Integrand size = 51, antiderivative size = 28 \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=-1-x+e^{\left (3+x^2\right )^2} \left (x+x^2\right ) \log (4)-\log (5 x) \]
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Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14, 45, 2326} \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=\frac {e^{x^4+6 x^2+9} \left (x^5+x^4+3 x^3+3 x^2\right ) \log (4)}{x^3+3 x}-x-\log (x) \]
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Rule 14
Rule 45
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-1-x}{x}+e^{9+6 x^2+x^4} \left (1+2 x+12 x^2+12 x^3+4 x^4+4 x^5\right ) \log (4)\right ) \, dx \\ & = \log (4) \int e^{9+6 x^2+x^4} \left (1+2 x+12 x^2+12 x^3+4 x^4+4 x^5\right ) \, dx+\int \frac {-1-x}{x} \, dx \\ & = \frac {e^{9+6 x^2+x^4} \left (3 x^2+3 x^3+x^4+x^5\right ) \log (4)}{3 x+x^3}+\int \left (-1-\frac {1}{x}\right ) \, dx \\ & = -x+\frac {e^{9+6 x^2+x^4} \left (3 x^2+3 x^3+x^4+x^5\right ) \log (4)}{3 x+x^3}-\log (x) \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=x \left (-1+e^{\left (3+x^2\right )^2} (1+x) \log (4)\right )-\log (x) \]
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
method | result | size |
risch | \(-x -\ln \left (x \right )+\left (2 x^{2} \ln \left (2\right )+2 x \ln \left (2\right )\right ) {\mathrm e}^{\left (x^{2}+3\right )^{2}}\) | \(31\) |
norman | \(-x +2 \ln \left (2\right ) {\mathrm e}^{x^{4}+6 x^{2}+9} x +2 \ln \left (2\right ) {\mathrm e}^{x^{4}+6 x^{2}+9} x^{2}-\ln \left (x \right )\) | \(43\) |
parallelrisch | \(-x +2 \ln \left (2\right ) {\mathrm e}^{x^{4}+6 x^{2}+9} x +2 \ln \left (2\right ) {\mathrm e}^{x^{4}+6 x^{2}+9} x^{2}-\ln \left (x \right )\) | \(43\) |
parts | \(-x +2 \ln \left (2\right ) {\mathrm e}^{x^{4}+6 x^{2}+9} x +2 \ln \left (2\right ) {\mathrm e}^{x^{4}+6 x^{2}+9} x^{2}-\ln \left (x \right )\) | \(43\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=2 \, {\left (x^{2} + x\right )} e^{\left (x^{4} + 6 \, x^{2} + 9\right )} \log \left (2\right ) - x - \log \left (x\right ) \]
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Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=- x + \left (2 x^{2} \log {\left (2 \right )} + 2 x \log {\left (2 \right )}\right ) e^{x^{4} + 6 x^{2} + 9} - \log {\left (x \right )} \]
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Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=2 \, {\left (x^{2} e^{9} \log \left (2\right ) + x e^{9} \log \left (2\right )\right )} e^{\left (x^{4} + 6 \, x^{2}\right )} - x - \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=2 \, x^{2} e^{\left (x^{4} + 6 \, x^{2} + 9\right )} \log \left (2\right ) + 2 \, x e^{\left (x^{4} + 6 \, x^{2} + 9\right )} \log \left (2\right ) - x - \log \left (x\right ) \]
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Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-1-x+e^{9+6 x^2+x^4} \left (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6\right ) \log (4)}{x} \, dx=2\,x\,{\mathrm {e}}^{x^4+6\,x^2+9}\,\ln \left (2\right )-\ln \left (x\right )-x+2\,x^2\,{\mathrm {e}}^{x^4+6\,x^2+9}\,\ln \left (2\right ) \]
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