Integrand size = 13, antiderivative size = 35 \[ \int \sqrt {5 x-9 x^2} \, dx=-\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}-\frac {25}{216} \arcsin \left (1-\frac {18 x}{5}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {626, 633, 222} \[ \int \sqrt {5 x-9 x^2} \, dx=-\frac {25}{216} \arcsin \left (1-\frac {18 x}{5}\right )-\frac {1}{36} \sqrt {5 x-9 x^2} (5-18 x) \]
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Rule 222
Rule 626
Rule 633
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}+\frac {25}{72} \int \frac {1}{\sqrt {5 x-9 x^2}} \, dx \\ & = -\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}-\frac {5}{216} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{25}}} \, dx,x,5-18 x\right ) \\ & = -\frac {1}{36} (5-18 x) \sqrt {5 x-9 x^2}-\frac {25}{216} \sin ^{-1}\left (1-\frac {18 x}{5}\right ) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{108} \sqrt {-x (-5+9 x)} \left (-15+54 x+\frac {25 \log \left (-3 \sqrt {x}+\sqrt {-5+9 x}\right )}{\sqrt {x} \sqrt {-5+9 x}}\right ) \]
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Time = 2.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {25 \arcsin \left (-1+\frac {18 x}{5}\right )}{216}-\frac {\left (5-18 x \right ) \sqrt {-9 x^{2}+5 x}}{36}\) | \(28\) |
risch | \(-\frac {\left (-5+18 x \right ) x \left (9 x -5\right )}{36 \sqrt {-x \left (9 x -5\right )}}+\frac {25 \arcsin \left (-1+\frac {18 x}{5}\right )}{216}\) | \(33\) |
pseudoelliptic | \(-\frac {25 \arctan \left (\frac {\sqrt {-9 x^{2}+5 x}}{3 x}\right )}{108}+\frac {\left (-5+18 x \right ) \sqrt {-9 x^{2}+5 x}}{36}\) | \(39\) |
meijerg | \(-\frac {25 i \left (-\frac {i \sqrt {\pi }\, \sqrt {x}\, \sqrt {5}\, \left (-\frac {54 x}{5}+3\right ) \sqrt {-\frac {9 x}{5}+1}}{10}+\frac {i \sqrt {\pi }\, \arcsin \left (\frac {3 \sqrt {5}\, \sqrt {x}}{5}\right )}{2}\right )}{54 \sqrt {\pi }}\) | \(47\) |
trager | \(\left (-\frac {5}{36}+\frac {x}{2}\right ) \sqrt {-9 x^{2}+5 x}-\frac {25 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (18 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+6 \sqrt {-9 x^{2}+5 x}\right )}{216}\) | \(59\) |
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Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{36} \, \sqrt {-9 \, x^{2} + 5 \, x} {\left (18 \, x - 5\right )} - \frac {25}{108} \, \arctan \left (\frac {\sqrt {-9 \, x^{2} + 5 \, x}}{3 \, x}\right ) \]
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \sqrt {5 x-9 x^2} \, dx=\left (\frac {x}{2} - \frac {5}{36}\right ) \sqrt {- 9 x^{2} + 5 x} + \frac {25 \operatorname {asin}{\left (\frac {18 x}{5} - 1 \right )}}{216} \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{2} \, \sqrt {-9 \, x^{2} + 5 \, x} x - \frac {5}{36} \, \sqrt {-9 \, x^{2} + 5 \, x} - \frac {25}{216} \, \arcsin \left (-\frac {18}{5} \, x + 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {1}{36} \, \sqrt {-9 \, x^{2} + 5 \, x} {\left (18 \, x - 5\right )} + \frac {25}{216} \, \arcsin \left (\frac {18}{5} \, x - 1\right ) \]
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Time = 9.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \sqrt {5 x-9 x^2} \, dx=\frac {25\,\mathrm {asin}\left (\frac {18\,x}{5}-1\right )}{216}+\left (\frac {x}{2}-\frac {5}{36}\right )\,\sqrt {5\,x-9\,x^2} \]
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