Integrand size = 37, antiderivative size = 18 \[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=e^{2 x}+x+\frac {-28+x^2}{1+x} \]
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Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {27, 6874, 2225, 697} \[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=2 x+e^{2 x}-\frac {27}{x+1} \]
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Rule 27
Rule 697
Rule 2225
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{(1+x)^2} \, dx \\ & = \int \left (2 e^{2 x}+\frac {29+4 x+2 x^2}{(1+x)^2}\right ) \, dx \\ & = 2 \int e^{2 x} \, dx+\int \frac {29+4 x+2 x^2}{(1+x)^2} \, dx \\ & = e^{2 x}+\int \left (2+\frac {27}{(1+x)^2}\right ) \, dx \\ & = e^{2 x}+2 x-\frac {27}{1+x} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=e^{2 x}-\frac {27}{1+x}+2 (1+x) \]
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Time = 0.87 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89
method | result | size |
risch | \(2 x -\frac {27}{1+x}+{\mathrm e}^{2 x}\) | \(16\) |
parts | \(2 x -\frac {27}{1+x}+{\mathrm e}^{2 x}\) | \(16\) |
derivativedivides | \(2 x -\frac {54}{2+2 x}+{\mathrm e}^{2 x}\) | \(18\) |
default | \(2 x -\frac {54}{2+2 x}+{\mathrm e}^{2 x}\) | \(18\) |
norman | \(\frac {x \,{\mathrm e}^{2 x}+2 x^{2}-29+{\mathrm e}^{2 x}}{1+x}\) | \(24\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{2 x}+2 x^{2}-29+{\mathrm e}^{2 x}}{1+x}\) | \(24\) |
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Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=\frac {2 \, x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )} + 2 \, x - 27}{x + 1} \]
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Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=2 x + e^{2 x} - \frac {27}{x + 1} \]
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\[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=\int { \frac {2 \, x^{2} + 2 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, x + 29}{x^{2} + 2 \, x + 1} \,d x } \]
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none
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=\frac {2 \, x^{2} + x e^{\left (2 \, x\right )} + 2 \, x + e^{\left (2 \, x\right )} - 27}{x + 1} \]
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Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {29+4 x+2 x^2+e^{2 x} \left (2+4 x+2 x^2\right )}{1+2 x+x^2} \, dx=2\,x+{\mathrm {e}}^{2\,x}-\frac {27}{x+1} \]
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