Integrand size = 15, antiderivative size = 86 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {15}{4} \sqrt {a} b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {43, 52, 65, 211} \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=-\frac {15}{4} \sqrt {a} b^2 \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )+\frac {15}{4} b^2 \sqrt {b x-a}-\frac {(b x-a)^{5/2}}{2 x^2}-\frac {5 b (b x-a)^{3/2}}{4 x} \]
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Rule 43
Rule 52
Rule 65
Rule 211
Rubi steps \begin{align*} \text {integral}& = -\frac {(-a+b x)^{5/2}}{2 x^2}+\frac {1}{4} (5 b) \int \frac {(-a+b x)^{3/2}}{x^2} \, dx \\ & = -\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}+\frac {1}{8} \left (15 b^2\right ) \int \frac {\sqrt {-a+b x}}{x} \, dx \\ & = \frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {1}{8} \left (15 a b^2\right ) \int \frac {1}{x \sqrt {-a+b x}} \, dx \\ & = \frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {1}{4} (15 a b) \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right ) \\ & = \frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.78 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {-a+b x} \left (-2 a^2+9 a b x+8 b^2 x^2\right )}{x^2}-15 \sqrt {a} b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )\right ) \]
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Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {-15 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) \sqrt {a}\, x^{2}-\left (-8 b^{2} x^{2}-9 a b x +2 a^{2}\right ) \sqrt {b x -a}}{4 x^{2}}\) | \(62\) |
risch | \(\frac {a \left (-b x +a \right ) \left (-9 b x +2 a \right )}{4 x^{2} \sqrt {b x -a}}-\frac {15 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) \sqrt {a}}{4}+2 b^{2} \sqrt {b x -a}\) | \(67\) |
derivativedivides | \(2 b^{2} \left (\sqrt {b x -a}-a \left (\frac {-\frac {9 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\) | \(70\) |
default | \(2 b^{2} \left (\sqrt {b x -a}-a \left (\frac {-\frac {9 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\) | \(70\) |
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Time = 0.23 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.62 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\left [\frac {15 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, x^{2}}, -\frac {15 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, x^{2}}\right ] \]
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Result contains complex when optimal does not.
Time = 3.62 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.10 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\begin {cases} - \frac {15 i \sqrt {a} b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {i a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {11 i a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {2 i b^{\frac {5}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} - 1}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {15 \sqrt {a} b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} + \frac {a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {11 a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {2 b^{\frac {5}{2}} \sqrt {x}}{\sqrt {- \frac {a}{b x} + 1}} & \text {otherwise} \end {cases} \]
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Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=-\frac {15}{4} \, \sqrt {a} b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + 2 \, \sqrt {b x - a} b^{2} + \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{2} + 7 \, \sqrt {b x - a} a^{2} b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} + 2 \, {\left (b x - a\right )} a + a^{2}\right )}} \]
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Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=-\frac {15 \, \sqrt {a} b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - 8 \, \sqrt {b x - a} b^{3} - \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{3} + 7 \, \sqrt {b x - a} a^{2} b^{3}}{b^{2} x^{2}}}{4 \, b} \]
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Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=2\,b^2\,\sqrt {b\,x-a}-\frac {15\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4}+\frac {9\,a\,{\left (b\,x-a\right )}^{3/2}}{4\,x^2}+\frac {7\,a^2\,\sqrt {b\,x-a}}{4\,x^2} \]
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