Integrand size = 59, antiderivative size = 22 \[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=6-\frac {3}{x+\frac {3}{-\frac {5}{8}+49 e^x x}} \]
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\[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=\int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {75+460992 e^{2 x} x^2-2352 e^x (12+17 x)}{\left (24-5 x+392 e^x x^2\right )^2} \, dx \\ & = \int \left (\frac {3}{x^2}-\frac {72 \left (-48-19 x+5 x^2\right )}{x^2 \left (24-5 x+392 e^x x^2\right )^2}-\frac {72 (3+x)}{x^2 \left (24-5 x+392 e^x x^2\right )}\right ) \, dx \\ & = -\frac {3}{x}-72 \int \frac {-48-19 x+5 x^2}{x^2 \left (24-5 x+392 e^x x^2\right )^2} \, dx-72 \int \frac {3+x}{x^2 \left (24-5 x+392 e^x x^2\right )} \, dx \\ & = -\frac {3}{x}-72 \int \left (\frac {5}{\left (24-5 x+392 e^x x^2\right )^2}-\frac {48}{x^2 \left (24-5 x+392 e^x x^2\right )^2}-\frac {19}{x \left (24-5 x+392 e^x x^2\right )^2}\right ) \, dx-72 \int \left (\frac {3}{x^2 \left (24-5 x+392 e^x x^2\right )}+\frac {1}{x \left (24-5 x+392 e^x x^2\right )}\right ) \, dx \\ & = -\frac {3}{x}-72 \int \frac {1}{x \left (24-5 x+392 e^x x^2\right )} \, dx-216 \int \frac {1}{x^2 \left (24-5 x+392 e^x x^2\right )} \, dx-360 \int \frac {1}{\left (24-5 x+392 e^x x^2\right )^2} \, dx+1368 \int \frac {1}{x \left (24-5 x+392 e^x x^2\right )^2} \, dx+3456 \int \frac {1}{x^2 \left (24-5 x+392 e^x x^2\right )^2} \, dx \\ \end{align*}
Time = 2.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=-\frac {3}{x}+\frac {72}{x \left (24-5 x+392 e^x x^2\right )} \]
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Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {15-1176 \,{\mathrm e}^{x} x}{392 \,{\mathrm e}^{x} x^{2}-5 x +24}\) | \(23\) |
parallelrisch | \(\frac {5880-460992 \,{\mathrm e}^{x} x}{153664 \,{\mathrm e}^{x} x^{2}-1960 x +9408}\) | \(24\) |
risch | \(-\frac {3}{x}+\frac {72}{x \left (392 \,{\mathrm e}^{x} x^{2}-5 x +24\right )}\) | \(26\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=-\frac {3 \, {\left (392 \, x e^{x} - 5\right )}}{392 \, x^{2} e^{x} - 5 \, x + 24} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=\frac {72}{392 x^{3} e^{x} - 5 x^{2} + 24 x} - \frac {3}{x} \]
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=-\frac {3 \, {\left (392 \, x e^{x} - 5\right )}}{392 \, x^{2} e^{x} - 5 \, x + 24} \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=-\frac {3 \, {\left (392 \, x e^{x} - 5\right )}}{392 \, x^{2} e^{x} - 5 \, x + 24} \]
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Time = 14.51 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {75+e^x (-28224-39984 x)+460992 e^{2 x} x^2}{576-240 x+25 x^2+153664 e^{2 x} x^4+e^x \left (18816 x^2-3920 x^3\right )} \, dx=-\frac {1176\,x\,{\mathrm {e}}^x-15}{392\,x^2\,{\mathrm {e}}^x-5\,x+24} \]
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