Integrand size = 17, antiderivative size = 92 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=\frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {15}{4} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {676, 678, 634, 212} \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=\frac {15}{4} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}+\frac {15}{4} a b \sqrt {a x+b x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3} \]
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Rule 212
Rule 634
Rule 676
Rule 678
Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+(5 b) \int \frac {\left (a x+b x^2\right )^{3/2}}{x^2} \, dx \\ & = \frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{4} (15 a b) \int \frac {\sqrt {a x+b x^2}}{x} \, dx \\ & = \frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{8} \left (15 a^2 b\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx \\ & = \frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{4} \left (15 a^2 b\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right ) \\ & = \frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {15}{4} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=\frac {\sqrt {a+b x} \left (\sqrt {a+b x} \left (-8 a^2+9 a b x+2 b^2 x^2\right )+30 a^2 \sqrt {b} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{4 \sqrt {x (a+b x)}} \]
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Time = 2.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75
method | result | size |
risch | \(-\frac {\left (b x +a \right ) \left (-2 b^{2} x^{2}-9 a b x +8 a^{2}\right )}{4 \sqrt {x \left (b x +a \right )}}+\frac {15 a^{2} \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8}\) | \(69\) |
pseudoelliptic | \(\frac {2 \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} x^{2}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a^{2} b x +9 a \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}-8 a^{2} \sqrt {x \left (b x +a \right )}\, \sqrt {b}}{4 x \sqrt {b}}\) | \(84\) |
default | \(-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{4}}+\frac {6 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{3}}-\frac {8 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{2}}-\frac {10 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\right )}{3 a}\right )}{a}\right )}{a}\) | \(182\) |
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Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=\left [\frac {15 \, a^{2} \sqrt {b} x \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{8 \, x}, -\frac {15 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{4 \, x}\right ] \]
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\[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{4}}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=\frac {15}{8} \, a^{2} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {15 \, \sqrt {b x^{2} + a x} a^{2}}{4 \, x} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{4 \, x^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{2 \, x^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=-\frac {15}{8} \, a^{2} \sqrt {b} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right ) + \frac {2 \, a^{3}}{\sqrt {b} x - \sqrt {b x^{2} + a x}} + \frac {1}{4} \, {\left (2 \, b^{2} x + 9 \, a b\right )} \sqrt {b x^{2} + a x} \]
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Timed out. \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{x^4} \,d x \]
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