Integrand size = 38, antiderivative size = 29 \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {25 e^5 (1+x) (\log (4)+\log (x))}{-x^2+x (x-\log (x))} \]
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {6820, 12, 6874, 2395, 2343, 2346, 2209, 2339, 30} \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=-\frac {25 e^5}{x}-\frac {25 e^5 \log (4)}{\log (x)}-\frac {25 e^5 \log (4)}{x \log (x)} \]
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Rule 12
Rule 30
Rule 2209
Rule 2339
Rule 2343
Rule 2346
Rule 2395
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {25 e^5 \left ((1+x) \log (4)+\log (4) \log (x)+\log ^2(x)\right )}{x^2 \log ^2(x)} \, dx \\ & = \left (25 e^5\right ) \int \frac {(1+x) \log (4)+\log (4) \log (x)+\log ^2(x)}{x^2 \log ^2(x)} \, dx \\ & = \left (25 e^5\right ) \int \left (\frac {1}{x^2}+\frac {(1+x) \log (4)}{x^2 \log ^2(x)}+\frac {\log (4)}{x^2 \log (x)}\right ) \, dx \\ & = -\frac {25 e^5}{x}+\left (25 e^5 \log (4)\right ) \int \frac {1+x}{x^2 \log ^2(x)} \, dx+\left (25 e^5 \log (4)\right ) \int \frac {1}{x^2 \log (x)} \, dx \\ & = -\frac {25 e^5}{x}+\left (25 e^5 \log (4)\right ) \int \left (\frac {1}{x^2 \log ^2(x)}+\frac {1}{x \log ^2(x)}\right ) \, dx+\left (25 e^5 \log (4)\right ) \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right ) \\ & = -\frac {25 e^5}{x}+25 e^5 \operatorname {ExpIntegralEi}(-\log (x)) \log (4)+\left (25 e^5 \log (4)\right ) \int \frac {1}{x^2 \log ^2(x)} \, dx+\left (25 e^5 \log (4)\right ) \int \frac {1}{x \log ^2(x)} \, dx \\ & = -\frac {25 e^5}{x}+25 e^5 \operatorname {ExpIntegralEi}(-\log (x)) \log (4)-\frac {25 e^5 \log (4)}{x \log (x)}-\left (25 e^5 \log (4)\right ) \int \frac {1}{x^2 \log (x)} \, dx+\left (25 e^5 \log (4)\right ) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right ) \\ & = -\frac {25 e^5}{x}+25 e^5 \operatorname {ExpIntegralEi}(-\log (x)) \log (4)-\frac {25 e^5 \log (4)}{\log (x)}-\frac {25 e^5 \log (4)}{x \log (x)}-\left (25 e^5 \log (4)\right ) \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right ) \\ & = -\frac {25 e^5}{x}-\frac {25 e^5 \log (4)}{\log (x)}-\frac {25 e^5 \log (4)}{x \log (x)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=25 e^5 \left (-\frac {1}{x}-\frac {(1+x) \log (4)}{x \log (x)}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
risch | \(-\frac {25 \,{\mathrm e}^{5}}{x}-\frac {50 \,{\mathrm e}^{5} \ln \left (2\right ) \left (1+x \right )}{x \ln \left (x \right )}\) | \(25\) |
norman | \(\frac {-50 \,{\mathrm e}^{5} \ln \left (2\right )-25 \,{\mathrm e}^{5} \ln \left (x \right )-50 x \,{\mathrm e}^{5} \ln \left (2\right )}{x \ln \left (x \right )}\) | \(29\) |
parallelrisch | \(-\frac {50 x \,{\mathrm e}^{5} \ln \left (2\right )+50 \,{\mathrm e}^{5} \ln \left (2\right )+25 \,{\mathrm e}^{5} \ln \left (x \right )}{x \ln \left (x \right )}\) | \(30\) |
parts | \(-\frac {25 \,{\mathrm e}^{5}}{x}-50 \,{\mathrm e}^{5} \ln \left (2\right ) \operatorname {Ei}_{1}\left (\ln \left (x \right )\right )+50 \,{\mathrm e}^{5} \ln \left (2\right ) \left (-\frac {1}{\ln \left (x \right )}-\frac {1}{x \ln \left (x \right )}+\operatorname {Ei}_{1}\left (\ln \left (x \right )\right )\right )\) | \(45\) |
default | \(-\frac {25 \,{\mathrm e}^{5}}{x}-50 \,{\mathrm e}^{5} \ln \left (2\right ) \operatorname {Ei}_{1}\left (\ln \left (x \right )\right )-\frac {50 \,{\mathrm e}^{5} \ln \left (2\right )}{\ln \left (x \right )}+50 \,{\mathrm e}^{5} \ln \left (2\right ) \left (-\frac {1}{x \ln \left (x \right )}+\operatorname {Ei}_{1}\left (\ln \left (x \right )\right )\right )\) | \(49\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=-\frac {25 \, {\left (2 \, {\left (x + 1\right )} e^{5} \log \left (2\right ) + e^{5} \log \left (x\right )\right )}}{x \log \left (x\right )} \]
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Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {- 50 x e^{5} \log {\left (2 \right )} - 50 e^{5} \log {\left (2 \right )}}{x \log {\left (x \right )}} - \frac {25 e^{5}}{x} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=50 \, {\rm Ei}\left (-\log \left (x\right )\right ) e^{5} \log \left (2\right ) - 50 \, e^{5} \Gamma \left (-1, \log \left (x\right )\right ) \log \left (2\right ) - \frac {50 \, e^{5} \log \left (2\right )}{\log \left (x\right )} - \frac {25 \, e^{5}}{x} \]
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Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=-\frac {25 \, {\left (2 \, x e^{5} \log \left (2\right ) + 2 \, e^{5} \log \left (2\right ) + e^{5} \log \left (x\right )\right )}}{x \log \left (x\right )} \]
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Time = 12.93 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx=-\frac {25\,{\mathrm {e}}^5}{x}-\frac {25\,{\mathrm {e}}^5\,\left (2\,\ln \left (2\right )+2\,x\,\ln \left (2\right )\right )}{x\,\ln \left (x\right )} \]
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