Integrand size = 13, antiderivative size = 114 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {105}{8} a b^3 \sqrt {a+b x}+\frac {35}{8} b^3 (a+b x)^{3/2}-\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}-\frac {105}{8} a^{3/2} b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.03 (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} \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=-\frac {105}{8} a^{3/2} b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {35}{8} b^3 (a+b x)^{3/2}+\frac {105}{8} a b^3 \sqrt {a+b x}-\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {(a+b x)^{9/2}}{3 x^3}-\frac {3 b (a+b x)^{7/2}}{4 x^2} \]
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Rule 43
Rule 52
Rule 65
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{9/2}}{3 x^3}+\frac {1}{2} (3 b) \int \frac {(a+b x)^{7/2}}{x^3} \, dx \\ & = -\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}+\frac {1}{8} \left (21 b^2\right ) \int \frac {(a+b x)^{5/2}}{x^2} \, dx \\ & = -\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}+\frac {1}{16} \left (105 b^3\right ) \int \frac {(a+b x)^{3/2}}{x} \, dx \\ & = \frac {35}{8} b^3 (a+b x)^{3/2}-\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}+\frac {1}{16} \left (105 a b^3\right ) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = \frac {105}{8} a b^3 \sqrt {a+b x}+\frac {35}{8} b^3 (a+b x)^{3/2}-\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}+\frac {1}{16} \left (105 a^2 b^3\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = \frac {105}{8} a b^3 \sqrt {a+b x}+\frac {35}{8} b^3 (a+b x)^{3/2}-\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}+\frac {1}{8} \left (105 a^2 b^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right ) \\ & = \frac {105}{8} a b^3 \sqrt {a+b x}+\frac {35}{8} b^3 (a+b x)^{3/2}-\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}-\frac {105}{8} a^{3/2} b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {1}{24} \left (\frac {\sqrt {a+b x} \left (-8 a^4-50 a^3 b x-165 a^2 b^2 x^2+208 a b^3 x^3+16 b^4 x^4\right )}{x^3}-315 a^{3/2} b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right ) \]
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Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68
method | result | size |
risch | \(-\frac {a^{2} \sqrt {b x +a}\, \left (165 b^{2} x^{2}+50 a b x +8 a^{2}\right )}{24 x^{3}}+\frac {b^{3} \left (\frac {32 \left (b x +a \right )^{\frac {3}{2}}}{3}+128 a \sqrt {b x +a}-210 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right )}{16}\) | \(78\) |
pseudoelliptic | \(-\frac {105 \left (\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{3} x^{3}-\frac {16 \left (\sqrt {a}\, b^{4} x^{4}+13 a^{\frac {3}{2}} b^{3} x^{3}-\frac {165 a^{\frac {5}{2}} b^{2} x^{2}}{16}-\frac {25 a^{\frac {7}{2}} b x}{8}-\frac {a^{\frac {9}{2}}}{2}\right ) \sqrt {b x +a}}{315}\right )}{8 \sqrt {a}\, x^{3}}\) | \(86\) |
derivativedivides | \(2 b^{3} \left (\frac {\left (b x +a \right )^{\frac {3}{2}}}{3}+4 a \sqrt {b x +a}-a^{2} \left (-\frac {-\frac {55 \left (b x +a \right )^{\frac {5}{2}}}{16}+\frac {35 a \left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {41 a^{2} \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {105 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\right )\) | \(89\) |
default | \(2 b^{3} \left (\frac {\left (b x +a \right )^{\frac {3}{2}}}{3}+4 a \sqrt {b x +a}-a^{2} \left (-\frac {-\frac {55 \left (b x +a \right )^{\frac {5}{2}}}{16}+\frac {35 a \left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {41 a^{2} \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {105 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\right )\) | \(89\) |
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Time = 0.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\left [\frac {315 \, a^{\frac {3}{2}} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (16 \, b^{4} x^{4} + 208 \, a b^{3} x^{3} - 165 \, a^{2} b^{2} x^{2} - 50 \, a^{3} b x - 8 \, a^{4}\right )} \sqrt {b x + a}}{48 \, x^{3}}, \frac {315 \, \sqrt {-a} a b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (16 \, b^{4} x^{4} + 208 \, a b^{3} x^{3} - 165 \, a^{2} b^{2} x^{2} - 50 \, a^{3} b x - 8 \, a^{4}\right )} \sqrt {b x + a}}{24 \, x^{3}}\right ] \]
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Time = 13.53 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=- \frac {105 a^{\frac {3}{2}} b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8} - \frac {a^{5}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {29 a^{4} \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {215 a^{3} b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {43 a^{2} b^{\frac {5}{2}}}{24 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {28 a b^{\frac {7}{2}} \sqrt {x}}{3 \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {9}{2}} x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x} + 1}} \]
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Time = 0.44 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {105}{16} \, a^{\frac {3}{2}} b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} + 8 \, \sqrt {b x + a} a b^{3} - \frac {165 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{3} - 280 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{3} + 123 \, \sqrt {b x + a} a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {\frac {315 \, a^{2} b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 16 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4} + 192 \, \sqrt {b x + a} a b^{4} - \frac {165 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{4} - 280 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{4} + 123 \, \sqrt {b x + a} a^{4} b^{4}}{b^{3} x^{3}}}{24 \, b} \]
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Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {2\,b^3\,{\left (a+b\,x\right )}^{3/2}}{3}+\frac {\frac {41\,a^4\,b^3\,\sqrt {a+b\,x}}{8}-\frac {35\,a^3\,b^3\,{\left (a+b\,x\right )}^{3/2}}{3}+\frac {55\,a^2\,b^3\,{\left (a+b\,x\right )}^{5/2}}{8}}{3\,a\,{\left (a+b\,x\right )}^2-3\,a^2\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^3+a^3}+8\,a\,b^3\,\sqrt {a+b\,x}+\frac {a^{3/2}\,b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,105{}\mathrm {i}}{8} \]
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