Integrand size = 73, antiderivative size = 31 \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=e^x+3 \left (e^x-\frac {x}{-2+2 x-x^2}+\log \left (\frac {x}{5}\right )\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(104\) vs. \(2(31)=62\).
Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.35, number of steps used = 26, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6873, 6874, 2225, 628, 631, 210, 754, 814, 648, 642, 652, 736, 752, 787} \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=-\frac {3 (2-x) x^2}{2 \left (x^2-2 x+2\right )}+\frac {15 (2-x) x}{2 \left (x^2-2 x+2\right )}+\frac {3 x}{x^2-2 x+2}+\frac {9 (1-x)}{x^2-2 x+2}-\frac {12 (2-x)}{x^2-2 x+2}-\frac {3 x}{2}+4 e^x+3 \log (x) \]
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Rule 210
Rule 628
Rule 631
Rule 642
Rule 648
Rule 652
Rule 736
Rule 752
Rule 754
Rule 787
Rule 814
Rule 2225
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{x \left (2-2 x+x^2\right )^2} \, dx \\ & = \int \left (4 e^x-\frac {18}{\left (2-2 x+x^2\right )^2}+\frac {12}{x \left (2-2 x+x^2\right )^2}+\frac {24 x}{\left (2-2 x+x^2\right )^2}-\frac {15 x^2}{\left (2-2 x+x^2\right )^2}+\frac {3 x^3}{\left (2-2 x+x^2\right )^2}\right ) \, dx \\ & = 3 \int \frac {x^3}{\left (2-2 x+x^2\right )^2} \, dx+4 \int e^x \, dx+12 \int \frac {1}{x \left (2-2 x+x^2\right )^2} \, dx-15 \int \frac {x^2}{\left (2-2 x+x^2\right )^2} \, dx-18 \int \frac {1}{\left (2-2 x+x^2\right )^2} \, dx+24 \int \frac {x}{\left (2-2 x+x^2\right )^2} \, dx \\ & = 4 e^x+\frac {9 (1-x)}{2-2 x+x^2}-\frac {12 (2-x)}{2-2 x+x^2}+\frac {3 x}{2-2 x+x^2}+\frac {15 (2-x) x}{2 \left (2-2 x+x^2\right )}-\frac {3 (2-x) x^2}{2 \left (2-2 x+x^2\right )}+\frac {3}{4} \int \frac {(8-2 x) x}{2-2 x+x^2} \, dx+\frac {3}{2} \int \frac {4+2 x}{x \left (2-2 x+x^2\right )} \, dx-9 \int \frac {1}{2-2 x+x^2} \, dx+12 \int \frac {1}{2-2 x+x^2} \, dx-15 \int \frac {1}{2-2 x+x^2} \, dx \\ & = 4 e^x-\frac {3 x}{2}+\frac {9 (1-x)}{2-2 x+x^2}-\frac {12 (2-x)}{2-2 x+x^2}+\frac {3 x}{2-2 x+x^2}+\frac {15 (2-x) x}{2 \left (2-2 x+x^2\right )}-\frac {3 (2-x) x^2}{2 \left (2-2 x+x^2\right )}+\frac {3}{4} \int \frac {4+4 x}{2-2 x+x^2} \, dx+\frac {3}{2} \int \left (\frac {2}{x}-\frac {2 (-3+x)}{2-2 x+x^2}\right ) \, dx-9 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-x\right )+12 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-x\right )-15 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-x\right ) \\ & = 4 e^x-\frac {3 x}{2}+\frac {9 (1-x)}{2-2 x+x^2}-\frac {12 (2-x)}{2-2 x+x^2}+\frac {3 x}{2-2 x+x^2}+\frac {15 (2-x) x}{2 \left (2-2 x+x^2\right )}-\frac {3 (2-x) x^2}{2 \left (2-2 x+x^2\right )}+12 \arctan (1-x)+3 \log (x)+\frac {3}{2} \int \frac {-2+2 x}{2-2 x+x^2} \, dx-3 \int \frac {-3+x}{2-2 x+x^2} \, dx+6 \int \frac {1}{2-2 x+x^2} \, dx \\ & = 4 e^x-\frac {3 x}{2}+\frac {9 (1-x)}{2-2 x+x^2}-\frac {12 (2-x)}{2-2 x+x^2}+\frac {3 x}{2-2 x+x^2}+\frac {15 (2-x) x}{2 \left (2-2 x+x^2\right )}-\frac {3 (2-x) x^2}{2 \left (2-2 x+x^2\right )}+12 \arctan (1-x)+3 \log (x)+\frac {3}{2} \log \left (2-2 x+x^2\right )-\frac {3}{2} \int \frac {-2+2 x}{2-2 x+x^2} \, dx+6 \int \frac {1}{2-2 x+x^2} \, dx+6 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-x\right ) \\ & = 4 e^x-\frac {3 x}{2}+\frac {9 (1-x)}{2-2 x+x^2}-\frac {12 (2-x)}{2-2 x+x^2}+\frac {3 x}{2-2 x+x^2}+\frac {15 (2-x) x}{2 \left (2-2 x+x^2\right )}-\frac {3 (2-x) x^2}{2 \left (2-2 x+x^2\right )}+6 \arctan (1-x)+3 \log (x)+6 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-x\right ) \\ & = 4 e^x-\frac {3 x}{2}+\frac {9 (1-x)}{2-2 x+x^2}-\frac {12 (2-x)}{2-2 x+x^2}+\frac {3 x}{2-2 x+x^2}+\frac {15 (2-x) x}{2 \left (2-2 x+x^2\right )}-\frac {3 (2-x) x^2}{2 \left (2-2 x+x^2\right )}+3 \log (x) \\ \end{align*}
Time = 1.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=4 e^x+3 \left (\frac {x}{2-2 x+x^2}+\log (x)\right ) \]
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Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {3 x}{x^{2}-2 x +2}+3 \ln \left (x \right )+4 \,{\mathrm e}^{x}\) | \(23\) |
parts | \(\frac {3 x}{x^{2}-2 x +2}+3 \ln \left (x \right )+4 \,{\mathrm e}^{x}\) | \(23\) |
norman | \(\frac {3 x -8 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x} x^{2}+8 \,{\mathrm e}^{x}}{x^{2}-2 x +2}+3 \ln \left (x \right )\) | \(37\) |
parallelrisch | \(\frac {3 x^{2} \ln \left (x \right )+4 \,{\mathrm e}^{x} x^{2}-6 x \ln \left (x \right )-8 \,{\mathrm e}^{x} x +6 \ln \left (x \right )+8 \,{\mathrm e}^{x}+3 x}{x^{2}-2 x +2}\) | \(48\) |
default | \(3 \ln \left (x \right )-\frac {9 \left (-2+2 x \right )}{2 \left (x^{2}-2 x +2\right )}+\frac {12 x -24}{x^{2}-2 x +2}+\frac {15}{x^{2}-2 x +2}+4 \,{\mathrm e}^{x}-\frac {16 \,{\mathrm e}^{x} \left (-2+x \right )}{x^{2}-2 x +2}-\frac {32 \,{\mathrm e}^{x}}{x^{2}-2 x +2}+\frac {16 \,{\mathrm e}^{x} x}{x^{2}-2 x +2}\) | \(102\) |
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=\frac {4 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 3 \, {\left (x^{2} - 2 \, x + 2\right )} \log \left (x\right ) + 3 \, x}{x^{2} - 2 \, x + 2} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=\frac {3 x}{x^{2} - 2 x + 2} + 4 e^{x} + 3 \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (24) = 48\).
Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=-\frac {9 \, {\left (x - 1\right )}}{x^{2} - 2 \, x + 2} + \frac {12 \, {\left (x - 2\right )}}{x^{2} - 2 \, x + 2} + \frac {15}{x^{2} - 2 \, x + 2} + 4 \, e^{x} + 3 \, \log \left (x\right ) \]
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Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=\frac {4 \, x^{2} e^{x} + 3 \, x^{2} \log \left (x\right ) - 8 \, x e^{x} - 6 \, x \log \left (x\right ) + 3 \, x + 8 \, e^{x} + 6 \, \log \left (x\right )}{x^{2} - 2 \, x + 2} \]
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Time = 8.90 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {12-18 x+24 x^2-15 x^3+3 x^4+e^x \left (16 x-32 x^2+32 x^3-16 x^4+4 x^5\right )}{4 x-8 x^2+8 x^3-4 x^4+x^5} \, dx=4\,{\mathrm {e}}^x+3\,\ln \left (x\right )+\frac {3\,x}{x^2-2\,x+2} \]
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