Integrand size = 15, antiderivative size = 86 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \]
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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 = {49, 52, 56, 221} \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\frac {15 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {15 \sqrt {x} \sqrt {b x+2}}{2 b^3}+\frac {5 x^{3/2} \sqrt {b x+2}}{2 b^2}-\frac {2 x^{5/2}}{b \sqrt {b x+2}} \]
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Rule 49
Rule 52
Rule 56
Rule 221
Rubi steps \begin{align*} \text {integral}& = -\frac {2 x^{5/2}}{b \sqrt {2+b x}}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx}{b} \\ & = -\frac {2 x^{5/2}}{b \sqrt {2+b x}}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}-\frac {15 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{2 b^2} \\ & = -\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{2 b^3} \\ & = -\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b^3} \\ & = -\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (-30-5 b x+b^2 x^2\right )}{2 b^3 \sqrt {2+b x}}-\frac {30 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{b^{7/2}} \]
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Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73
method | result | size |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-\frac {7}{2} b^{2} x^{2}+\frac {35}{2} b x +105\right )}{14 \sqrt {\frac {b x}{2}+1}}+15 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\) | \(63\) |
risch | \(\frac {\left (b x -7\right ) \sqrt {x}\, \sqrt {b x +2}}{2 b^{3}}+\frac {\left (\frac {15 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 b^{\frac {7}{2}}}-\frac {8 \sqrt {b \left (x +\frac {2}{b}\right )^{2}-2 x -\frac {4}{b}}}{b^{4} \left (x +\frac {2}{b}\right )}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\) | \(106\) |
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Time = 0.24 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.77 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (b x + 2\right )} \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}, -\frac {30 \, {\left (b x + 2\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}\right ] \]
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Time = 7.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\frac {x^{\frac {5}{2}}}{2 b \sqrt {b x + 2}} - \frac {5 x^{\frac {3}{2}}}{2 b^{2} \sqrt {b x + 2}} - \frac {15 \sqrt {x}}{b^{3} \sqrt {b x + 2}} + \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} \]
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Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=-\frac {8 \, b^{2} - \frac {25 \, {\left (b x + 2\right )} b}{x} + \frac {15 \, {\left (b x + 2\right )}^{2}}{x^{2}}}{\frac {\sqrt {b x + 2} b^{5}}{\sqrt {x}} - \frac {2 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}} - \frac {15 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, b^{\frac {7}{2}}} \]
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Time = 1.56 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\frac {{\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (\frac {b x + 2}{b^{3}} - \frac {9}{b^{3}}\right )} - \frac {15 \, \log \left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{b^{\frac {5}{2}}} - \frac {64}{{\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )} b^{\frac {3}{2}}}\right )} {\left | b \right |}}{2 \, b^{2}} \]
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Timed out. \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\int \frac {x^{5/2}}{{\left (b\,x+2\right )}^{3/2}} \,d x \]
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