Integrand size = 58, antiderivative size = 22 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-4+e^5-e^x-\frac {4}{e^{10} x^2}+3 x\right ) \]
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\[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {-8-4 x-e^{10} \left (7-e^5\right ) x^3+3 e^{10} x^4}{x \left (4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3\right )}\right ) \, dx \\ & = x+\int \frac {-8-4 x-e^{10} \left (7-e^5\right ) x^3+3 e^{10} x^4}{x \left (4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3\right )} \, dx \\ & = x+\int \left (\frac {e^{10} \left (-7+e^5\right ) x^2}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3}+\frac {3 e^{10} x^3}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3}+\frac {4}{-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3}+\frac {8}{x \left (-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3\right )}\right ) \, dx \\ & = x+4 \int \frac {1}{-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3} \, dx+8 \int \frac {1}{x \left (-4-e^{10+x} x^2-4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2+3 e^{10} x^3\right )} \, dx+\left (3 e^{10}\right ) \int \frac {x^3}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3} \, dx-\left (e^{10} \left (7-e^5\right )\right ) \int \frac {x^2}{4+e^{10+x} x^2+4 e^{10} \left (1-\frac {e^5}{4}\right ) x^2-3 e^{10} x^3} \, dx \\ \end{align*}
Time = 5.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=-2 \log (x)+\log \left (4-e^{15} x^2+e^{10+x} x^2+e^{10} (4-3 x) x^2\right ) \]
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Time = 0.92 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55
method | result | size |
risch | \(\ln \left (\frac {-3 x^{3}-x^{2} {\mathrm e}^{5}+4 x^{2}+{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{-10}}{x^{2}}\right )\) | \(34\) |
norman | \(-2 \ln \left (x \right )+\ln \left (x^{2} {\mathrm e}^{15}-{\mathrm e}^{x} {\mathrm e}^{10} x^{2}+3 x^{3} {\mathrm e}^{10}-4 x^{2} {\mathrm e}^{10}-4\right )\) | \(46\) |
parallelrisch | \(-2 \ln \left (x \right )+\ln \left (\frac {\left (x^{2} {\mathrm e}^{15}-{\mathrm e}^{x} {\mathrm e}^{10} x^{2}+3 x^{3} {\mathrm e}^{10}-4 x^{2} {\mathrm e}^{10}-4\right ) {\mathrm e}^{-10}}{3}\right )\) | \(52\) |
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Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-\frac {x^{2} e^{15} - x^{2} e^{\left (x + 10\right )} + {\left (3 \, x^{3} - 4 \, x^{2}\right )} e^{10} - 4}{x^{2}}\right ) \]
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Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log {\left (e^{x} + \frac {- 3 x^{3} e^{10} - x^{2} e^{15} + 4 x^{2} e^{10} + 4}{x^{2} e^{10}} \right )} \]
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Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-\frac {{\left (3 \, x^{3} e^{10} + x^{2} {\left (e^{15} - 4 \, e^{10}\right )} - x^{2} e^{\left (x + 10\right )} - 4\right )} e^{\left (-10\right )}}{x^{2}}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\log \left (-3 \, x^{3} e^{10} - x^{2} e^{15} + 4 \, x^{2} e^{10} + x^{2} e^{\left (x + 10\right )} + 4\right ) - 2 \, \log \left (x\right ) \]
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Time = 15.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {-8-3 e^{10} x^3+e^{10+x} x^3}{4 x-e^{15} x^3+e^{10+x} x^3+e^{10} \left (4 x^3-3 x^4\right )} \, dx=\ln \left (\frac {x^2\,{\mathrm {e}}^5}{3}-\frac {x^2\,{\mathrm {e}}^x}{3}-\frac {4\,{\mathrm {e}}^{-10}}{3}-\frac {4\,x^2}{3}+x^3\right )-2\,\ln \left (x\right ) \]
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