Integrand size = 74, antiderivative size = 31 \[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=e^{e^{x^2} \left (2+\frac {-5-x}{x}\right )^2-x}+(2+x)^2 \]
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\[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=\int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (2 (2+x)+\frac {e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x} \left (-50 e^{x^2}+10 e^{x^2} x+50 e^{x^2} x^2-x^3-20 e^{x^2} x^3+2 e^{x^2} x^4\right )}{x^3}\right ) \, dx \\ & = (2+x)^2+\int \frac {e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x} \left (-50 e^{x^2}+10 e^{x^2} x+50 e^{x^2} x^2-x^3-20 e^{x^2} x^3+2 e^{x^2} x^4\right )}{x^3} \, dx \\ & = (2+x)^2+\int \frac {e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x} \left (-x^3+2 e^{x^2} \left (-25+5 x+25 x^2-10 x^3+x^4\right )\right )}{x^3} \, dx \\ & = (2+x)^2+\int \left (-e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x}+\frac {2 e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2} (-5+x) \left (5-5 x^2+x^3\right )}{x^3}\right ) \, dx \\ & = (2+x)^2+2 \int \frac {e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2} (-5+x) \left (5-5 x^2+x^3\right )}{x^3} \, dx-\int e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x} \, dx \\ & = (2+x)^2+2 \int \left (-10 e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2}-\frac {25 e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2}}{x^3}+\frac {5 e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2}}{x^2}+\frac {25 e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2}}{x}+e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2} x\right ) \, dx-\int e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x} \, dx \\ & = (2+x)^2+2 \int e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2} x \, dx+10 \int \frac {e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2}}{x^2} \, dx-20 \int e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2} \, dx-50 \int \frac {e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2}}{x^3} \, dx+50 \int \frac {e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x+x^2}}{x} \, dx-\int e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x} \, dx \\ \end{align*}
Time = 2.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=e^{\frac {e^{x^2} (-5+x)^2}{x^2}-x}+4 x+x^2 \]
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Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(x^{2}+4 x +{\mathrm e}^{\frac {\left (x^{2}-10 x +25\right ) {\mathrm e}^{x^{2}}-x^{3}}{x^{2}}}\) | \(32\) |
risch | \(x^{2}+4 x +{\mathrm e}^{-\frac {-x^{2} {\mathrm e}^{x^{2}}+x^{3}+10 \,{\mathrm e}^{x^{2}} x -25 \,{\mathrm e}^{x^{2}}}{x^{2}}}\) | \(40\) |
norman | \(\frac {x^{4}+x^{2} {\mathrm e}^{\frac {\left (x^{2}-10 x +25\right ) {\mathrm e}^{x^{2}}-x^{3}}{x^{2}}}+4 x^{3}}{x^{2}}\) | \(42\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=x^{2} + 4 \, x + e^{\left (-\frac {x^{3} - {\left (x^{2} - 10 \, x + 25\right )} e^{\left (x^{2}\right )}}{x^{2}}\right )} \]
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Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=x^{2} + 4 x + e^{\frac {- x^{3} + \left (x^{2} - 10 x + 25\right ) e^{x^{2}}}{x^{2}}} \]
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Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=x^{2} + 4 \, x + e^{\left (-x - \frac {10 \, e^{\left (x^{2}\right )}}{x} + \frac {25 \, e^{\left (x^{2}\right )}}{x^{2}} + e^{\left (x^{2}\right )}\right )} \]
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\[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=\int { \frac {2 \, x^{4} + 4 \, x^{3} - {\left (x^{3} - 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2} + 5 \, x - 25\right )} e^{\left (x^{2}\right )}\right )} e^{\left (-\frac {x^{3} - {\left (x^{2} - 10 \, x + 25\right )} e^{\left (x^{2}\right )}}{x^{2}}\right )}}{x^{3}} \,d x } \]
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Time = 7.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {4 x^3+2 x^4+e^{\frac {-x^3+e^{x^2} \left (25-10 x+x^2\right )}{x^2}} \left (-x^3+e^{x^2} \left (-50+10 x+50 x^2-20 x^3+2 x^4\right )\right )}{x^3} \, dx=4\,x+x^2+{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{-\frac {10\,{\mathrm {e}}^{x^2}}{x}}\,{\mathrm {e}}^{\frac {25\,{\mathrm {e}}^{x^2}}{x^2}} \]
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