Integrand size = 135, antiderivative size = 33 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=(3-x) \left (e^x-\log \left (-\frac {5 e^{x^2}}{16}+\frac {4}{x}-x^2\right )\right ) \]
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\[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=\int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (2 e^x-6 x-e^x x+2 x^2-\frac {32 \left (6-2 x+12 x^2-x^3-x^4-3 x^5+x^6\right )}{x \left (-64+5 e^{x^2} x+16 x^3\right )}+\log \left (-\frac {-64+5 e^{x^2} x+16 x^3}{16 x}\right )\right ) \, dx \\ & = -3 x^2+\frac {2 x^3}{3}+2 \int e^x \, dx-32 \int \frac {6-2 x+12 x^2-x^3-x^4-3 x^5+x^6}{x \left (-64+5 e^{x^2} x+16 x^3\right )} \, dx-\int e^x x \, dx+\int \log \left (-\frac {-64+5 e^{x^2} x+16 x^3}{16 x}\right ) \, dx \\ & = 2 e^x-e^x x-3 x^2+\frac {2 x^3}{3}+x \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )-32 \int \left (-\frac {2}{-64+5 e^{x^2} x+16 x^3}+\frac {6}{x \left (-64+5 e^{x^2} x+16 x^3\right )}+\frac {12 x}{-64+5 e^{x^2} x+16 x^3}-\frac {x^2}{-64+5 e^{x^2} x+16 x^3}-\frac {x^3}{-64+5 e^{x^2} x+16 x^3}-\frac {3 x^4}{-64+5 e^{x^2} x+16 x^3}+\frac {x^5}{-64+5 e^{x^2} x+16 x^3}\right ) \, dx+\int e^x \, dx-\int \frac {64+2 \left (16+5 e^{x^2}\right ) x^3}{5 e^{x^2} x+16 \left (-4+x^3\right )} \, dx \\ & = 3 e^x-e^x x-3 x^2+\frac {2 x^3}{3}+x \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )+32 \int \frac {x^2}{-64+5 e^{x^2} x+16 x^3} \, dx+32 \int \frac {x^3}{-64+5 e^{x^2} x+16 x^3} \, dx-32 \int \frac {x^5}{-64+5 e^{x^2} x+16 x^3} \, dx+64 \int \frac {1}{-64+5 e^{x^2} x+16 x^3} \, dx+96 \int \frac {x^4}{-64+5 e^{x^2} x+16 x^3} \, dx-192 \int \frac {1}{x \left (-64+5 e^{x^2} x+16 x^3\right )} \, dx-384 \int \frac {x}{-64+5 e^{x^2} x+16 x^3} \, dx-\int \left (2 x^2-\frac {32 \left (-2-4 x^2-x^3+x^5\right )}{-64+5 e^{x^2} x+16 x^3}\right ) \, dx \\ & = 3 e^x-e^x x-3 x^2+x \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )+32 \int \frac {x^2}{-64+5 e^{x^2} x+16 x^3} \, dx+32 \int \frac {x^3}{-64+5 e^{x^2} x+16 x^3} \, dx-32 \int \frac {x^5}{-64+5 e^{x^2} x+16 x^3} \, dx+32 \int \frac {-2-4 x^2-x^3+x^5}{-64+5 e^{x^2} x+16 x^3} \, dx+64 \int \frac {1}{-64+5 e^{x^2} x+16 x^3} \, dx+96 \int \frac {x^4}{-64+5 e^{x^2} x+16 x^3} \, dx-192 \int \frac {1}{x \left (-64+5 e^{x^2} x+16 x^3\right )} \, dx-384 \int \frac {x}{-64+5 e^{x^2} x+16 x^3} \, dx \\ & = 3 e^x-e^x x-3 x^2+x \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )+32 \int \frac {x^2}{-64+5 e^{x^2} x+16 x^3} \, dx+32 \int \frac {x^3}{-64+5 e^{x^2} x+16 x^3} \, dx-32 \int \frac {x^5}{-64+5 e^{x^2} x+16 x^3} \, dx+32 \int \left (-\frac {2}{-64+5 e^{x^2} x+16 x^3}-\frac {4 x^2}{-64+5 e^{x^2} x+16 x^3}-\frac {x^3}{-64+5 e^{x^2} x+16 x^3}+\frac {x^5}{-64+5 e^{x^2} x+16 x^3}\right ) \, dx+64 \int \frac {1}{-64+5 e^{x^2} x+16 x^3} \, dx+96 \int \frac {x^4}{-64+5 e^{x^2} x+16 x^3} \, dx-192 \int \frac {1}{x \left (-64+5 e^{x^2} x+16 x^3\right )} \, dx-384 \int \frac {x}{-64+5 e^{x^2} x+16 x^3} \, dx \\ & = 3 e^x-e^x x-3 x^2+x \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )+32 \int \frac {x^2}{-64+5 e^{x^2} x+16 x^3} \, dx+96 \int \frac {x^4}{-64+5 e^{x^2} x+16 x^3} \, dx-128 \int \frac {x^2}{-64+5 e^{x^2} x+16 x^3} \, dx-192 \int \frac {1}{x \left (-64+5 e^{x^2} x+16 x^3\right )} \, dx-384 \int \frac {x}{-64+5 e^{x^2} x+16 x^3} \, dx \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=e^x (3-x)+3 \log (x)+x \log \left (-\frac {5 e^{x^2}}{16}+\frac {4}{x}-x^2\right )-3 \log \left (64-5 e^{x^2} x-16 x^3\right ) \]
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Time = 7.62 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.58
method | result | size |
parallelrisch | \(-{\mathrm e}^{x} x +\ln \left (-\frac {16 x^{3}+5 \,{\mathrm e}^{x^{2}} x -64}{16 x}\right ) x +3 \ln \left (x \right )-3 \ln \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )+3 \,{\mathrm e}^{x}\) | \(52\) |
default | \(-3 \ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right )+\ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right ) x -{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}\) | \(55\) |
parts | \(-3 \ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right )+\ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right ) x -{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}\) | \(55\) |
risch | \(x \ln \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )-x \ln \left (x \right )+\frac {i \pi x \,\operatorname {csgn}\left (i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{3}}{2}-i \pi x {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+i \pi x -{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}-3 \ln \left ({\mathrm e}^{x^{2}}+\frac {\frac {16 x^{3}}{5}-\frac {64}{5}}{x}\right )\) | \(228\) |
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=-{\left (x - 3\right )} e^{x} + {\left (x - 3\right )} \log \left (-\frac {16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64}{16 \, x}\right ) \]
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Time = 0.52 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=x \log {\left (\frac {- x^{3} - \frac {5 x e^{x^{2}}}{16} + 4}{x} \right )} + \left (3 - x\right ) e^{x} - 3 \log {\left (e^{x^{2}} + \frac {16 x^{3} - 64}{5 x} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).
Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=-{\left (x - 3\right )} e^{x} - 4 \, x \log \left (2\right ) + x \log \left (-16 \, x^{3} - 5 \, x e^{\left (x^{2}\right )} + 64\right ) - x \log \left (x\right ) - 3 \, \log \left (\frac {16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64}{5 \, x}\right ) \]
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Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=-x e^{x} + x \log \left (-\frac {16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64}{16 \, x}\right ) + 3 \, e^{x} - 3 \, \log \left (16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64\right ) + 3 \, \log \left (x\right ) \]
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Time = 9.35 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=x\,\ln \left (-\frac {\frac {5\,x\,{\mathrm {e}}^{x^2}}{16}+x^3-4}{x}\right )-{\mathrm {e}}^x\,\left (x-3\right )-3\,\ln \left (\frac {5\,x\,{\mathrm {e}}^{x^2}+16\,x^3-64}{x}\right ) \]
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