Integrand size = 18, antiderivative size = 54 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b}-2 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {81, 52, 65, 214} \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=-2 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b} \]
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Rule 52
Rule 65
Rule 81
Rule 214
Rubi steps \begin{align*} \text {integral}& = \frac {2 B (a+b x)^{3/2}}{3 b}+A \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = 2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b}+(a A) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = 2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b}+\frac {(2 a A) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b} \\ & = 2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b}-2 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=\frac {2 \sqrt {a+b x} (3 A b+B (a+b x))}{3 b}-2 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 1.35 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}-2 A \sqrt {a}\, b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\) | \(46\) |
default | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}-2 A \sqrt {a}\, b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\) | \(46\) |
pseudoelliptic | \(\frac {-2 A \sqrt {a}\, b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \left (\left (\frac {B x}{3}+A \right ) b +\frac {B a}{3}\right ) \sqrt {b x +a}}{b}\) | \(47\) |
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Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.06 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=\left [\frac {3 \, A \sqrt {a} b \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (B b x + B a + 3 \, A b\right )} \sqrt {b x + a}}{3 \, b}, \frac {2 \, {\left (3 \, A \sqrt {-a} b \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (B b x + B a + 3 \, A b\right )} \sqrt {b x + a}\right )}}{3 \, b}\right ] \]
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Time = 1.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=\begin {cases} \frac {2 A a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A \sqrt {a + b x} + \frac {2 B \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} \left (A \log {\left (B x \right )} + B x\right ) & \text {otherwise} \end {cases} \]
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Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=A \sqrt {a} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B + 3 \, \sqrt {b x + a} A b\right )}}{3 \, b} \]
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Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=\frac {2 \, A a \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B b^{2} + 3 \, \sqrt {b x + a} A b^{3}\right )}}{3 \, b^{3}} \]
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Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx=2\,A\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{3/2}}{3\,b}+A\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]
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