Integrand size = 138, antiderivative size = 24 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=e^{7-e^{\frac {x}{x-\log (10+5 x)}}} x \]
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Leaf count is larger than twice the leaf count of optimal. \(121\) vs. \(2(24)=48\).
Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 5.04, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.007, Rules used = {2326} \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=-\frac {e^{7-e^{\frac {x}{x-\log (5 x+10)}}} \left (x^2-\left (x^2+2 x\right ) \log (5 x+10)\right )}{\left (\frac {x \left (1-\frac {1}{x+2}\right )}{(x-\log (5 x+10))^2}-\frac {1}{x-\log (5 x+10)}\right ) \left (x^3+2 x^2-2 \left (x^2+2 x\right ) \log (5 x+10)+(x+2) \log ^2(5 x+10)\right )} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = -\frac {e^{7-e^{\frac {x}{x-\log (10+5 x)}}} \left (x^2-\left (2 x+x^2\right ) \log (10+5 x)\right )}{\left (\frac {x \left (1-\frac {1}{2+x}\right )}{(x-\log (10+5 x))^2}-\frac {1}{x-\log (10+5 x)}\right ) \left (2 x^2+x^3-2 \left (2 x+x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)\right )} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=e^{7-e^{-\frac {x}{-x+\log (5 (2+x))}}} x \]
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Time = 2.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x \,{\mathrm e}^{-{\mathrm e}^{\frac {x}{x -\ln \left (5 x +10\right )}}+7}\) | \(23\) |
parallelrisch | \(x \,{\mathrm e}^{-{\mathrm e}^{-\frac {x}{\ln \left (5 x +10\right )-x}}+7}\) | \(24\) |
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none
Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=x e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )} \]
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Exception generated. \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=\text {Exception raised: TypeError} \]
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\[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=\int { \frac {{\left (x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - {\left (x^{2} - {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )}}{x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )} \,d x } \]
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\[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=\int { \frac {{\left (x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - {\left (x^{2} - {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )}}{x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )} \,d x } \]
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Time = 12.76 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {x}{x-\ln \left (5\,x+10\right )}}}\,{\mathrm {e}}^7 \]
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