Integrand size = 25, antiderivative size = 27 \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=5 \left (x+9 \left (e^{e^2}-e^{1+x}+x-\frac {\log (x)}{x^2}\right )\right ) \]
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Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {14, 2225, 2341} \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=-\frac {45 \log (x)}{x^2}+50 x-45 e^{x+1} \]
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Rule 14
Rule 2225
Rule 2341
Rubi steps \begin{align*} \text {integral}& = \int \left (-45 e^{1+x}+\frac {5 \left (-9+10 x^3+18 \log (x)\right )}{x^3}\right ) \, dx \\ & = 5 \int \frac {-9+10 x^3+18 \log (x)}{x^3} \, dx-45 \int e^{1+x} \, dx \\ & = -45 e^{1+x}+5 \int \left (\frac {-9+10 x^3}{x^3}+\frac {18 \log (x)}{x^3}\right ) \, dx \\ & = -45 e^{1+x}+5 \int \frac {-9+10 x^3}{x^3} \, dx+90 \int \frac {\log (x)}{x^3} \, dx \\ & = -45 e^{1+x}-\frac {45}{2 x^2}-\frac {45 \log (x)}{x^2}+5 \int \left (10-\frac {9}{x^3}\right ) \, dx \\ & = -45 e^{1+x}+50 x-\frac {45 \log (x)}{x^2} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=-45 e^{1+x}+50 x-\frac {45 \log (x)}{x^2} \]
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Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67
method | result | size |
default | \(50 x -\frac {45 \ln \left (x \right )}{x^{2}}-45 \,{\mathrm e}^{1+x}\) | \(18\) |
risch | \(50 x -\frac {45 \ln \left (x \right )}{x^{2}}-45 \,{\mathrm e}^{1+x}\) | \(18\) |
parts | \(50 x -\frac {45 \ln \left (x \right )}{x^{2}}-45 \,{\mathrm e}^{1+x}\) | \(18\) |
norman | \(\frac {50 x^{3}-45 x^{2} {\mathrm e}^{1+x}-45 \ln \left (x \right )}{x^{2}}\) | \(24\) |
parallelrisch | \(-\frac {-50 x^{3}+45 x^{2} {\mathrm e}^{1+x}+45 \ln \left (x \right )}{x^{2}}\) | \(25\) |
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=\frac {5 \, {\left (10 \, x^{3} - 9 \, x^{2} e^{\left (x + 1\right )} - 9 \, \log \left (x\right )\right )}}{x^{2}} \]
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Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=50 x - 45 e^{x + 1} - \frac {45 \log {\left (x \right )}}{x^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=50 \, x - \frac {45 \, \log \left (x\right )}{x^{2}} - 45 \, e^{\left (x + 1\right )} \]
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=\frac {5 \, {\left (10 \, x^{3} - 9 \, x^{2} e^{\left (x + 1\right )} - 9 \, \log \left (x\right )\right )}}{x^{2}} \]
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Time = 15.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx=50\,x-\frac {45\,\ln \left (x\right )}{x^2}-45\,\mathrm {e}\,{\mathrm {e}}^x \]
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