Integrand size = 20, antiderivative size = 78 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11 \arcsin (x)}{2} \]
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Time = 0.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {99, 155, 152, 41, 222} \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\frac {11 \arcsin (x)}{2}+\frac {2 \sqrt {x+1} x^3}{3 (1-x)^{3/2}}-\frac {13 \sqrt {x+1} x^2}{3 \sqrt {1-x}}-\frac {1}{6} \sqrt {1-x} \sqrt {x+1} (33 x+52) \]
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Rule 41
Rule 99
Rule 152
Rule 155
Rule 222
Rubi steps \begin{align*} \text {integral}& = \frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {2}{3} \int \frac {x^2 \left (3+\frac {7 x}{2}\right )}{(1-x)^{3/2} \sqrt {1+x}} \, dx \\ & = -\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {2}{3} \int \frac {\left (-13-\frac {33 x}{2}\right ) x}{\sqrt {1-x} \sqrt {1+x}} \, dx \\ & = -\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11}{2} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx \\ & = -\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11}{2} \sin ^{-1}(x) \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {\sqrt {1+x} \left (52-71 x+12 x^2+3 x^3\right )}{6 (1-x)^{3/2}}+11 \arctan \left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \]
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Time = 0.60 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {\left (3 x^{4}+15 x^{3}-59 x^{2}-19 x +52\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{6 \left (-1+x \right ) \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}+\frac {11 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {1-x}\, \sqrt {1+x}}\) | \(89\) |
default | \(\frac {\left (-3 x^{3} \sqrt {-x^{2}+1}+33 \arcsin \left (x \right ) x^{2}-12 x^{2} \sqrt {-x^{2}+1}-66 \arcsin \left (x \right ) x +71 x \sqrt {-x^{2}+1}+33 \arcsin \left (x \right )-52 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{6 \left (-1+x \right )^{2} \sqrt {-x^{2}+1}}\) | \(97\) |
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Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {52 \, x^{2} + {\left (3 \, x^{3} + 12 \, x^{2} - 71 \, x + 52\right )} \sqrt {x + 1} \sqrt {-x + 1} + 66 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 104 \, x + 52}{6 \, {\left (x^{2} - 2 \, x + 1\right )}} \]
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\[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\int \frac {x^{3} \sqrt {x + 1}}{\left (1 - x\right )^{\frac {5}{2}}}\, dx \]
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Timed out. \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\text {Timed out} \]
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Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.63 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {{\left ({\left (3 \, {\left (x + 2\right )} {\left (x + 1\right )} - 86\right )} {\left (x + 1\right )} + 132\right )} \sqrt {x + 1} \sqrt {-x + 1}}{6 \, {\left (x - 1\right )}^{2}} + 11 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]
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Timed out. \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\int \frac {x^3\,\sqrt {x+1}}{{\left (1-x\right )}^{5/2}} \,d x \]
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