Integrand size = 59, antiderivative size = 16 \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx=-x+5 x (e+x)^{\frac {1}{e^2}} \log (4) \]
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Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 1607, 6820, 34} \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx=5 x (x+e)^{\frac {1}{e^2}} \log (4)-x \]
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Rule 12
Rule 34
Rule 1607
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e x+x^2} \, dx}{e^2} \\ & = \frac {\int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{x (e+x)} \, dx}{e^2} \\ & = \frac {\int \left (-e^2+5 (e+x)^{-1+\frac {1}{e^2}} \left (e^3+\left (1+e^2\right ) x\right ) \log (4)\right ) \, dx}{e^2} \\ & = -x+\frac {(5 \log (4)) \int (e+x)^{-1+\frac {1}{e^2}} \left (e^3+\left (1+e^2\right ) x\right ) \, dx}{e^2} \\ & = -x+5 x (e+x)^{\frac {1}{e^2}} \log (4) \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx=x \left (-1+5 (e+x)^{\frac {1}{e^2}} \log (4)\right ) \]
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Time = 0.58 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06
method | result | size |
risch | \(-x +10 x \ln \left (2\right ) \left (x +{\mathrm e}\right )^{{\mathrm e}^{-2}}\) | \(17\) |
norman | \(-x +{\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \left (2\right )\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}\) | \(26\) |
parts | \(-x +{\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \left (2\right )\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}\) | \(26\) |
default | \({\mathrm e}^{-2} \left ({\mathrm e}^{2} {\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \left (2\right )\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}-{\mathrm e}^{2} x \right )\) | \(36\) |
parallelrisch | \({\mathrm e}^{-2} \left (2 \,{\mathrm e} \,{\mathrm e}^{2}-{\mathrm e}^{2} x +{\mathrm e}^{2} {\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \left (2\right )\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}\right )\) | \(42\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx=-x + e^{\left ({\left (e^{2} \log \left (10 \, x \log \left (2\right )\right ) + \log \left (x + e\right )\right )} e^{\left (-2\right )}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (17) = 34\).
Time = 10.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 7.75 \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx=\frac {\frac {10 x \left (x + e\right )^{e^{-2}} e^{2} \log {\left (2 \right )}}{1 + e^{2}} + \frac {10 x \left (x + e\right )^{e^{-2}} e^{4} \log {\left (2 \right )}}{1 + e^{2}} - x e^{2} - \frac {10 \left (x + e\right )^{e^{-2}} e^{7} \log {\left (2 \right )}}{1 + e^{2}} - \frac {10 \left (x + e\right )^{e^{-2}} e^{5} \log {\left (2 \right )}}{1 + e^{2}} + 10 \left (x + e\right )^{e^{-2}} e^{5} \log {\left (2 \right )}}{e^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (20) = 40\).
Time = 0.22 (sec) , antiderivative size = 173, normalized size of antiderivative = 10.81 \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx={\left ({\left (e \log \left (x + e\right ) - x\right )} e^{2} + 10 \, {\left (\frac {{\left (x + e\right )} e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 4\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 7\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )}}{e^{2} + 1}\right )} \log \left (2\right ) + 10 \, {\left (\frac {{\left (x + e\right )} e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 2\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 3\right )}}{e^{2} + 1}\right )} \log \left (2\right ) + 10 \, e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )} \log \left (2\right ) - e^{3} \log \left (x + e\right )\right )} e^{\left (-2\right )} \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx={\left (10 \, {\left (x + e\right )}^{e^{\left (-2\right )}} x e^{2} \log \left (2\right ) - x e^{2}\right )} e^{\left (-2\right )} \]
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Time = 15.54 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e^2 \left (e x+x^2\right )} \, dx=x\,\left (10\,\ln \left (2\right )\,{\left (x+\mathrm {e}\right )}^{{\mathrm {e}}^{-2}}-1\right ) \]
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