Integrand size = 15, antiderivative size = 59 \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=-\frac {\sqrt {2+b x}}{5 x^{5/2}}+\frac {2 b \sqrt {2+b x}}{15 x^{3/2}}-\frac {2 b^2 \sqrt {2+b x}}{15 \sqrt {x}} \]
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Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {47, 37} \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=-\frac {2 b^2 \sqrt {b x+2}}{15 \sqrt {x}}+\frac {2 b \sqrt {b x+2}}{15 x^{3/2}}-\frac {\sqrt {b x+2}}{5 x^{5/2}} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {2+b x}}{5 x^{5/2}}-\frac {1}{5} (2 b) \int \frac {1}{x^{5/2} \sqrt {2+b x}} \, dx \\ & = -\frac {\sqrt {2+b x}}{5 x^{5/2}}+\frac {2 b \sqrt {2+b x}}{15 x^{3/2}}+\frac {1}{15} \left (2 b^2\right ) \int \frac {1}{x^{3/2} \sqrt {2+b x}} \, dx \\ & = -\frac {\sqrt {2+b x}}{5 x^{5/2}}+\frac {2 b \sqrt {2+b x}}{15 x^{3/2}}-\frac {2 b^2 \sqrt {2+b x}}{15 \sqrt {x}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=\frac {\sqrt {2+b x} \left (-3+2 b x-2 b^2 x^2\right )}{15 x^{5/2}} \]
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Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.46
method | result | size |
gosper | \(-\frac {\sqrt {b x +2}\, \left (2 b^{2} x^{2}-2 b x +3\right )}{15 x^{\frac {5}{2}}}\) | \(27\) |
meijerg | \(-\frac {\sqrt {2}\, \left (\frac {2}{3} b^{2} x^{2}-\frac {2}{3} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{5 x^{\frac {5}{2}}}\) | \(31\) |
risch | \(-\frac {2 b^{3} x^{3}+2 b^{2} x^{2}-b x +6}{15 x^{\frac {5}{2}} \sqrt {b x +2}}\) | \(35\) |
default | \(-\frac {\sqrt {b x +2}}{5 x^{\frac {5}{2}}}-\frac {2 b \left (-\frac {\sqrt {b x +2}}{3 x^{\frac {3}{2}}}+\frac {b \sqrt {b x +2}}{3 \sqrt {x}}\right )}{5}\) | \(43\) |
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Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=-\frac {{\left (2 \, b^{2} x^{2} - 2 \, b x + 3\right )} \sqrt {b x + 2}}{15 \, x^{\frac {5}{2}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (53) = 106\).
Time = 3.87 (sec) , antiderivative size = 224, normalized size of antiderivative = 3.80 \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=- \frac {2 b^{\frac {17}{2}} x^{4} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {6 b^{\frac {15}{2}} x^{3} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {3 b^{\frac {13}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {4 b^{\frac {11}{2}} x \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {12 b^{\frac {9}{2}} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} \]
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Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=-\frac {\sqrt {b x + 2} b^{2}}{4 \, \sqrt {x}} + \frac {{\left (b x + 2\right )}^{\frac {3}{2}} b}{6 \, x^{\frac {3}{2}}} - \frac {{\left (b x + 2\right )}^{\frac {5}{2}}}{20 \, x^{\frac {5}{2}}} \]
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Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=-\frac {{\left (15 \, b^{5} + 2 \, {\left ({\left (b x + 2\right )} b^{5} - 5 \, b^{5}\right )} {\left (b x + 2\right )}\right )} \sqrt {b x + 2} b}{15 \, {\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac {5}{2}} {\left | b \right |}} \]
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Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx=-\frac {\sqrt {b\,x+2}\,\left (\frac {2\,b^2\,x^2}{15}-\frac {2\,b\,x}{15}+\frac {1}{5}\right )}{x^{5/2}} \]
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