Integrand size = 13, antiderivative size = 45 \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=-\frac {2 a^3}{\sqrt {x}}+6 a^2 b \sqrt {x}+2 a b^2 x^{3/2}+\frac {2}{5} b^3 x^{5/2} \]
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Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=-\frac {2 a^3}{\sqrt {x}}+6 a^2 b \sqrt {x}+2 a b^2 x^{3/2}+\frac {2}{5} b^3 x^{5/2} \]
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Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^3}{x^{3/2}}+\frac {3 a^2 b}{\sqrt {x}}+3 a b^2 \sqrt {x}+b^3 x^{3/2}\right ) \, dx \\ & = -\frac {2 a^3}{\sqrt {x}}+6 a^2 b \sqrt {x}+2 a b^2 x^{3/2}+\frac {2}{5} b^3 x^{5/2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=-\frac {2 \left (5 a^3-15 a^2 b x-5 a b^2 x^2-b^3 x^3\right )}{5 \sqrt {x}} \]
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Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80
method | result | size |
gosper | \(-\frac {2 \left (-b^{3} x^{3}-5 a \,b^{2} x^{2}-15 a^{2} b x +5 a^{3}\right )}{5 \sqrt {x}}\) | \(36\) |
derivativedivides | \(2 a \,b^{2} x^{\frac {3}{2}}+\frac {2 b^{3} x^{\frac {5}{2}}}{5}-\frac {2 a^{3}}{\sqrt {x}}+6 a^{2} b \sqrt {x}\) | \(36\) |
default | \(2 a \,b^{2} x^{\frac {3}{2}}+\frac {2 b^{3} x^{\frac {5}{2}}}{5}-\frac {2 a^{3}}{\sqrt {x}}+6 a^{2} b \sqrt {x}\) | \(36\) |
trager | \(-\frac {2 \left (-b^{3} x^{3}-5 a \,b^{2} x^{2}-15 a^{2} b x +5 a^{3}\right )}{5 \sqrt {x}}\) | \(36\) |
risch | \(-\frac {2 \left (-b^{3} x^{3}-5 a \,b^{2} x^{2}-15 a^{2} b x +5 a^{3}\right )}{5 \sqrt {x}}\) | \(36\) |
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Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} + 5 \, a b^{2} x^{2} + 15 \, a^{2} b x - 5 \, a^{3}\right )}}{5 \, \sqrt {x}} \]
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Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=- \frac {2 a^{3}}{\sqrt {x}} + 6 a^{2} b \sqrt {x} + 2 a b^{2} x^{\frac {3}{2}} + \frac {2 b^{3} x^{\frac {5}{2}}}{5} \]
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Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=\frac {2}{5} \, b^{3} x^{\frac {5}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} + 6 \, a^{2} b \sqrt {x} - \frac {2 \, a^{3}}{\sqrt {x}} \]
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Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=\frac {2}{5} \, b^{3} x^{\frac {5}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} + 6 \, a^{2} b \sqrt {x} - \frac {2 \, a^{3}}{\sqrt {x}} \]
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Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^3}{x^{3/2}} \, dx=\frac {2\,b^3\,x^{5/2}}{5}-\frac {2\,a^3}{\sqrt {x}}+6\,a^2\,b\,\sqrt {x}+2\,a\,b^2\,x^{3/2} \]
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