Integrand size = 45, antiderivative size = 21 \[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=\frac {2 \left (-10+e^x\right ) (5-x)}{x \log (25 x)} \]
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Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(21)=42\).
Time = 0.54 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {6874, 2395, 2343, 2346, 2209, 2339, 30, 2326} \[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-\frac {100}{x \log (25 x)}+\frac {20}{\log (25 x)} \]
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Rule 30
Rule 2209
Rule 2326
Rule 2339
Rule 2343
Rule 2346
Rule 2395
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {20 (-5+x-5 \log (25 x))}{x^2 \log ^2(25 x)}-\frac {2 e^x \left (5-x+5 \log (25 x)-5 x \log (25 x)+x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}\right ) \, dx \\ & = -\left (2 \int \frac {e^x \left (5-x+5 \log (25 x)-5 x \log (25 x)+x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)} \, dx\right )-20 \int \frac {-5+x-5 \log (25 x)}{x^2 \log ^2(25 x)} \, dx \\ & = \frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \left (\frac {-5+x}{x^2 \log ^2(25 x)}-\frac {5}{x^2 \log (25 x)}\right ) \, dx \\ & = \frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \frac {-5+x}{x^2 \log ^2(25 x)} \, dx+100 \int \frac {1}{x^2 \log (25 x)} \, dx \\ & = \frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \left (-\frac {5}{x^2 \log ^2(25 x)}+\frac {1}{x \log ^2(25 x)}\right ) \, dx+2500 \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (25 x)\right ) \\ & = 2500 \operatorname {ExpIntegralEi}(-\log (25 x))+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \frac {1}{x \log ^2(25 x)} \, dx+100 \int \frac {1}{x^2 \log ^2(25 x)} \, dx \\ & = 2500 \operatorname {ExpIntegralEi}(-\log (25 x))-\frac {100}{x \log (25 x)}+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (25 x)\right )-100 \int \frac {1}{x^2 \log (25 x)} \, dx \\ & = 2500 \operatorname {ExpIntegralEi}(-\log (25 x))+\frac {20}{\log (25 x)}-\frac {100}{x \log (25 x)}+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-2500 \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (25 x)\right ) \\ & = \frac {20}{\log (25 x)}-\frac {100}{x \log (25 x)}+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)} \\ \end{align*}
Time = 0.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=-\frac {2 \left (-10+e^x\right ) (-5+x)}{x \log (25 x)} \]
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Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19
method | result | size |
norman | \(\frac {-100+20 x -2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}}{x \ln \left (25 x \right )}\) | \(25\) |
risch | \(-\frac {2 \left ({\mathrm e}^{x} x -10 x -5 \,{\mathrm e}^{x}+50\right )}{x \ln \left (25 x \right )}\) | \(25\) |
parallelrisch | \(-\frac {100+2 \,{\mathrm e}^{x} x -20 x -10 \,{\mathrm e}^{x}}{x \ln \left (25 x \right )}\) | \(26\) |
default | \(\frac {-2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}}{\ln \left (25 x \right ) x}-\frac {100}{x \ln \left (25 x \right )}+\frac {20}{\ln \left (25 x \right )}\) | \(41\) |
parts | \(\frac {-2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}}{\ln \left (25 x \right ) x}-\frac {100}{x \ln \left (25 x \right )}+\frac {20}{\ln \left (25 x \right )}\) | \(41\) |
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=-\frac {2 \, {\left ({\left (x - 5\right )} e^{x} - 10 \, x + 50\right )}}{x \log \left (25 \, x\right )} \]
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Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=\frac {\left (10 - 2 x\right ) e^{x}}{x \log {\left (25 x \right )}} + \frac {20 x - 100}{x \log {\left (25 x \right )}} \]
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\[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=\int { \frac {2 \, {\left ({\left (x - 5\right )} e^{x} - {\left ({\left (x^{2} - 5 \, x + 5\right )} e^{x} - 50\right )} \log \left (25 \, x\right ) - 10 \, x + 50\right )}}{x^{2} \log \left (25 \, x\right )^{2}} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=-\frac {2 \, {\left (x e^{x} - 10 \, x - 5 \, e^{x} + 50\right )}}{x \log \left (25 \, x\right )} \]
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Time = 13.80 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {100-20 x+e^x (-10+2 x)+\left (100+e^x \left (-10+10 x-2 x^2\right )\right ) \log (25 x)}{x^2 \log ^2(25 x)} \, dx=-\frac {2\,\left ({\mathrm {e}}^x-10\right )\,\left (x-5\right )}{x\,\ln \left (25\,x\right )} \]
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