Integrand size = 13, antiderivative size = 87 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 b^2}{4 a^3 \sqrt {a+b x}}-\frac {1}{2 a x^2 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
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Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {44, 53, 65, 214} \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}}+\frac {15 b^2}{4 a^3 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}-\frac {1}{2 a x^2 \sqrt {a+b x}} \]
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Rule 44
Rule 53
Rule 65
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2 a x^2 \sqrt {a+b x}}-\frac {(5 b) \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx}{4 a} \\ & = -\frac {1}{2 a x^2 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}+\frac {\left (15 b^2\right ) \int \frac {1}{x (a+b x)^{3/2}} \, dx}{8 a^2} \\ & = \frac {15 b^2}{4 a^3 \sqrt {a+b x}}-\frac {1}{2 a x^2 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}+\frac {\left (15 b^2\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{8 a^3} \\ & = \frac {15 b^2}{4 a^3 \sqrt {a+b x}}-\frac {1}{2 a x^2 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}+\frac {(15 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{4 a^3} \\ & = \frac {15 b^2}{4 a^3 \sqrt {a+b x}}-\frac {1}{2 a x^2 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}-\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {-2 a^2+5 a b x+15 b^2 x^2}{4 a^3 x^2 \sqrt {a+b x}}-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
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Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (-7 b x +2 a \right )}{4 a^{3} x^{2}}+\frac {b^{2} \left (-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {16}{\sqrt {b x +a}}\right )}{8 a^{3}}\) | \(60\) |
pseudoelliptic | \(-\frac {15 \left (\sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-\sqrt {a}\, b^{2} x^{2}-\frac {a^{\frac {3}{2}} b x}{3}+\frac {2 a^{\frac {5}{2}}}{15}\right )}{4 \sqrt {b x +a}\, a^{\frac {7}{2}} x^{2}}\) | \(66\) |
derivativedivides | \(2 b^{2} \left (\frac {1}{a^{3} \sqrt {b x +a}}-\frac {\frac {-\frac {7 \left (b x +a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\) | \(68\) |
default | \(2 b^{2} \left (\frac {1}{a^{3} \sqrt {b x +a}}-\frac {\frac {-\frac {7 \left (b x +a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\) | \(68\) |
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Time = 0.24 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.17 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{8 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, \frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{4 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \]
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Time = 5.55 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=- \frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}} \]
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Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 \, {\left (b x + a\right )}^{2} b^{2} - 25 \, {\left (b x + a\right )} a b^{2} + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x + a} a^{5}\right )}} + \frac {15 \, b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]
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Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3}} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]
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Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {\frac {2\,b^2}{a}+\frac {15\,b^2\,{\left (a+b\,x\right )}^2}{4\,a^3}-\frac {25\,b^2\,\left (a+b\,x\right )}{4\,a^2}}{{\left (a+b\,x\right )}^{5/2}-2\,a\,{\left (a+b\,x\right )}^{3/2}+a^2\,\sqrt {a+b\,x}}-\frac {15\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{7/2}} \]
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