Integrand size = 13, antiderivative size = 114 \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=\frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}-\frac {63}{4} a^{5/2} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 214} \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=-\frac {63}{4} a^{5/2} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {63}{20} b^2 (a+b x)^{5/2}+\frac {21}{4} a b^2 (a+b x)^{3/2}-\frac {(a+b x)^{9/2}}{2 x^2}-\frac {9 b (a+b x)^{7/2}}{4 x} \]
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Rule 43
Rule 52
Rule 65
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{4} (9 b) \int \frac {(a+b x)^{7/2}}{x^2} \, dx \\ & = -\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 b^2\right ) \int \frac {(a+b x)^{5/2}}{x} \, dx \\ & = \frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 a b^2\right ) \int \frac {(a+b x)^{3/2}}{x} \, dx \\ & = \frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 a^2 b^2\right ) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = \frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 a^3 b^2\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = \frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{4} \left (63 a^3 b\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right ) \\ & = \frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}-\frac {63}{4} a^{5/2} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=\frac {\sqrt {a+b x} \left (-10 a^4-85 a^3 b x+288 a^2 b^2 x^2+56 a b^3 x^3+8 b^4 x^4\right )}{20 x^2}-\frac {63}{4} a^{5/2} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {a^{3} \sqrt {b x +a}\, \left (17 b x +2 a \right )}{4 x^{2}}+\frac {b^{2} \left (\frac {16 \left (b x +a \right )^{\frac {5}{2}}}{5}+16 a \left (b x +a \right )^{\frac {3}{2}}+96 a^{2} \sqrt {b x +a}-126 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right )}{8}\) | \(79\) |
pseudoelliptic | \(-\frac {63 \left (\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{3} b^{2} x^{2}-\frac {8 \left (\sqrt {a}\, b^{4} x^{4}+7 a^{\frac {3}{2}} b^{3} x^{3}+36 a^{\frac {5}{2}} b^{2} x^{2}-\frac {85 a^{\frac {7}{2}} b x}{8}-\frac {5 a^{\frac {9}{2}}}{4}\right ) \sqrt {b x +a}}{315}\right )}{4 \sqrt {a}\, x^{2}}\) | \(86\) |
derivativedivides | \(2 b^{2} \left (\frac {\left (b x +a \right )^{\frac {5}{2}}}{5}+a \left (b x +a \right )^{\frac {3}{2}}+6 a^{2} \sqrt {b x +a}-a^{3} \left (\frac {\frac {17 \left (b x +a \right )^{\frac {3}{2}}}{8}-\frac {15 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {63 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\) | \(87\) |
default | \(2 b^{2} \left (\frac {\left (b x +a \right )^{\frac {5}{2}}}{5}+a \left (b x +a \right )^{\frac {3}{2}}+6 a^{2} \sqrt {b x +a}-a^{3} \left (\frac {\frac {17 \left (b x +a \right )^{\frac {3}{2}}}{8}-\frac {15 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {63 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\) | \(87\) |
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Time = 0.24 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=\left [\frac {315 \, a^{\frac {5}{2}} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{4} x^{4} + 56 \, a b^{3} x^{3} + 288 \, a^{2} b^{2} x^{2} - 85 \, a^{3} b x - 10 \, a^{4}\right )} \sqrt {b x + a}}{40 \, x^{2}}, \frac {315 \, \sqrt {-a} a^{2} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (8 \, b^{4} x^{4} + 56 \, a b^{3} x^{3} + 288 \, a^{2} b^{2} x^{2} - 85 \, a^{3} b x - 10 \, a^{4}\right )} \sqrt {b x + a}}{20 \, x^{2}}\right ] \]
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Time = 13.37 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=- \frac {63 a^{\frac {5}{2}} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {a^{5}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {19 a^{4} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {203 a^{3} b^{\frac {3}{2}}}{20 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {86 a^{2} b^{\frac {5}{2}} \sqrt {x}}{5 \sqrt {\frac {a}{b x} + 1}} + \frac {16 a b^{\frac {7}{2}} x^{\frac {3}{2}}}{5 \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {9}{2}} x^{\frac {5}{2}}}{5 \sqrt {\frac {a}{b x} + 1}} \]
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Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=\frac {63}{8} \, a^{\frac {5}{2}} b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} b^{2} + 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{2} + 12 \, \sqrt {b x + a} a^{2} b^{2} - \frac {17 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{2} - 15 \, \sqrt {b x + a} a^{4} b^{2}}{4 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a + a^{2}\right )}} \]
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Time = 0.31 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=\frac {\frac {315 \, a^{3} b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 8 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} + 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 240 \, \sqrt {b x + a} a^{2} b^{3} - \frac {5 \, {\left (17 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{3} - 15 \, \sqrt {b x + a} a^{4} b^{3}\right )}}{b^{2} x^{2}}}{20 \, b} \]
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Time = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^{9/2}}{x^3} \, dx=\frac {2\,b^2\,{\left (a+b\,x\right )}^{5/2}}{5}+\frac {\frac {15\,a^4\,b^2\,\sqrt {a+b\,x}}{4}-\frac {17\,a^3\,b^2\,{\left (a+b\,x\right )}^{3/2}}{4}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}+12\,a^2\,b^2\,\sqrt {a+b\,x}+2\,a\,b^2\,{\left (a+b\,x\right )}^{3/2}+\frac {a^{5/2}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,63{}\mathrm {i}}{4} \]
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