Optimal. Leaf size=169 \[ \frac{3 a^3 \tan ^5(c+d x)}{5 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}-\frac{a^3 \tan (c+d x)}{d}+\frac{19 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac{a^3 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac{3 a^3 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{17 a^3 \tan (c+d x) \sec (c+d x)}{16 d}+a^3 x \]
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Rubi [A] time = 0.224364, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3886, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ \frac{3 a^3 \tan ^5(c+d x)}{5 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}-\frac{a^3 \tan (c+d x)}{d}+\frac{19 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 \tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac{a^3 \tan (c+d x) \sec ^3(c+d x)}{8 d}+\frac{3 a^3 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{17 a^3 \tan (c+d x) \sec (c+d x)}{16 d}+a^3 x \]
Antiderivative was successfully verified.
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Rule 3886
Rule 3473
Rule 8
Rule 2611
Rule 3770
Rule 2607
Rule 30
Rule 3768
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^3 \tan ^4(c+d x) \, dx &=\int \left (a^3 \tan ^4(c+d x)+3 a^3 \sec (c+d x) \tan ^4(c+d x)+3 a^3 \sec ^2(c+d x) \tan ^4(c+d x)+a^3 \sec ^3(c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a^3 \int \tan ^4(c+d x) \, dx+a^3 \int \sec ^3(c+d x) \tan ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sec (c+d x) \tan ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \tan ^4(c+d x) \, dx\\ &=\frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}-\frac{1}{2} a^3 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx-a^3 \int \tan ^2(c+d x) \, dx-\frac{1}{4} \left (9 a^3\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{a^3 \tan (c+d x)}{d}-\frac{9 a^3 \sec (c+d x) \tan (c+d x)}{8 d}-\frac{a^3 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac{3 a^3 \tan ^5(c+d x)}{5 d}+\frac{1}{8} a^3 \int \sec ^3(c+d x) \, dx+a^3 \int 1 \, dx+\frac{1}{8} \left (9 a^3\right ) \int \sec (c+d x) \, dx\\ &=a^3 x+\frac{9 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{a^3 \tan (c+d x)}{d}-\frac{17 a^3 \sec (c+d x) \tan (c+d x)}{16 d}-\frac{a^3 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac{3 a^3 \tan ^5(c+d x)}{5 d}+\frac{1}{16} a^3 \int \sec (c+d x) \, dx\\ &=a^3 x+\frac{19 a^3 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{a^3 \tan (c+d x)}{d}-\frac{17 a^3 \sec (c+d x) \tan (c+d x)}{16 d}-\frac{a^3 \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{3 a^3 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{a^3 \sec ^3(c+d x) \tan ^3(c+d x)}{6 d}+\frac{3 a^3 \tan ^5(c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 1.40425, size = 303, normalized size = 1.79 \[ \frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (\sec (c) (210 \sin (2 c+d x)-1440 \sin (c+2 d x)+1200 \sin (3 c+2 d x)-865 \sin (2 c+3 d x)-865 \sin (4 c+3 d x)-1296 \sin (3 c+4 d x)-240 \sin (5 c+4 d x)-435 \sin (4 c+5 d x)-435 \sin (6 c+5 d x)-176 \sin (5 c+6 d x)+2400 d x \cos (c)+1800 d x \cos (c+2 d x)+1800 d x \cos (3 c+2 d x)+720 d x \cos (3 c+4 d x)+720 d x \cos (5 c+4 d x)+120 d x \cos (5 c+6 d x)+120 d x \cos (7 c+6 d x)+1760 \sin (c)+210 \sin (d x))-9120 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{61440 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.049, size = 193, normalized size = 1.1 \begin{align*}{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}+{a}^{3}x+{\frac{{a}^{3}c}{d}}+{\frac{19\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{24\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{19\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{48\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{19\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{48\,d}}-{\frac{19\,{a}^{3}\sin \left ( dx+c \right ) }{16\,d}}+{\frac{19\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.68206, size = 284, normalized size = 1.68 \begin{align*} \frac{288 \, a^{3} \tan \left (d x + c\right )^{5} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - 5 \, a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{5} + 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 90 \, a^{3}{\left (\frac{2 \,{\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.23788, size = 404, normalized size = 2.39 \begin{align*} \frac{480 \, a^{3} d x \cos \left (d x + c\right )^{6} + 285 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 285 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (176 \, a^{3} \cos \left (d x + c\right )^{5} + 435 \, a^{3} \cos \left (d x + c\right )^{4} + 208 \, a^{3} \cos \left (d x + c\right )^{3} - 110 \, a^{3} \cos \left (d x + c\right )^{2} - 144 \, a^{3} \cos \left (d x + c\right ) - 40 \, a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\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^{3} \left (\int 3 \tan ^{4}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 3 \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.71284, size = 221, normalized size = 1.31 \begin{align*} \frac{240 \,{\left (d x + c\right )} a^{3} + 285 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 285 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 95 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 366 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1746 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3135 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 525 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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