Optimal. Leaf size=125 \[ \frac{a^3 (4 c+3 d) \tan ^3(e+f x)}{12 f}+\frac{a^3 (4 c+3 d) \tan (e+f x)}{f}+\frac{5 a^3 (4 c+3 d) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{3 a^3 (4 c+3 d) \tan (e+f x) \sec (e+f x)}{8 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^3}{4 f} \]
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Rubi [A] time = 0.152805, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4001, 3791, 3770, 3767, 8, 3768} \[ \frac{a^3 (4 c+3 d) \tan ^3(e+f x)}{12 f}+\frac{a^3 (4 c+3 d) \tan (e+f x)}{f}+\frac{5 a^3 (4 c+3 d) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{3 a^3 (4 c+3 d) \tan (e+f x) \sec (e+f x)}{8 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^3}{4 f} \]
Antiderivative was successfully verified.
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Rule 4001
Rule 3791
Rule 3770
Rule 3767
Rule 8
Rule 3768
Rubi steps
\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \, dx &=\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{4} (4 c+3 d) \int \sec (e+f x) (a+a \sec (e+f x))^3 \, dx\\ &=\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{4} (4 c+3 d) \int \left (a^3 \sec (e+f x)+3 a^3 \sec ^2(e+f x)+3 a^3 \sec ^3(e+f x)+a^3 \sec ^4(e+f x)\right ) \, dx\\ &=\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{4} \left (a^3 (4 c+3 d)\right ) \int \sec (e+f x) \, dx+\frac{1}{4} \left (a^3 (4 c+3 d)\right ) \int \sec ^4(e+f x) \, dx+\frac{1}{4} \left (3 a^3 (4 c+3 d)\right ) \int \sec ^2(e+f x) \, dx+\frac{1}{4} \left (3 a^3 (4 c+3 d)\right ) \int \sec ^3(e+f x) \, dx\\ &=\frac{a^3 (4 c+3 d) \tanh ^{-1}(\sin (e+f x))}{4 f}+\frac{3 a^3 (4 c+3 d) \sec (e+f x) \tan (e+f x)}{8 f}+\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{8} \left (3 a^3 (4 c+3 d)\right ) \int \sec (e+f x) \, dx-\frac{\left (a^3 (4 c+3 d)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{4 f}-\frac{\left (3 a^3 (4 c+3 d)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{4 f}\\ &=\frac{5 a^3 (4 c+3 d) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^3 (4 c+3 d) \tan (e+f x)}{f}+\frac{3 a^3 (4 c+3 d) \sec (e+f x) \tan (e+f x)}{8 f}+\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{a^3 (4 c+3 d) \tan ^3(e+f x)}{12 f}\\ \end{align*}
Mathematica [B] time = 1.30439, size = 273, normalized size = 2.18 \[ -\frac{a^3 (\cos (e+f x)+1)^3 \sec ^6\left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x) \left (120 (4 c+3 d) \cos ^4(e+f x) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-\sec (e) (-24 (11 c+9 d) \sin (e)+(36 c+69 d) \sin (f x)+36 c \sin (2 e+f x)+280 c \sin (e+2 f x)-72 c \sin (3 e+2 f x)+36 c \sin (2 e+3 f x)+36 c \sin (4 e+3 f x)+88 c \sin (3 e+4 f x)+69 d \sin (2 e+f x)+264 d \sin (e+2 f x)-24 d \sin (3 e+2 f x)+45 d \sin (2 e+3 f x)+45 d \sin (4 e+3 f x)+72 d \sin (3 e+4 f x))\right )}{1536 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.047, size = 188, normalized size = 1.5 \begin{align*}{\frac{5\,{a}^{3}c\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+3\,{\frac{{a}^{3}d\tan \left ( fx+e \right ) }{f}}+{\frac{11\,{a}^{3}c\tan \left ( fx+e \right ) }{3\,f}}+{\frac{15\,{a}^{3}d\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{15\,{a}^{3}d\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{3\,{a}^{3}c\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{3}d\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{{a}^{3}c\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}}+{\frac{{a}^{3}d\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 0.99547, size = 354, normalized size = 2.83 \begin{align*} \frac{16 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c + 48 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} d - 3 \, a^{3} d{\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 )} - 36 \, a^{3} 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 )} - 36 \, a^{3} d{\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^{3} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 144 \, a^{3} c \tan \left (f x + e\right ) + 48 \, a^{3} d \tan \left (f x + e\right )}{48 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.499817, size = 393, normalized size = 3.14 \begin{align*} \frac{15 \,{\left (4 \, a^{3} c + 3 \, a^{3} d\right )} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left (4 \, a^{3} c + 3 \, a^{3} d\right )} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (6 \, a^{3} d + 8 \,{\left (11 \, a^{3} c + 9 \, a^{3} d\right )} \cos \left (f x + e\right )^{3} + 9 \,{\left (4 \, a^{3} c + 5 \, a^{3} d\right )} \cos \left (f x + e\right )^{2} + 8 \,{\left (a^{3} c + 3 \, a^{3} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \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} \left (\int c \sec{\left (e + f x \right )}\, dx + \int 3 c \sec ^{2}{\left (e + f x \right )}\, dx + \int 3 c \sec ^{3}{\left (e + f x \right )}\, dx + \int c \sec ^{4}{\left (e + f x \right )}\, dx + \int d \sec ^{2}{\left (e + f x \right )}\, dx + \int 3 d \sec ^{3}{\left (e + f x \right )}\, dx + \int 3 d \sec ^{4}{\left (e + f x \right )}\, dx + \int d \sec ^{5}{\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 = 1.20533, size = 301, normalized size = 2.41 \begin{align*} \frac{15 \,{\left (4 \, a^{3} c + 3 \, a^{3} d\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 15 \,{\left (4 \, a^{3} c + 3 \, a^{3} d\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (60 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 45 \, a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 220 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 165 \, a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 292 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 219 \, a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 132 \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 147 \, a^{3} d \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 )}^{4}}}{24 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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