Integrand size = 47, antiderivative size = 21 \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=\frac {2 \left (1+e^{5+e}\right )}{4+\frac {1}{x^2}+\frac {2}{x}} \]
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Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {1694, 12, 1828, 8} \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=-\frac {2 \left (1+e^{5+e}\right ) (2 x+1)}{16 \left (x+\frac {1}{4}\right )^2+3} \]
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Rule 8
Rule 12
Rule 1694
Rule 1828
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {4 \left (1+e^{5+e}\right ) \left (-3+8 x+16 x^2\right )}{\left (3+16 x^2\right )^2} \, dx,x,\frac {1}{4}+x\right ) \\ & = \left (4 \left (1+e^{5+e}\right )\right ) \text {Subst}\left (\int \frac {-3+8 x+16 x^2}{\left (3+16 x^2\right )^2} \, dx,x,\frac {1}{4}+x\right ) \\ & = -\frac {2 \left (1+e^{5+e}\right ) (1+2 x)}{3+(1+4 x)^2}-\frac {1}{3} \left (2 \left (1+e^{5+e}\right )\right ) \text {Subst}\left (\int 0 \, dx,x,\frac {1}{4}+x\right ) \\ & = -\frac {2 \left (1+e^{5+e}\right ) (1+2 x)}{3+(1+4 x)^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=\frac {\left (1+e^{5+e}\right ) (-1-2 x)}{2 \left (1+2 x+4 x^2\right )} \]
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Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {\left (4 \,{\mathrm e}^{{\mathrm e}+5}+4\right ) \left (-\frac {x}{16}-\frac {1}{32}\right )}{x^{2}+\frac {1}{2} x +\frac {1}{4}}\) | \(26\) |
gosper | \(-\frac {\left (1+2 x \right ) \left ({\mathrm e}^{{\mathrm e}+5}+1\right )}{2 \left (4 x^{2}+2 x +1\right )}\) | \(27\) |
risch | \(\frac {\left (-\frac {{\mathrm e}^{{\mathrm e}+5}}{4}-\frac {1}{4}\right ) x -\frac {{\mathrm e}^{{\mathrm e}+5}}{8}-\frac {1}{8}}{x^{2}+\frac {1}{2} x +\frac {1}{4}}\) | \(32\) |
norman | \(\frac {\left (-{\mathrm e}^{{\mathrm e}} {\mathrm e}^{5}-1\right ) x -\frac {{\mathrm e}^{{\mathrm e}} {\mathrm e}^{5}}{2}-\frac {1}{2}}{4 x^{2}+2 x +1}\) | \(34\) |
parallelrisch | \(-\frac {2+4 \,{\mathrm e}^{{\mathrm e}+5} x +2 \,{\mathrm e}^{{\mathrm e}+5}+4 x}{4 \left (4 x^{2}+2 x +1\right )}\) | \(35\) |
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Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=-\frac {{\left (2 \, x + 1\right )} e^{\left (e + 5\right )} + 2 \, x + 1}{2 \, {\left (4 \, x^{2} + 2 \, x + 1\right )}} \]
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Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=\frac {x \left (- 2 e^{5} e^{e} - 2\right ) - e^{5} e^{e} - 1}{8 x^{2} + 4 x + 2} \]
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Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=-\frac {2 \, x {\left (e^{\left (e + 5\right )} + 1\right )} + e^{\left (e + 5\right )} + 1}{2 \, {\left (4 \, x^{2} + 2 \, x + 1\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=-\frac {2 \, x e^{\left (e + 5\right )} + 2 \, x + e^{\left (e + 5\right )} + 1}{2 \, {\left (4 \, x^{2} + 2 \, x + 1\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {4 x+4 x^2+e^{5+e} \left (4 x+4 x^2\right )}{1+4 x+12 x^2+16 x^3+16 x^4} \, dx=-\frac {\left (2\,x+1\right )\,\left ({\mathrm {e}}^{\mathrm {e}+5}+1\right )}{2\,\left (4\,x^2+2\,x+1\right )} \]
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