Integrand size = 17, antiderivative size = 65 \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=\frac {15}{2} \sqrt {1-x} \sqrt {1+x}+\frac {5}{2} \sqrt {1-x} (1+x)^{3/2}+\frac {2 (1+x)^{5/2}}{\sqrt {1-x}}-\frac {15 \arcsin (x)}{2} \]
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Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {49, 52, 41, 222} \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=-\frac {15 \arcsin (x)}{2}+\frac {2 (x+1)^{5/2}}{\sqrt {1-x}}+\frac {5}{2} \sqrt {1-x} (x+1)^{3/2}+\frac {15}{2} \sqrt {1-x} \sqrt {x+1} \]
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Rule 41
Rule 49
Rule 52
Rule 222
Rubi steps \begin{align*} \text {integral}& = \frac {2 (1+x)^{5/2}}{\sqrt {1-x}}-5 \int \frac {(1+x)^{3/2}}{\sqrt {1-x}} \, dx \\ & = \frac {5}{2} \sqrt {1-x} (1+x)^{3/2}+\frac {2 (1+x)^{5/2}}{\sqrt {1-x}}-\frac {15}{2} \int \frac {\sqrt {1+x}}{\sqrt {1-x}} \, dx \\ & = \frac {15}{2} \sqrt {1-x} \sqrt {1+x}+\frac {5}{2} \sqrt {1-x} (1+x)^{3/2}+\frac {2 (1+x)^{5/2}}{\sqrt {1-x}}-\frac {15}{2} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx \\ & = \frac {15}{2} \sqrt {1-x} \sqrt {1+x}+\frac {5}{2} \sqrt {1-x} (1+x)^{3/2}+\frac {2 (1+x)^{5/2}}{\sqrt {1-x}}-\frac {15}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = \frac {15}{2} \sqrt {1-x} \sqrt {1+x}+\frac {5}{2} \sqrt {1-x} (1+x)^{3/2}+\frac {2 (1+x)^{5/2}}{\sqrt {1-x}}-\frac {15}{2} \sin ^{-1}(x) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=\frac {\sqrt {1-x^2} \left (-24+7 x+x^2\right )}{2 (-1+x)}+15 \arctan \left (\frac {\sqrt {1-x^2}}{-1+x}\right ) \]
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Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18
method | result | size |
risch | \(-\frac {\left (x^{3}+8 x^{2}-17 x -24\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}-\frac {15 \sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{2 \sqrt {1+x}\, \sqrt {1-x}}\) | \(77\) |
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none
Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=\frac {{\left (x^{2} + 7 \, x - 24\right )} \sqrt {x + 1} \sqrt {-x + 1} + 30 \, {\left (x - 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + 24 \, x - 24}{2 \, {\left (x - 1\right )}} \]
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Result contains complex when optimal does not.
Time = 6.99 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.12 \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=\begin {cases} 15 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} + \frac {i \left (x + 1\right )^{\frac {5}{2}}}{2 \sqrt {x - 1}} + \frac {5 i \left (x + 1\right )^{\frac {3}{2}}}{2 \sqrt {x - 1}} - \frac {15 i \sqrt {x + 1}}{\sqrt {x - 1}} & \text {for}\: \left |{x + 1}\right | > 2 \\- 15 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} - \frac {\left (x + 1\right )^{\frac {5}{2}}}{2 \sqrt {1 - x}} - \frac {5 \left (x + 1\right )^{\frac {3}{2}}}{2 \sqrt {1 - x}} + \frac {15 \sqrt {x + 1}}{\sqrt {1 - x}} & \text {otherwise} \end {cases} \]
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none
Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=-\frac {x^{3}}{2 \, \sqrt {-x^{2} + 1}} - \frac {4 \, x^{2}}{\sqrt {-x^{2} + 1}} + \frac {17 \, x}{2 \, \sqrt {-x^{2} + 1}} + \frac {12}{\sqrt {-x^{2} + 1}} - \frac {15}{2} \, \arcsin \left (x\right ) \]
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Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=\frac {{\left ({\left (x + 6\right )} {\left (x + 1\right )} - 30\right )} \sqrt {x + 1} \sqrt {-x + 1}}{2 \, {\left (x - 1\right )}} - 15 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]
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Timed out. \[ \int \frac {(1+x)^{5/2}}{(1-x)^{3/2}} \, dx=\int \frac {{\left (x+1\right )}^{5/2}}{{\left (1-x\right )}^{3/2}} \,d x \]
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