Integrand size = 13, antiderivative size = 103 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \]
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Time = 0.02 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 44, 65, 214} \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {(a+b x)^{5/2}}{4 x^4}-\frac {5 b (a+b x)^{3/2}}{24 x^3} \]
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Rule 43
Rule 44
Rule 65
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{5/2}}{4 x^4}+\frac {1}{8} (5 b) \int \frac {(a+b x)^{3/2}}{x^4} \, dx \\ & = -\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {1}{16} \left (5 b^2\right ) \int \frac {\sqrt {a+b x}}{x^3} \, dx \\ & = -\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {1}{64} \left (5 b^3\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx \\ & = -\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}-\frac {\left (5 b^4\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a} \\ & = -\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a} \\ & = -\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {\sqrt {a+b x} \left (48 a^3+136 a^2 b x+118 a b^2 x^2+15 b^3 x^3\right )}{192 a x^4}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \]
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Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.65
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (15 b^{3} x^{3}+118 a \,b^{2} x^{2}+136 a^{2} b x +48 a^{3}\right )}{192 x^{4} a}+\frac {5 b^{4} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64 a^{\frac {3}{2}}}\) | \(67\) |
pseudoelliptic | \(-\frac {5 \left (-\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}+\sqrt {b x +a}\, \left (\sqrt {a}\, b^{3} x^{3}+\frac {118 a^{\frac {3}{2}} b^{2} x^{2}}{15}+\frac {136 a^{\frac {5}{2}} b x}{15}+\frac {16 a^{\frac {7}{2}}}{5}\right )\right )}{64 a^{\frac {3}{2}} x^{4}}\) | \(72\) |
derivativedivides | \(2 b^{4} \left (-\frac {\frac {5 \left (b x +a \right )^{\frac {7}{2}}}{128 a}+\frac {73 \left (b x +a \right )^{\frac {5}{2}}}{384}-\frac {55 a \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {5 a^{2} \sqrt {b x +a}}{128}}{b^{4} x^{4}}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}\right )\) | \(76\) |
default | \(2 b^{4} \left (-\frac {\frac {5 \left (b x +a \right )^{\frac {7}{2}}}{128 a}+\frac {73 \left (b x +a \right )^{\frac {5}{2}}}{384}-\frac {55 a \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {5 a^{2} \sqrt {b x +a}}{128}}{b^{4} x^{4}}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}\right )\) | \(76\) |
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Time = 0.24 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=\left [\frac {15 \, \sqrt {a} b^{4} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, -\frac {15 \, \sqrt {-a} b^{4} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \]
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Time = 9.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=- \frac {a^{3}}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {23 a^{2} \sqrt {b}}{24 x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {127 a b^{\frac {3}{2}}}{96 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {133 b^{\frac {5}{2}}}{192 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 b^{\frac {7}{2}}}{64 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {3}{2}}} \]
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Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {5 \, b^{4} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{128 \, a^{\frac {3}{2}}} - \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{4} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{4} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {b x + a} a^{3} b^{4}}{192 \, {\left ({\left (b x + a\right )}^{4} a - 4 \, {\left (b x + a\right )}^{3} a^{2} + 6 \, {\left (b x + a\right )}^{2} a^{3} - 4 \, {\left (b x + a\right )} a^{4} + a^{5}\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {\frac {15 \, b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x + a} a^{3} b^{5}}{a b^{4} x^{4}}}{192 \, b} \]
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Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=\frac {55\,a\,{\left (a+b\,x\right )}^{3/2}}{192\,x^4}-\frac {5\,a^2\,\sqrt {a+b\,x}}{64\,x^4}-\frac {5\,{\left (a+b\,x\right )}^{7/2}}{64\,a\,x^4}-\frac {73\,{\left (a+b\,x\right )}^{5/2}}{192\,x^4}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{64\,a^{3/2}} \]
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