Integrand size = 150, antiderivative size = 25 \[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\left (e^{\frac {x^2 (x+\log (4))}{5+e^4+\log (x)}}-x\right ) x \]
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Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(25)=50\).
Time = 1.66 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.96, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6, 6873, 6874, 2326} \[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=-x^2-\frac {4^{\frac {x^2}{\log (x)+e^4+5}} e^{\frac {x^3}{\log (x)+e^4+5}} \left (x^2 \log (16) \log (x)+\left (9+2 e^4\right ) x^2 \log (4)\right )}{\log (4) \left (\log (x)+e^4+5\right )^2 \left (\frac {x}{\left (\log (x)+e^4+5\right )^2}-\frac {2 x}{\log (x)+e^4+5}\right )} \]
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Rule 6
Rule 2326
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 e^8 x+\left (-50-20 e^4\right ) x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx \\ & = \int \frac {\left (-50-20 e^4-2 e^8\right ) x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx \\ & = \int \frac {\left (-50-20 e^4-2 e^8\right ) x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{\left (5 \left (1+\frac {e^4}{5}\right )+\log (x)\right )^2} \, dx \\ & = \int \left (-2 x+\frac {4^{\frac {x^2}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} e^{\frac {x^3}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} \left (25 \left (1+\frac {1}{25} e^4 \left (10+e^4\right )\right )+14 \left (1+\frac {3 e^4}{14}\right ) x^3+9 x^2 \log (4) \left (1+\frac {e^4 \log (16)}{9 \log (4)}\right )+10 \left (1+\frac {e^4}{5}\right ) \log (x)+3 x^3 \log (x)+x^2 \log (16) \log (x)+\log ^2(x)\right )}{\left (5 \left (1+\frac {e^4}{5}\right )+\log (x)\right )^2}\right ) \, dx \\ & = -x^2+\int \frac {4^{\frac {x^2}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} e^{\frac {x^3}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} \left (25 \left (1+\frac {1}{25} e^4 \left (10+e^4\right )\right )+14 \left (1+\frac {3 e^4}{14}\right ) x^3+9 x^2 \log (4) \left (1+\frac {e^4 \log (16)}{9 \log (4)}\right )+10 \left (1+\frac {e^4}{5}\right ) \log (x)+3 x^3 \log (x)+x^2 \log (16) \log (x)+\log ^2(x)\right )}{\left (5 \left (1+\frac {e^4}{5}\right )+\log (x)\right )^2} \, dx \\ & = -x^2-\frac {4^{\frac {x^2}{5+e^4+\log (x)}} e^{\frac {x^3}{5+e^4+\log (x)}} \left (x^2 \left (9 \log (4)+e^4 \log (16)\right )+x^2 \log (16) \log (x)\right )}{\log (4) \left (5+e^4+\log (x)\right )^2 \left (\frac {x}{\left (5+e^4+\log (x)\right )^2}-\frac {2 x}{5+e^4+\log (x)}\right )} \\ \end{align*}
\[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx \]
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Time = 4.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
risch | \(-x^{2}+{\mathrm e}^{\frac {\left (x +2 \ln \left (2\right )\right ) x^{2}}{\ln \left (x \right )+5+{\mathrm e}^{4}}} x\) | \(28\) |
parallelrisch | \(-x^{2}+{\mathrm e}^{\frac {\left (x +2 \ln \left (2\right )\right ) x^{2}}{\ln \left (x \right )+5+{\mathrm e}^{4}}} x\) | \(28\) |
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Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=-x^{2} + x e^{\left (\frac {x^{3} + 2 \, x^{2} \log \left (2\right )}{e^{4} + \log \left (x\right ) + 5}\right )} \]
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Exception generated. \[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
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Exception generated. \[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.95 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=-x^{2} + x e^{\left (\frac {x^{3} + 2 \, x^{2} \log \left (2\right )}{e^{4} + \log \left (x\right ) + 5}\right )} \]
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Timed out. \[ \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=-\int \frac {50\,x+2\,x\,{\ln \left (x\right )}^2+20\,x\,{\mathrm {e}}^4+2\,x\,{\mathrm {e}}^8+\ln \left (x\right )\,\left (20\,x+4\,x\,{\mathrm {e}}^4\right )-{\mathrm {e}}^{\frac {x^3+2\,\ln \left (2\right )\,x^2}{{\mathrm {e}}^4+\ln \left (x\right )+5}}\,\left ({\mathrm {e}}^8+\ln \left (x\right )\,\left (3\,x^3+4\,\ln \left (2\right )\,x^2+2\,{\mathrm {e}}^4+10\right )+{\ln \left (x\right )}^2+{\mathrm {e}}^4\,\left (3\,x^3+10\right )+14\,x^3+2\,\ln \left (2\right )\,\left (2\,x^2\,{\mathrm {e}}^4+9\,x^2\right )+25\right )}{{\ln \left (x\right )}^2+\left (2\,{\mathrm {e}}^4+10\right )\,\ln \left (x\right )+10\,{\mathrm {e}}^4+{\mathrm {e}}^8+25} \,d x \]
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