Integrand size = 21, antiderivative size = 86 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{15 a^3 x^{3/2}} \]
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Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2041, 2039} \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {16 b^2 \sqrt {a x^2+b x^3}}{15 a^3 x^{3/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}} \]
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Rule 2039
Rule 2041
Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}-\frac {(4 b) \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx}{5 a} \\ & = -\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}+\frac {\left (8 b^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a x^2+b x^3}} \, dx}{15 a^2} \\ & = -\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{15 a^3 x^{3/2}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \sqrt {x^2 (a+b x)} \left (3 a^2-4 a b x+8 b^2 x^2\right )}{15 a^3 x^{7/2}} \]
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Time = 1.82 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51
method | result | size |
risch | \(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 \sqrt {x^{2} \left (b x +a \right )}\, x^{\frac {3}{2}} a^{3}}\) | \(44\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 x^{\frac {3}{2}} a^{3} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(46\) |
default | \(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 x^{\frac {3}{2}} a^{3} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(46\) |
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Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \, {\left (8 \, b^{2} x^{2} - 4 \, a b x + 3 \, a^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, a^{3} x^{\frac {7}{2}}} \]
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\[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^{\frac {5}{2}} \sqrt {x^{2} \left (a + b x\right )}}\, dx \]
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\[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x^{2}} x^{\frac {5}{2}}} \,d x } \]
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Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\frac {32 \, {\left (10 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 5 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + a^{2}\right )} b^{\frac {5}{2}}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5} \mathrm {sgn}\left (x\right )} \]
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Timed out. \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^{5/2}\,\sqrt {b\,x^3+a\,x^2}} \,d x \]
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