Integrand size = 16, antiderivative size = 56 \[ \int x \sqrt {8+2 x-x^2} \, dx=-\frac {1}{2} (1-x) \sqrt {8+2 x-x^2}-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}-\frac {9}{2} \arcsin \left (\frac {1-x}{3}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {654, 626, 633, 222} \[ \int x \sqrt {8+2 x-x^2} \, dx=-\frac {9}{2} \arcsin \left (\frac {1-x}{3}\right )-\frac {1}{3} \left (-x^2+2 x+8\right )^{3/2}-\frac {1}{2} (1-x) \sqrt {-x^2+2 x+8} \]
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Rule 222
Rule 626
Rule 633
Rule 654
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}+\int \sqrt {8+2 x-x^2} \, dx \\ & = -\frac {1}{2} (1-x) \sqrt {8+2 x-x^2}-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}+\frac {9}{2} \int \frac {1}{\sqrt {8+2 x-x^2}} \, dx \\ & = -\frac {1}{2} (1-x) \sqrt {8+2 x-x^2}-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{36}}} \, dx,x,2-2 x\right ) \\ & = -\frac {1}{2} (1-x) \sqrt {8+2 x-x^2}-\frac {1}{3} \left (8+2 x-x^2\right )^{3/2}-\frac {9}{2} \sin ^{-1}\left (\frac {1-x}{3}\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int x \sqrt {8+2 x-x^2} \, dx=\frac {1}{6} \sqrt {8+2 x-x^2} \left (-19-x+2 x^2\right )-9 \arctan \left (\frac {\sqrt {8+2 x-x^2}}{2+x}\right ) \]
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Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75
method | result | size |
risch | \(-\frac {\left (2 x^{2}-x -19\right ) \left (x^{2}-2 x -8\right )}{6 \sqrt {-x^{2}+2 x +8}}+\frac {9 \arcsin \left (-\frac {1}{3}+\frac {x}{3}\right )}{2}\) | \(42\) |
default | \(-\frac {\left (-x^{2}+2 x +8\right )^{\frac {3}{2}}}{3}-\frac {\left (2-2 x \right ) \sqrt {-x^{2}+2 x +8}}{4}+\frac {9 \arcsin \left (-\frac {1}{3}+\frac {x}{3}\right )}{2}\) | \(43\) |
trager | \(\left (\frac {1}{3} x^{2}-\frac {1}{6} x -\frac {19}{6}\right ) \sqrt {-x^{2}+2 x +8}+\frac {9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +\sqrt {-x^{2}+2 x +8}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{2}\) | \(62\) |
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Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int x \sqrt {8+2 x-x^2} \, dx=\frac {1}{6} \, {\left (2 \, x^{2} - x - 19\right )} \sqrt {-x^{2} + 2 \, x + 8} - \frac {9}{2} \, \arctan \left (\frac {\sqrt {-x^{2} + 2 \, x + 8} {\left (x - 1\right )}}{x^{2} - 2 \, x - 8}\right ) \]
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Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.61 \[ \int x \sqrt {8+2 x-x^2} \, dx=\sqrt {- x^{2} + 2 x + 8} \left (\frac {x^{2}}{3} - \frac {x}{6} - \frac {19}{6}\right ) + \frac {9 \operatorname {asin}{\left (\frac {x}{3} - \frac {1}{3} \right )}}{2} \]
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Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int x \sqrt {8+2 x-x^2} \, dx=-\frac {1}{3} \, {\left (-x^{2} + 2 \, x + 8\right )}^{\frac {3}{2}} + \frac {1}{2} \, \sqrt {-x^{2} + 2 \, x + 8} x - \frac {1}{2} \, \sqrt {-x^{2} + 2 \, x + 8} - \frac {9}{2} \, \arcsin \left (-\frac {1}{3} \, x + \frac {1}{3}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.57 \[ \int x \sqrt {8+2 x-x^2} \, dx=\frac {1}{6} \, {\left ({\left (2 \, x - 1\right )} x - 19\right )} \sqrt {-x^{2} + 2 \, x + 8} + \frac {9}{2} \, \arcsin \left (\frac {1}{3} \, x - \frac {1}{3}\right ) \]
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Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.84 \[ \int x \sqrt {8+2 x-x^2} \, dx=-\frac {\sqrt {-x^2+2\,x+8}\,\left (-8\,x^2+4\,x+76\right )}{24}-\frac {\ln \left (x-1-\sqrt {-x^2+2\,x+8}\,1{}\mathrm {i}\right )\,9{}\mathrm {i}}{2} \]
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