Integrand size = 43, antiderivative size = 25 \[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=x-e^{8-2 x} x+\frac {x (x+\log (7))}{x+x^2} \]
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Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {27, 6873, 6874, 2207, 2225, 697} \[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=\frac {1}{2} e^{8-2 x} (1-2 x)-\frac {1}{2} e^{8-2 x}+x-\frac {1-\log (7)}{x+1} \]
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Rule 27
Rule 697
Rule 2207
Rule 2225
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{(1+x)^2} \, dx \\ & = \int \frac {2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )+2 \left (1-\frac {\log (7)}{2}\right )}{(1+x)^2} \, dx \\ & = \int \left (e^{8-2 x} (-1+2 x)+\frac {2+2 x+x^2-\log (7)}{(1+x)^2}\right ) \, dx \\ & = \int e^{8-2 x} (-1+2 x) \, dx+\int \frac {2+2 x+x^2-\log (7)}{(1+x)^2} \, dx \\ & = \frac {1}{2} e^{8-2 x} (1-2 x)+\int e^{8-2 x} \, dx+\int \left (1+\frac {1-\log (7)}{(1+x)^2}\right ) \, dx \\ & = -\frac {1}{2} e^{8-2 x}+\frac {1}{2} e^{8-2 x} (1-2 x)+x-\frac {1-\log (7)}{1+x} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=x-e^{8-2 x} x+\frac {-1+\log (7)}{1+x} \]
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Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
risch | \(x +\frac {\ln \left (7\right )}{1+x}-\frac {1}{1+x}-x \,{\mathrm e}^{-2 x +8}\) | \(27\) |
parts | \(-4 \,{\mathrm e}^{-2 x +8}+{\mathrm e}^{-2 x +8} \left (-x +4\right )+x +\frac {\ln \left (7\right )-1}{1+x}\) | \(37\) |
norman | \(\frac {x^{2}-x \,{\mathrm e}^{-2 x +8}-x^{2} {\mathrm e}^{-2 x +8}-2+\ln \left (7\right )}{1+x}\) | \(38\) |
parallelrisch | \(\frac {x^{2}-x \,{\mathrm e}^{-2 x +8}-x^{2} {\mathrm e}^{-2 x +8}-2+\ln \left (7\right )}{1+x}\) | \(38\) |
derivativedivides | \(x -4+\frac {1}{-1-x}-\frac {\ln \left (7\right )}{-1-x}-4 \,{\mathrm e}^{-2 x +8}+{\mathrm e}^{-2 x +8} \left (-x +4\right )\) | \(46\) |
default | \(x -4+\frac {1}{-1-x}-\frac {\ln \left (7\right )}{-1-x}-4 \,{\mathrm e}^{-2 x +8}+{\mathrm e}^{-2 x +8} \left (-x +4\right )\) | \(46\) |
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Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=\frac {x^{2} - {\left (x^{2} + x\right )} e^{\left (-2 \, x + 8\right )} + x + \log \left (7\right ) - 1}{x + 1} \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=- x e^{8 - 2 x} + x + \frac {-1 + \log {\left (7 \right )}}{x + 1} \]
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\[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=\int { \frac {x^{2} + {\left (2 \, x^{3} + 3 \, x^{2} - 1\right )} e^{\left (-2 \, x + 8\right )} + 2 \, x - \log \left (7\right ) + 2}{x^{2} + 2 \, x + 1} \,d x } \]
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=-\frac {x^{2} e^{\left (-2 \, x + 8\right )} - x^{2} + x e^{\left (-2 \, x + 8\right )} - x - \log \left (7\right ) + 1}{x + 1} \]
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Time = 10.76 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {2+2 x+x^2+e^{8-2 x} \left (-1+3 x^2+2 x^3\right )-\log (7)}{1+2 x+x^2} \, dx=\frac {\ln \left (7\right )-1}{x+1}-x\,\left ({\mathrm {e}}^{8-2\,x}-1\right ) \]
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