Integrand size = 11, antiderivative size = 68 \[ \int \frac {1}{x^5 (a+b x)} \, dx=-\frac {1}{4 a x^4}+\frac {b}{3 a^2 x^3}-\frac {b^2}{2 a^3 x^2}+\frac {b^3}{a^4 x}+\frac {b^4 \log (x)}{a^5}-\frac {b^4 \log (a+b x)}{a^5} \]
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Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{x^5 (a+b x)} \, dx=\frac {b^4 \log (x)}{a^5}-\frac {b^4 \log (a+b x)}{a^5}+\frac {b^3}{a^4 x}-\frac {b^2}{2 a^3 x^2}+\frac {b}{3 a^2 x^3}-\frac {1}{4 a x^4} \]
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Rule 46
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{a x^5}-\frac {b}{a^2 x^4}+\frac {b^2}{a^3 x^3}-\frac {b^3}{a^4 x^2}+\frac {b^4}{a^5 x}-\frac {b^5}{a^5 (a+b x)}\right ) \, dx \\ & = -\frac {1}{4 a x^4}+\frac {b}{3 a^2 x^3}-\frac {b^2}{2 a^3 x^2}+\frac {b^3}{a^4 x}+\frac {b^4 \log (x)}{a^5}-\frac {b^4 \log (a+b x)}{a^5} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 (a+b x)} \, dx=-\frac {1}{4 a x^4}+\frac {b}{3 a^2 x^3}-\frac {b^2}{2 a^3 x^2}+\frac {b^3}{a^4 x}+\frac {b^4 \log (x)}{a^5}-\frac {b^4 \log (a+b x)}{a^5} \]
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Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {1}{4 a \,x^{4}}+\frac {b}{3 a^{2} x^{3}}-\frac {b^{2}}{2 a^{3} x^{2}}+\frac {b^{3}}{a^{4} x}+\frac {b^{4} \ln \left (x \right )}{a^{5}}-\frac {b^{4} \ln \left (b x +a \right )}{a^{5}}\) | \(63\) |
norman | \(\frac {\frac {b^{3} x^{3}}{a^{4}}-\frac {1}{4 a}+\frac {b x}{3 a^{2}}-\frac {b^{2} x^{2}}{2 a^{3}}}{x^{4}}+\frac {b^{4} \ln \left (x \right )}{a^{5}}-\frac {b^{4} \ln \left (b x +a \right )}{a^{5}}\) | \(63\) |
risch | \(\frac {\frac {b^{3} x^{3}}{a^{4}}-\frac {1}{4 a}+\frac {b x}{3 a^{2}}-\frac {b^{2} x^{2}}{2 a^{3}}}{x^{4}}-\frac {b^{4} \ln \left (b x +a \right )}{a^{5}}+\frac {b^{4} \ln \left (-x \right )}{a^{5}}\) | \(65\) |
parallelrisch | \(\frac {12 b^{4} \ln \left (x \right ) x^{4}-12 b^{4} \ln \left (b x +a \right ) x^{4}+12 a \,b^{3} x^{3}-6 a^{2} b^{2} x^{2}+4 a^{3} b x -3 a^{4}}{12 a^{5} x^{4}}\) | \(66\) |
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none
Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^5 (a+b x)} \, dx=-\frac {12 \, b^{4} x^{4} \log \left (b x + a\right ) - 12 \, b^{4} x^{4} \log \left (x\right ) - 12 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 3 \, a^{4}}{12 \, a^{5} x^{4}} \]
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Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^5 (a+b x)} \, dx=\frac {- 3 a^{3} + 4 a^{2} b x - 6 a b^{2} x^{2} + 12 b^{3} x^{3}}{12 a^{4} x^{4}} + \frac {b^{4} \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{5}} \]
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Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^5 (a+b x)} \, dx=-\frac {b^{4} \log \left (b x + a\right )}{a^{5}} + \frac {b^{4} \log \left (x\right )}{a^{5}} + \frac {12 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 4 \, a^{2} b x - 3 \, a^{3}}{12 \, a^{4} x^{4}} \]
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Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^5 (a+b x)} \, dx=-\frac {b^{4} \log \left ({\left | b x + a \right |}\right )}{a^{5}} + \frac {b^{4} \log \left ({\left | x \right |}\right )}{a^{5}} + \frac {12 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x - 3 \, a^{4}}{12 \, a^{5} x^{4}} \]
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Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^5 (a+b x)} \, dx=-\frac {\frac {a^4}{4}-\frac {a^3\,b\,x}{3}+\frac {a^2\,b^2\,x^2}{2}-a\,b^3\,x^3}{a^5\,x^4}-\frac {2\,b^4\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^5} \]
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