Integrand size = 21, antiderivative size = 51 \[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {a+a \sec (c+d x)}}{d} \]
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Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3965, 52, 65, 213} \[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=\frac {2 \sqrt {a \sec (c+d x)+a}}{d}-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d} \]
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Rule 52
Rule 65
Rule 213
Rule 3965
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {a \text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d} \\ & = -\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {a+a \sec (c+d x)}}{d} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=\frac {\sqrt {a (1+\sec (c+d x))} \left (-2 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+2 \sqrt {1+\sec (c+d x)}\right )}{d \sqrt {1+\sec (c+d x)}} \]
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Time = 0.93 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {2 \sqrt {a +a \sec \left (d x +c \right )}-2 \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\) | \(42\) |
default | \(\frac {2 \sqrt {a +a \sec \left (d x +c \right )}-2 \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\) | \(42\) |
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none
Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 3.61 \[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=\left [\frac {\sqrt {a} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, d}, \frac {\sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{d}\right ] \]
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\[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \tan {\left (c + d x \right )}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.31 \[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=\frac {\sqrt {a} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 2 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{d} \]
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\[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \tan \left (d x + c\right ) \,d x } \]
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Time = 14.58 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx=\frac {2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{d}-\frac {2\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{\sqrt {a}}\right )}{d} \]
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