Integrand size = 21, antiderivative size = 97 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d} \]
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Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3965, 52, 65, 213} \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a \sec (c+d x)+a}}{d}+\frac {2 a (a \sec (c+d x)+a)^{3/2}}{3 d}+\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 d} \]
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Rule 52
Rule 65
Rule 213
Rule 3965
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+a x)^{5/2}}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {a \text {Subst}\left (\int \frac {(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {a^2 \text {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {a^3 \text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {\left (2 a^2\right ) \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^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {2 (a (1+\sec (c+d x)))^{5/2} \left (-15 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (23+11 \sec (c+d x)+3 \sec ^2(c+d x)\right )\right )}{15 d (1+\sec (c+d x))^{5/2}} \]
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Time = 7.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {a +a \sec \left (d x +c \right )}\, a^{2}-2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\) | \(74\) |
default | \(\frac {\frac {2 \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {a +a \sec \left (d x +c \right )}\, a^{2}-2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\) | \(74\) |
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Time = 0.34 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.91 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\left [\frac {15 \, a^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (23 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{30 \, d \cos \left (d x + c\right )^{2}}, \frac {15 \, \sqrt {-a} a^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \, {\left (23 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}}\right ] \]
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\[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \tan {\left (c + d x \right )}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {15 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 6 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} + 10 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a + 30 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}}{15 \, d} \]
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\[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right ) \,d x } \]
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Time = 13.96 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{5\,d}+\frac {2\,a\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{3\,d}+\frac {2\,a^2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{d}+\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i}}{d} \]
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