Integrand size = 23, antiderivative size = 169 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d} \]
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Time = 0.17 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3965, 90, 52, 65, 213} \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 (a \sec (c+d x)+a)^{11/2}}{11 a^4 d}-\frac {2 (a \sec (c+d x)+a)^{9/2}}{3 a^3 d}+\frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a^2 d}+\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a d}+\frac {2 (a \sec (c+d x)+a)^{3/2}}{3 d}+\frac {2 a \sqrt {a \sec (c+d x)+a}}{d} \]
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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 (a+a x)^{7/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (-3 a^2 (a+a x)^{7/2}+\frac {a^2 (a+a x)^{7/2}}{x}+a (a+a x)^{9/2}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d} \\ & = -\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d}+\frac {\text {Subst}\left (\int \frac {(a+a x)^{7/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = \frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d}+\frac {\text {Subst}\left (\int \frac {(a+a x)^{5/2}}{x} \, dx,x,\sec (c+d x)\right )}{a d} \\ & = \frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d}+\frac {\text {Subst}\left (\int \frac {(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d}+\frac {a \text {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 a \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d}+\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 a \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d} \\ & = -\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d} \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\frac {2 (a (1+\sec (c+d x)))^{3/2} \left (-1155 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (1656+327 \sec (c+d x)-534 \sec ^2(c+d x)-325 \sec ^3(c+d x)+140 \sec ^4(c+d x)+105 \sec ^5(c+d x)\right )\right )}{1155 d (1+\sec (c+d x))^{3/2}} \]
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Time = 4.93 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66
method | result | size |
default | \(\frac {2 a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (1155 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+1656+327 \sec \left (d x +c \right )-534 \sec \left (d x +c \right )^{2}-325 \sec \left (d x +c \right )^{3}+140 \sec \left (d x +c \right )^{4}+105 \sec \left (d x +c \right )^{5}\right )}{1155 d}\) | \(112\) |
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Time = 0.36 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.98 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\left [\frac {1155 \, a^{\frac {3}{2}} \cos \left (d x + c\right )^{5} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (1656 \, a \cos \left (d x + c\right )^{5} + 327 \, a \cos \left (d x + c\right )^{4} - 534 \, a \cos \left (d x + c\right )^{3} - 325 \, a \cos \left (d x + c\right )^{2} + 140 \, a \cos \left (d x + c\right ) + 105 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2310 \, d \cos \left (d x + c\right )^{5}}, \frac {1155 \, \sqrt {-a} a \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{5} + 2 \, {\left (1656 \, a \cos \left (d x + c\right )^{5} + 327 \, a \cos \left (d x + c\right )^{4} - 534 \, a \cos \left (d x + c\right )^{3} - 325 \, a \cos \left (d x + c\right )^{2} + 140 \, a \cos \left (d x + c\right ) + 105 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{1155 \, d \cos \left (d x + c\right )^{5}}\right ] \]
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\[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \tan ^{5}{\left (c + d x \right )}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.96 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\frac {1155 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 770 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} + \frac {210 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {11}{2}}}{a^{4}} - \frac {770 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{a^{3}} + \frac {330 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a^{2}} + \frac {462 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a} + 2310 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a}{1155 \, d} \]
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\[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{5} \,d x } \]
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Timed out. \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]
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