Integrand size = 49, antiderivative size = 26 \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=5 x \left (-x+\left (x+\frac {e^x+x}{x}\right )^2+\frac {\log (x)}{x}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {14, 2207, 2225, 2228} \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=5 x^3+5 x^2+5 x-10 e^x+10 e^x (x+2)+\frac {5 e^{2 x}}{x}+5 \log (x) \]
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Rule 14
Rule 2207
Rule 2225
Rule 2228
Rubi steps \begin{align*} \text {integral}& = \int \left (10 e^x (2+x)+\frac {5 e^{2 x} (-1+2 x)}{x^2}+\frac {5 \left (1+x+2 x^2+3 x^3\right )}{x}\right ) \, dx \\ & = 5 \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx+5 \int \frac {1+x+2 x^2+3 x^3}{x} \, dx+10 \int e^x (2+x) \, dx \\ & = \frac {5 e^{2 x}}{x}+10 e^x (2+x)+5 \int \left (1+\frac {1}{x}+2 x+3 x^2\right ) \, dx-10 \int e^x \, dx \\ & = -10 e^x+\frac {5 e^{2 x}}{x}+5 x+5 x^2+5 x^3+10 e^x (2+x)+5 \log (x) \\ \end{align*}
Time = 1.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=5 \left (\frac {e^{2 x}}{x}+x+x^2+x^3+e^x (2+2 x)+\log (x)\right ) \]
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Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38
method | result | size |
risch | \(5 x^{2}+5 x +5 \ln \left (x \right )+5 x^{3}+\frac {5 \,{\mathrm e}^{2 x}}{x}+\left (10 x +10\right ) {\mathrm e}^{x}\) | \(36\) |
default | \(5 x^{2}+5 x +5 \ln \left (x \right )+5 x^{3}+10 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+\frac {5 \,{\mathrm e}^{2 x}}{x}\) | \(37\) |
parts | \(5 x^{2}+5 x +5 \ln \left (x \right )+5 x^{3}+10 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}+\frac {5 \,{\mathrm e}^{2 x}}{x}\) | \(37\) |
norman | \(\frac {5 x^{2}+5 x^{3}+5 x^{4}+5 \,{\mathrm e}^{2 x}+10 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x} x^{2}}{x}+5 \ln \left (x \right )\) | \(44\) |
parallelrisch | \(\frac {5 x^{4}+5 x^{3}+10 \,{\mathrm e}^{x} x^{2}+5 x \ln \left (x \right )+5 x^{2}+10 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{2 x}}{x}\) | \(44\) |
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=\frac {5 \, {\left (x^{4} + x^{3} + x^{2} + 2 \, {\left (x^{2} + x\right )} e^{x} + x \log \left (x\right ) + e^{\left (2 \, x\right )}\right )}}{x} \]
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Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=5 x^{3} + 5 x^{2} + 5 x + 5 \log {\left (x \right )} + \frac {\left (10 x^{2} + 10 x\right ) e^{x} + 5 e^{2 x}}{x} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=5 \, x^{3} + 5 \, x^{2} + 10 \, {\left (x - 1\right )} e^{x} + 5 \, x + 10 \, {\rm Ei}\left (2 \, x\right ) + 20 \, e^{x} - 10 \, \Gamma \left (-1, -2 \, x\right ) + 5 \, \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=\frac {5 \, {\left (x^{4} + x^{3} + 2 \, x^{2} e^{x} + x^{2} + 2 \, x e^{x} + x \log \left (x\right ) + e^{\left (2 \, x\right )}\right )}}{x} \]
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Time = 7.88 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {5 x+5 x^2+10 x^3+15 x^4+e^{2 x} (-5+10 x)+e^x \left (20 x^2+10 x^3\right )}{x^2} \, dx=10\,{\mathrm {e}}^x+5\,\ln \left (x\right )+x\,\left (10\,{\mathrm {e}}^x+5\right )+\frac {5\,{\mathrm {e}}^{2\,x}}{x}+5\,x^2+5\,x^3 \]
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