Integrand size = 25, antiderivative size = 55 \[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {\arcsin (a x)}{a^3} \]
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Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1653, 12, 807, 222} \[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=-\frac {\arcsin (a x)}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (a x+1)}-\frac {\sqrt {1-a^2 x^2}}{a^3} \]
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Rule 12
Rule 222
Rule 807
Rule 1653
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\int \frac {a^3 x}{(1+a x) \sqrt {1-a^2 x^2}} \, dx}{a^4} \\ & = -\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\int \frac {x}{(1+a x) \sqrt {1-a^2 x^2}} \, dx}{a} \\ & = -\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^2} \\ & = -\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {\sin ^{-1}(a x)}{a^3} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.07 \[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\frac {(-2-a x) \sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {2 \arctan \left (\frac {a x}{-1+\sqrt {1-a^2 x^2}}\right )}{a^3} \]
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Time = 0.37 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.53
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}}{a^{3}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2}+2 \left (x +\frac {1}{a}\right ) a}}{a^{4} \left (x +\frac {1}{a}\right )}\) | \(84\) |
risch | \(\frac {a^{2} x^{2}-1}{a^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2}+2 \left (x +\frac {1}{a}\right ) a}}{a^{4} \left (x +\frac {1}{a}\right )}\) | \(92\) |
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Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=-\frac {2 \, a x - 2 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x + 2\right )} + 2}{a^{4} x + a^{3}} \]
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\[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\int \frac {x^{2}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4} x + a^{3}} - \frac {\arcsin \left (a x\right )}{a^{3}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{3}} \]
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Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.27 \[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=-\frac {\arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{a^{2} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{3}} + \frac {2}{a^{2} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]
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Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.53 \[ \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\frac {\sqrt {1-a^2\,x^2}}{\left (a\,\sqrt {-a^2}+a^2\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^2\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a^3} \]
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