Integrand size = 16, antiderivative size = 106 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b}+\frac {5}{8} x^{3/2} \sqrt {2-b x}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{3/2}} \]
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Time = 0.02 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{3/2}}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {5}{8} x^{3/2} \sqrt {2-b x}-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b} \]
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Rule 52
Rule 56
Rule 222
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5}{4} \int \sqrt {x} (2-b x)^{3/2} \, dx \\ & = \frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5}{4} \int \sqrt {x} \sqrt {2-b x} \, dx \\ & = \frac {5}{8} x^{3/2} \sqrt {2-b x}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5}{8} \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b}+\frac {5}{8} x^{3/2} \sqrt {2-b x}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{8 b} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b}+\frac {5}{8} x^{3/2} \sqrt {2-b x}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{4 b} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b}+\frac {5}{8} x^{3/2} \sqrt {2-b x}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{3/2}} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\frac {\sqrt {x} \sqrt {2-b x} \left (-15+59 b x-34 b^2 x^2+6 b^3 x^3\right )}{24 b}-\frac {5 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{2 b^{3/2}} \]
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Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.84
method | result | size |
meijerg | \(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {3}{2}} \left (-6 b^{3} x^{3}+34 b^{2} x^{2}-59 b x +15\right ) \sqrt {-\frac {b x}{2}+1}}{24 b}-\frac {5 \sqrt {\pi }\, \left (-b \right )^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {3}{2}}}}{\sqrt {-b}\, \sqrt {\pi }\, b}\) | \(89\) |
default | \(\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {5}{2}}}{4}+\frac {5 x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {3}{2}}}{12}+\frac {5 x^{\frac {3}{2}} \sqrt {-b x +2}}{8}-\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{8 b}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\) | \(107\) |
risch | \(-\frac {\left (6 b^{3} x^{3}-34 b^{2} x^{2}+59 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{24 b \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\) | \(115\) |
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Time = 0.24 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.33 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\left [\frac {{\left (6 \, b^{4} x^{3} - 34 \, b^{3} x^{2} + 59 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{24 \, b^{2}}, \frac {{\left (6 \, b^{4} x^{3} - 34 \, b^{3} x^{2} + 59 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{24 \, b^{2}}\right ] \]
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Result contains complex when optimal does not.
Time = 9.92 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.39 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\begin {cases} \frac {i b^{3} x^{\frac {9}{2}}}{4 \sqrt {b x - 2}} - \frac {23 i b^{2} x^{\frac {7}{2}}}{12 \sqrt {b x - 2}} + \frac {127 i b x^{\frac {5}{2}}}{24 \sqrt {b x - 2}} - \frac {133 i x^{\frac {3}{2}}}{24 \sqrt {b x - 2}} + \frac {5 i \sqrt {x}}{4 b \sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {3}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b^{3} x^{\frac {9}{2}}}{4 \sqrt {- b x + 2}} + \frac {23 b^{2} x^{\frac {7}{2}}}{12 \sqrt {- b x + 2}} - \frac {127 b x^{\frac {5}{2}}}{24 \sqrt {- b x + 2}} + \frac {133 x^{\frac {3}{2}}}{24 \sqrt {- b x + 2}} - \frac {5 \sqrt {x}}{4 b \sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
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Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.37 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\frac {\frac {15 \, \sqrt {-b x + 2} b^{3}}{\sqrt {x}} + \frac {55 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} + \frac {73 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}}{12 \, {\left (b^{5} - \frac {4 \, {\left (b x - 2\right )} b^{4}}{x} + \frac {6 \, {\left (b x - 2\right )}^{2} b^{3}}{x^{2}} - \frac {4 \, {\left (b x - 2\right )}^{3} b^{2}}{x^{3}} + \frac {{\left (b x - 2\right )}^{4} b}{x^{4}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{4 \, b^{\frac {3}{2}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (73) = 146\).
Time = 23.34 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.35 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\frac {{\left ({\left ({\left (b x - 2\right )} {\left (2 \, {\left (b x - 2\right )} {\left (\frac {3 \, {\left (b x - 2\right )}}{b^{3}} + \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} + \frac {279}{b^{3}}\right )} \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b^{2}}\right )} b {\left | b \right |} - 24 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left ({\left (b x - 2\right )} {\left (\frac {2 \, {\left (b x - 2\right )}}{b^{2}} + \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} - \frac {30 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b}\right )} {\left | b \right |} + \frac {144 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} {\left (b x + 3\right )} \sqrt {-b x + 2} - \frac {6 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} {\left | b \right |}}{b^{2}} + \frac {192 \, {\left (\frac {2 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}} - \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2}\right )} {\left | b \right |}}{b^{2}}}{24 \, b} \]
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Timed out. \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\int \sqrt {x}\,{\left (2-b\,x\right )}^{5/2} \,d x \]
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