Integrand size = 23, antiderivative size = 100 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \sqrt {a+a \sec (c+d x)}}{a^2 d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{a^3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^4 d} \]
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Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3965, 90, 52, 65, 213} \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a^4 d}-\frac {2 (a \sec (c+d x)+a)^{3/2}}{a^3 d}+\frac {2 \sqrt {a \sec (c+d x)+a}}{a^2 d} \]
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Rule 52
Rule 65
Rule 90
Rule 213
Rule 3965
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(-a+a x)^2 \sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (-3 a^2 \sqrt {a+a x}+\frac {a^2 \sqrt {a+a x}}{x}+a (a+a x)^{3/2}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d} \\ & = -\frac {2 (a+a \sec (c+d x))^{3/2}}{a^3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^4 d}+\frac {\text {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = \frac {2 \sqrt {a+a \sec (c+d x)}}{a^2 d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{a^3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^4 d}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{a d} \\ & = \frac {2 \sqrt {a+a \sec (c+d x)}}{a^2 d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{a^3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^4 d}+\frac {2 \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{a^2 d} \\ & = -\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \sqrt {a+a \sec (c+d x)}}{a^2 d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{a^3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^4 d} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (1-2 \sec (c+d x)-2 \sec ^2(c+d x)+\sec ^3(c+d x)-5 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \sqrt {1+\sec (c+d x)}\right )}{5 a d \sqrt {a (1+\sec (c+d x))}} \]
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Time = 3.59 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+1-3 \sec \left (d x +c \right )+\sec \left (d x +c \right )^{2}\right )}{5 d \,a^{2}}\) | \(82\) |
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Time = 0.31 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.61 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {5 \, \sqrt {a} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{10 \, a^{2} d \cos \left (d x + c\right )^{2}}, \frac {5 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \, {\left (\cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{5 \, a^{2} d \cos \left (d x + c\right )^{2}}\right ] \]
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\[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {5 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a^{4}} - \frac {10 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a^{3}} + \frac {10 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{a^{2}}}{5 \, d} \]
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Time = 2.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.54 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (\frac {5 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {\sqrt {2} {\left (5 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} + 10 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a + 4 \, a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )}}{5 \, d} \]
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Timed out. \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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