Integrand size = 8, antiderivative size = 18 \[ \int \frac {\log (1+x)}{x^2} \, dx=\log (x)-\log (1+x)-\frac {\log (1+x)}{x} \]
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Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2442, 36, 29, 31} \[ \int \frac {\log (1+x)}{x^2} \, dx=\log (x)-\frac {\log (x+1)}{x}-\log (x+1) \]
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Rule 29
Rule 31
Rule 36
Rule 2442
Rubi steps \begin{align*} \text {integral}& = -\frac {\log (1+x)}{x}+\int \frac {1}{x (1+x)} \, dx \\ & = -\frac {\log (1+x)}{x}+\int \frac {1}{x} \, dx-\int \frac {1}{1+x} \, dx \\ & = \log (x)-\log (1+x)-\frac {\log (1+x)}{x} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1+x)}{x^2} \, dx=\log (x)-\log (1+x)-\frac {\log (1+x)}{x} \]
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Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\ln \left (x \right )-\frac {\ln \left (1+x \right ) \left (1+x \right )}{x}\) | \(16\) |
default | \(\ln \left (x \right )-\frac {\ln \left (1+x \right ) \left (1+x \right )}{x}\) | \(16\) |
meijerg | \(\ln \left (x \right )-\frac {\left (2 x +2\right ) \ln \left (1+x \right )}{2 x}\) | \(18\) |
risch | \(\ln \left (x \right )-\ln \left (1+x \right )-\frac {\ln \left (1+x \right )}{x}\) | \(19\) |
parts | \(\ln \left (x \right )-\ln \left (1+x \right )-\frac {\ln \left (1+x \right )}{x}\) | \(19\) |
norman | \(\frac {-\ln \left (1+x \right ) x -\ln \left (1+x \right )}{x}+\ln \left (x \right )\) | \(22\) |
parallelrisch | \(\frac {x \ln \left (x \right )-\ln \left (1+x \right ) x -\ln \left (1+x \right )}{x}\) | \(23\) |
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none
Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {\log (1+x)}{x^2} \, dx=-\frac {{\left (x + 1\right )} \log \left (x + 1\right ) - x \log \left (x\right )}{x} \]
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Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\log (1+x)}{x^2} \, dx=\log {\left (x \right )} - \log {\left (x + 1 \right )} - \frac {\log {\left (x + 1 \right )}}{x} \]
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none
Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1+x)}{x^2} \, dx=-\frac {\log \left (x + 1\right )}{x} - \log \left (x + 1\right ) + \log \left (x\right ) \]
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none
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {\log (1+x)}{x^2} \, dx=-\frac {\log \left (x + 1\right )}{x} - \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]
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Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1+x)}{x^2} \, dx=-\ln \left (\frac {1}{x}+1\right )-\frac {\ln \left (x+1\right )}{x} \]
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