Integrand size = 17, antiderivative size = 84 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=-\frac {1}{4 a^2 x^4}+\frac {2 b}{3 a^3 x^3}-\frac {3 b^2}{2 a^4 x^2}+\frac {4 b^3}{a^5 x}+\frac {b^4}{a^5 (a+b x)}+\frac {5 b^4 \log (x)}{a^6}-\frac {5 b^4 \log (a+b x)}{a^6} \]
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Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 46} \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {5 b^4 \log (x)}{a^6}-\frac {5 b^4 \log (a+b x)}{a^6}+\frac {b^4}{a^5 (a+b x)}+\frac {4 b^3}{a^5 x}-\frac {3 b^2}{2 a^4 x^2}+\frac {2 b}{3 a^3 x^3}-\frac {1}{4 a^2 x^4} \]
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Rule 46
Rule 1598
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^5 (a+b x)^2} \, dx \\ & = \int \left (\frac {1}{a^2 x^5}-\frac {2 b}{a^3 x^4}+\frac {3 b^2}{a^4 x^3}-\frac {4 b^3}{a^5 x^2}+\frac {5 b^4}{a^6 x}-\frac {b^5}{a^5 (a+b x)^2}-\frac {5 b^5}{a^6 (a+b x)}\right ) \, dx \\ & = -\frac {1}{4 a^2 x^4}+\frac {2 b}{3 a^3 x^3}-\frac {3 b^2}{2 a^4 x^2}+\frac {4 b^3}{a^5 x}+\frac {b^4}{a^5 (a+b x)}+\frac {5 b^4 \log (x)}{a^6}-\frac {5 b^4 \log (a+b x)}{a^6} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {\frac {a \left (-3 a^4+5 a^3 b x-10 a^2 b^2 x^2+30 a b^3 x^3+60 b^4 x^4\right )}{x^4 (a+b x)}+60 b^4 \log (x)-60 b^4 \log (a+b x)}{12 a^6} \]
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Time = 4.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94
method | result | size |
default | \(-\frac {1}{4 a^{2} x^{4}}+\frac {2 b}{3 a^{3} x^{3}}-\frac {3 b^{2}}{2 a^{4} x^{2}}+\frac {4 b^{3}}{a^{5} x}+\frac {b^{4}}{a^{5} \left (b x +a \right )}+\frac {5 b^{4} \ln \left (x \right )}{a^{6}}-\frac {5 b^{4} \ln \left (b x +a \right )}{a^{6}}\) | \(79\) |
norman | \(\frac {-\frac {5 b^{5} x^{5}}{a^{6}}-\frac {1}{4 a}+\frac {5 b x}{12 a^{2}}-\frac {5 b^{2} x^{2}}{6 a^{3}}+\frac {5 b^{3} x^{3}}{2 a^{4}}}{x^{4} \left (b x +a \right )}+\frac {5 b^{4} \ln \left (x \right )}{a^{6}}-\frac {5 b^{4} \ln \left (b x +a \right )}{a^{6}}\) | \(83\) |
risch | \(\frac {\frac {5 b^{4} x^{4}}{a^{5}}+\frac {5 b^{3} x^{3}}{2 a^{4}}-\frac {5 b^{2} x^{2}}{6 a^{3}}+\frac {5 b x}{12 a^{2}}-\frac {1}{4 a}}{x^{4} \left (b x +a \right )}+\frac {5 b^{4} \ln \left (-x \right )}{a^{6}}-\frac {5 b^{4} \ln \left (b x +a \right )}{a^{6}}\) | \(85\) |
parallelrisch | \(\frac {60 \ln \left (x \right ) x^{5} b^{5}-60 \ln \left (b x +a \right ) x^{5} b^{5}+60 \ln \left (x \right ) x^{4} a \,b^{4}-60 \ln \left (b x +a \right ) x^{4} a \,b^{4}-60 b^{5} x^{5}+30 a^{2} b^{3} x^{3}-10 a^{3} b^{2} x^{2}+5 a^{4} b x -3 a^{5}}{12 a^{6} x^{4} \left (b x +a \right )}\) | \(109\) |
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Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {60 \, a b^{4} x^{4} + 30 \, a^{2} b^{3} x^{3} - 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x - 3 \, a^{5} - 60 \, {\left (b^{5} x^{5} + a b^{4} x^{4}\right )} \log \left (b x + a\right ) + 60 \, {\left (b^{5} x^{5} + a b^{4} x^{4}\right )} \log \left (x\right )}{12 \, {\left (a^{6} b x^{5} + a^{7} x^{4}\right )}} \]
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Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {- 3 a^{4} + 5 a^{3} b x - 10 a^{2} b^{2} x^{2} + 30 a b^{3} x^{3} + 60 b^{4} x^{4}}{12 a^{6} x^{4} + 12 a^{5} b x^{5}} + \frac {5 b^{4} \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{6}} \]
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Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {60 \, b^{4} x^{4} + 30 \, a b^{3} x^{3} - 10 \, a^{2} b^{2} x^{2} + 5 \, a^{3} b x - 3 \, a^{4}}{12 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}} - \frac {5 \, b^{4} \log \left (b x + a\right )}{a^{6}} + \frac {5 \, b^{4} \log \left (x\right )}{a^{6}} \]
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Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=-\frac {5 \, b^{4} \log \left ({\left | b x + a \right |}\right )}{a^{6}} + \frac {5 \, b^{4} \log \left ({\left | x \right |}\right )}{a^{6}} + \frac {60 \, a b^{4} x^{4} + 30 \, a^{2} b^{3} x^{3} - 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x - 3 \, a^{5}}{12 \, {\left (b x + a\right )} a^{6} x^{4}} \]
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Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {\frac {5\,b^3\,x^3}{2\,a^4}-\frac {5\,b^2\,x^2}{6\,a^3}-\frac {1}{4\,a}+\frac {5\,b^4\,x^4}{a^5}+\frac {5\,b\,x}{12\,a^2}}{b\,x^5+a\,x^4}-\frac {10\,b^4\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^6} \]
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