Integrand size = 13, antiderivative size = 51 \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=3 b \sqrt {a+b x}-\frac {(a+b x)^{3/2}}{x}-3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 214} \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=-3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\frac {(a+b x)^{3/2}}{x}+3 b \sqrt {a+b x} \]
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Rule 43
Rule 52
Rule 65
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{3/2}}{x}+\frac {1}{2} (3 b) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = 3 b \sqrt {a+b x}-\frac {(a+b x)^{3/2}}{x}+\frac {1}{2} (3 a b) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = 3 b \sqrt {a+b x}-\frac {(a+b x)^{3/2}}{x}+(3 a) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right ) \\ & = 3 b \sqrt {a+b x}-\frac {(a+b x)^{3/2}}{x}-3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=-\frac {(a-2 b x) \sqrt {a+b x}}{x}-3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-\frac {a \sqrt {b x +a}}{x}+\frac {b \left (4 \sqrt {b x +a}-6 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}\right )}{2}\) | \(45\) |
derivativedivides | \(2 b \left (\sqrt {b x +a}-a \left (\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\) | \(48\) |
default | \(2 b \left (\sqrt {b x +a}-a \left (\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\) | \(48\) |
pseudoelliptic | \(\frac {2 b x \sqrt {b x +a}\, \sqrt {a}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a b x -\sqrt {b x +a}\, a^{\frac {3}{2}}}{x \sqrt {a}}\) | \(52\) |
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Time = 0.22 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.00 \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=\left [\frac {3 \, \sqrt {a} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, b x - a\right )} \sqrt {b x + a}}{2 \, x}, \frac {3 \, \sqrt {-a} b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, b x - a\right )} \sqrt {b x + a}}{x}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (44) = 88\).
Time = 1.97 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.80 \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=- 3 \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {a^{2}}{\sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {a \sqrt {b}}{\sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {3}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} + 1}} \]
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Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=\frac {3}{2} \, \sqrt {a} b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + 2 \, \sqrt {b x + a} b - \frac {\sqrt {b x + a} a}{x} \]
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Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=\frac {\frac {3 \, a b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {b x + a} b^{2} - \frac {\sqrt {b x + a} a b}{x}}{b} \]
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Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^{3/2}}{x^2} \, dx=2\,b\,\sqrt {a+b\,x}-3\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )-\frac {a\,\sqrt {a+b\,x}}{x} \]
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