Integrand size = 74, antiderivative size = 32 \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=-e^3 \left (x+x \left (-x^2+\frac {5 (4+x) \log (3)}{1+x}\right )\right )+\log (3+x) \]
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Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2099} \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=e^3 x^3-e^3 x (1+\log (243))+\log (x+3)+\frac {15 e^3 \log (3)}{x+1} \]
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Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (3 e^3 x^2+\frac {1}{3+x}-\frac {15 e^3 \log (3)}{(1+x)^2}-e^3 (1+\log (243))\right ) \, dx \\ & = e^3 x^3+\frac {15 e^3 \log (3)}{1+x}-e^3 x (1+\log (243))+\log (3+x) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=-3 e^3 (1+x)^2+e^3 (1+x)^3+e^3 (1+x) (2-5 \log (3))+\frac {e^3 (25 \log (3)-2 \log (243))}{1+x}+\log (3+x) \]
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Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
default | \(x^{3} {\mathrm e}^{3}-5 x \,{\mathrm e}^{3} \ln \left (3\right )-x \,{\mathrm e}^{3}+\frac {15 \,{\mathrm e}^{3} \ln \left (3\right )}{1+x}+\ln \left (3+x \right )\) | \(35\) |
risch | \(x^{3} {\mathrm e}^{3}-5 x \,{\mathrm e}^{3} \ln \left (3\right )-x \,{\mathrm e}^{3}+\frac {15 \,{\mathrm e}^{3} \ln \left (3\right )}{1+x}+\ln \left (3+x \right )\) | \(35\) |
norman | \(\frac {x^{3} {\mathrm e}^{3}+x^{4} {\mathrm e}^{3}+\left (-5 \,{\mathrm e}^{3} \ln \left (3\right )-{\mathrm e}^{3}\right ) x^{2}+20 \,{\mathrm e}^{3} \ln \left (3\right )+{\mathrm e}^{3}}{1+x}+\ln \left (3+x \right )\) | \(48\) |
parallelrisch | \(-\frac {-x^{4} {\mathrm e}^{3}+5 x^{2} {\mathrm e}^{3} \ln \left (3\right )-x^{3} {\mathrm e}^{3}+x^{2} {\mathrm e}^{3}-20 \,{\mathrm e}^{3} \ln \left (3\right )-x \ln \left (3+x \right )-{\mathrm e}^{3}-\ln \left (3+x \right )}{1+x}\) | \(61\) |
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Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=-\frac {5 \, {\left (x^{2} + x - 3\right )} e^{3} \log \left (3\right ) - {\left (x^{4} + x^{3} - x^{2} - x\right )} e^{3} - {\left (x + 1\right )} \log \left (x + 3\right )}{x + 1} \]
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Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=x^{3} e^{3} + x \left (- 5 e^{3} \log {\left (3 \right )} - e^{3}\right ) + \log {\left (x + 3 \right )} + \frac {15 e^{3} \log {\left (3 \right )}}{x + 1} \]
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Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=x^{3} e^{3} - {\left (5 \, e^{3} \log \left (3\right ) + e^{3}\right )} x + \frac {15 \, e^{3} \log \left (3\right )}{x + 1} + \log \left (x + 3\right ) \]
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=x^{3} e^{3} - 5 \, x e^{3} \log \left (3\right ) - x e^{3} + \frac {15 \, e^{3} \log \left (3\right )}{x + 1} + \log \left ({\left | x + 3 \right |}\right ) \]
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Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {1+2 x+x^2+e^3 \left (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5\right )+e^3 \left (-60-50 x-25 x^2-5 x^3\right ) \log (3)}{3+7 x+5 x^2+x^3} \, dx=\ln \left (x+3\right )+x^3\,{\mathrm {e}}^3-x\,\left ({\mathrm {e}}^3+5\,{\mathrm {e}}^3\,\ln \left (3\right )\right )+\frac {15\,{\mathrm {e}}^3\,\ln \left (3\right )}{x+1} \]
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