2 c^4\right ) \int \sec ^3(e+f x) \tan ^4(e+f x) \, dx-\left (2 a^2 c^4\right ) \int \sec ^2(e+f x) \tan ^4(e+f x) \, dx\\ &=\frac{a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac{a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac{1}{2} \left (a^2 c^4\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx-\frac{1}{4} \left (3 a^2 c^4\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx-\frac{\left (2 a^2 c^4\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{3 a^2 c^4 \sec (e+f x) \tan (e+f x)}{8 f}-\frac{a^2 c^4 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac{a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac{a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac{1}{8} \left (a^2 c^4\right ) \int \sec ^3(e+f x) \, dx+\frac{1}{8} \left (3 a^2 c^4\right ) \int \sec (e+f x) \, dx\\ &=\frac{3 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{5 a^2 c^4 \sec (e+f x) \tan (e+f x)}{16 f}-\frac{a^2 c^4 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac{a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac{a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f}+\frac{1}{16} \left (a^2 c^4\right ) \int \sec (e+f x) \, dx\\ &=\frac{7 a^2 c^4 \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac{5 a^2 c^4 \sec (e+f x) \tan (e+f x)}{16 f}-\frac{a^2 c^4 \sec ^3(e+f x) \tan (e+f x)}{8 f}+\frac{a^2 c^4 \sec (e+f x) \tan ^3(e+f x)}{4 f}+\frac{a^2 c^4 \sec ^3(e+f x) \tan ^3(e+f x)}{6 f}-\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f}\\ \end{align*}
Mathematica [A] time = 1.33315, size = 91, normalized size = 0.61 \[ \frac{a^2 c^4 \left (1680 \tanh ^{-1}(\sin (e+f x))+(330 \sin (e+f x)-240 \sin (2 (e+f x))-445 \sin (3 (e+f x))+192 \sin (4 (e+f x))-135 \sin (5 (e+f x))-48 \sin (6 (e+f x))) \sec ^6(e+f x)\right )}{3840 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.03, size = 167, normalized size = 1.1 \begin{align*} -{\frac{{a}^{2}{c}^{4} \left ( \sec \left ( fx+e \right ) \right ) ^{3}\tan \left ( fx+e \right ) }{24\,f}}-{\frac{9\,{a}^{2}{c}^{4}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{16\,f}}+{\frac{7\,{a}^{2}{c}^{4}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{16\,f}}-{\frac{2\,{a}^{2}{c}^{4}\tan \left ( fx+e \right ) }{5\,f}}+{\frac{4\,{a}^{2}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{5\,f}}-{\frac{2\,{a}^{2}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5\,f}}+{\frac{{a}^{2}{c}^{4}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{5}}{6\,f}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 0.993228, size = 433, normalized size = 2.89 \begin{align*} -\frac{64 \,{\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^{2} c^{4} - 640 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{4} + 5 \, a^{2} c^{4}{\left (\frac{2 \,{\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 30 \, a^{2} c^{4}{\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^{2} c^{4}{\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 )} - 480 \, a^{2} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 960 \, a^{2} c^{4} \tan \left (f x + e\right )}{480 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.507252, size = 404, normalized size = 2.69 \begin{align*} \frac{105 \, a^{2} c^{4} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, a^{2} c^{4} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (96 \, a^{2} c^{4} \cos \left (f x + e\right )^{5} + 135 \, a^{2} c^{4} \cos \left (f x + e\right )^{4} - 192 \, a^{2} c^{4} \cos \left (f x + e\right )^{3} + 10 \, a^{2} c^{4} \cos \left (f x + e\right )^{2} + 96 \, a^{2} c^{4} \cos \left (f x + e\right ) - 40 \, a^{2} c^{4}\right )} \sin \left (f x + e\right )}{480 \, f \cos \left (f x + e\right )^{6}} \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^{2} c^{4} \left (\int \sec{\left (e + f x \right )}\, dx + \int - 2 \sec ^{2}{\left (e + f x \right )}\, dx + \int - \sec ^{3}{\left (e + f x \right )}\, dx + \int 4 \sec ^{4}{\left (e + f x \right )}\, dx + \int - \sec ^{5}{\left (e + f x \right )}\, dx + \int - 2 \sec ^{6}{\left (e + f x \right )}\, dx + \int \sec ^{7}{\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 = 1.35006, size = 252, normalized size = 1.68 \begin{align*} \frac{105 \, a^{2} c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 105 \, a^{2} c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (105 \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{11} - 595 \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 1686 \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 1386 \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 595 \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 105 \, a^{2} c^{4} \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 )}^{6}}}{240 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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