Integrand size = 18, antiderivative size = 95 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=(3 A b+2 a B) \sqrt {a+b x}+\frac {(3 A b+2 a B) (a+b x)^{3/2}}{3 a}-\frac {A (a+b x)^{5/2}}{a x}-\sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 52, 65, 214} \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=-\sqrt {a} (2 a B+3 A b) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {(a+b x)^{3/2} (2 a B+3 A b)}{3 a}+\sqrt {a+b x} (2 a B+3 A b)-\frac {A (a+b x)^{5/2}}{a x} \]
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Rule 52
Rule 65
Rule 79
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {A (a+b x)^{5/2}}{a x}+\frac {\left (\frac {3 A b}{2}+a B\right ) \int \frac {(a+b x)^{3/2}}{x} \, dx}{a} \\ & = \frac {(3 A b+2 a B) (a+b x)^{3/2}}{3 a}-\frac {A (a+b x)^{5/2}}{a x}+\frac {1}{2} (3 A b+2 a B) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = (3 A b+2 a B) \sqrt {a+b x}+\frac {(3 A b+2 a B) (a+b x)^{3/2}}{3 a}-\frac {A (a+b x)^{5/2}}{a x}+\frac {1}{2} (a (3 A b+2 a B)) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = (3 A b+2 a B) \sqrt {a+b x}+\frac {(3 A b+2 a B) (a+b x)^{3/2}}{3 a}-\frac {A (a+b x)^{5/2}}{a x}+\frac {(a (3 A b+2 a B)) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b} \\ & = (3 A b+2 a B) \sqrt {a+b x}+\frac {(3 A b+2 a B) (a+b x)^{3/2}}{3 a}-\frac {A (a+b x)^{5/2}}{a x}-\sqrt {a} (3 A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\frac {\sqrt {a+b x} (2 b x (3 A+B x)+a (-3 A+8 B x))}{3 x}-\sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 1.38 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(-\frac {3 \left (a x \left (A b +\frac {2 B a}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\frac {2 \sqrt {b x +a}\, \left (\left (\frac {4 B x}{3}-\frac {A}{2}\right ) a^{\frac {3}{2}}+b x \sqrt {a}\, \left (\frac {B x}{3}+A \right )\right )}{3}\right )}{\sqrt {a}\, x}\) | \(67\) |
risch | \(-\frac {a A \sqrt {b x +a}}{x}+\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-\left (3 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}\) | \(74\) |
derivativedivides | \(\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-2 a \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (3 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\) | \(77\) |
default | \(\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-2 a \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (3 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\) | \(77\) |
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Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\left [\frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, B b x^{2} - 3 \, A a + 2 \, {\left (4 \, B a + 3 \, A b\right )} x\right )} \sqrt {b x + a}}{6 \, x}, \frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {-a} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, B b x^{2} - 3 \, A a + 2 \, {\left (4 \, B a + 3 \, A b\right )} x\right )} \sqrt {b x + a}}{3 \, x}\right ] \]
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Time = 11.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.79 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=- A \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {A a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} + A b \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + B a \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} \frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\frac {1}{6} \, {\left (\frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{b} - \frac {6 \, \sqrt {b x + a} A a}{b x} + \frac {4 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B + 3 \, {\left (B a + A b\right )} \sqrt {b x + a}\right )}}{b}\right )} b \]
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Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} B b + 6 \, \sqrt {b x + a} B a b + 6 \, \sqrt {b x + a} A b^{2} - \frac {3 \, \sqrt {b x + a} A a b}{x} + \frac {3 \, {\left (2 \, B a^{2} b + 3 \, A a b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}}{3 \, b} \]
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Time = 0.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\left (2\,A\,b+2\,B\,a\right )\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{3/2}}{3}+2\,\mathrm {atan}\left (\frac {2\,\left (3\,A\,b+2\,B\,a\right )\,\sqrt {-\frac {a}{4}}\,\sqrt {a+b\,x}}{2\,B\,a^2+3\,A\,b\,a}\right )\,\left (3\,A\,b+2\,B\,a\right )\,\sqrt {-\frac {a}{4}}-\frac {A\,a\,\sqrt {a+b\,x}}{x} \]
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