Integrand size = 56, antiderivative size = 24 \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=\frac {8 x^2 \left (e^{x/5}+\log (x)\right )}{\frac {4}{5}-x} \]
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Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(24)=48\).
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.38, number of steps used = 20, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {27, 6874, 2230, 2225, 2207, 2208, 2209, 45, 2404, 2332, 2351, 31} \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=-8 e^{x/5} x-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}+\frac {32 x \log (x)}{4-5 x}-8 x \log (x) \]
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Rule 27
Rule 31
Rule 45
Rule 2207
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rule 2332
Rule 2351
Rule 2404
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{(-4+5 x)^2} \, dx \\ & = \int \left (-\frac {8 e^{x/5} x \left (-40+21 x+5 x^2\right )}{(-4+5 x)^2}-\frac {40 x (-4+5 x-8 \log (x)+5 x \log (x))}{(-4+5 x)^2}\right ) \, dx \\ & = -\left (8 \int \frac {e^{x/5} x \left (-40+21 x+5 x^2\right )}{(-4+5 x)^2} \, dx\right )-40 \int \frac {x (-4+5 x-8 \log (x)+5 x \log (x))}{(-4+5 x)^2} \, dx \\ & = -\left (8 \int \left (\frac {29 e^{x/5}}{25}+\frac {1}{5} e^{x/5} x-\frac {16 e^{x/5}}{(-4+5 x)^2}+\frac {16 e^{x/5}}{25 (-4+5 x)}\right ) \, dx\right )-40 \int \left (\frac {x}{-4+5 x}+\frac {x (-8+5 x) \log (x)}{(-4+5 x)^2}\right ) \, dx \\ & = -\left (\frac {8}{5} \int e^{x/5} x \, dx\right )-\frac {128}{25} \int \frac {e^{x/5}}{-4+5 x} \, dx-\frac {232}{25} \int e^{x/5} \, dx-40 \int \frac {x}{-4+5 x} \, dx-40 \int \frac {x (-8+5 x) \log (x)}{(-4+5 x)^2} \, dx+128 \int \frac {e^{x/5}}{(-4+5 x)^2} \, dx \\ & = -\frac {232 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-\frac {128}{125} e^{4/25} \operatorname {ExpIntegralEi}\left (\frac {1}{25} (-4+5 x)\right )+\frac {128}{25} \int \frac {e^{x/5}}{-4+5 x} \, dx+8 \int e^{x/5} \, dx-40 \int \left (\frac {1}{5}+\frac {4}{5 (-4+5 x)}\right ) \, dx-40 \int \left (\frac {\log (x)}{5}-\frac {16 \log (x)}{5 (-4+5 x)^2}\right ) \, dx \\ & = -\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 x-8 e^{x/5} x-\frac {32}{5} \log (4-5 x)-8 \int \log (x) \, dx+128 \int \frac {\log (x)}{(-4+5 x)^2} \, dx \\ & = -\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-\frac {32}{5} \log (4-5 x)-8 x \log (x)+\frac {32 x \log (x)}{4-5 x}+32 \int \frac {1}{-4+5 x} \, dx \\ & = -\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-8 x \log (x)+\frac {32 x \log (x)}{4-5 x} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=-\frac {40 x^2 \left (e^{x/5}+\log (x)\right )}{-4+5 x} \]
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Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
norman | \(\frac {-40 x^{2} \ln \left (x \right )-40 \,{\mathrm e}^{\frac {x}{5}} x^{2}}{5 x -4}\) | \(26\) |
parallelrisch | \(\frac {-800 x^{2} \ln \left (x \right )-800 \,{\mathrm e}^{\frac {x}{5}} x^{2}}{100 x -80}\) | \(27\) |
default | \(-8 x \ln \left (x \right )-\frac {32 \ln \left (x \right ) x}{5 x -4}-\frac {128 \,{\mathrm e}^{\frac {x}{5}}}{125 \left (\frac {x}{5}-\frac {4}{25}\right )}-\frac {32 \,{\mathrm e}^{\frac {x}{5}}}{5}-8 x \,{\mathrm e}^{\frac {x}{5}}\) | \(45\) |
parts | \(-8 x \ln \left (x \right )-\frac {32 \ln \left (x \right ) x}{5 x -4}-\frac {128 \,{\mathrm e}^{\frac {x}{5}}}{125 \left (\frac {x}{5}-\frac {4}{25}\right )}-\frac {32 \,{\mathrm e}^{\frac {x}{5}}}{5}-8 x \,{\mathrm e}^{\frac {x}{5}}\) | \(45\) |
risch | \(-\frac {8 \left (25 x^{2}-20 x +16\right ) \ln \left (x \right )}{5 \left (5 x -4\right )}-\frac {8 \left (25 \,{\mathrm e}^{\frac {x}{5}} x^{2}+20 x \ln \left (x \right )-16 \ln \left (x \right )\right )}{5 \left (5 x -4\right )}\) | \(51\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=-\frac {40 \, {\left (x^{2} e^{\left (\frac {1}{5} \, x\right )} + x^{2} \log \left (x\right )\right )}}{5 \, x - 4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (17) = 34\).
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=- \frac {40 x^{2} e^{\frac {x}{5}}}{5 x - 4} - \frac {32 \log {\left (x \right )}}{5} + \frac {\left (- 200 x^{2} + 160 x - 128\right ) \log {\left (x \right )}}{25 x - 20} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (19) = 38\).
Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=-8 \, x - \frac {8 \, {\left (25 \, x^{2} e^{\left (\frac {1}{5} \, x\right )} - 25 \, x^{2} + {\left (25 \, x^{2} - 20 \, x + 16\right )} \log \left (x\right ) + 20 \, x\right )}}{5 \, {\left (5 \, x - 4\right )}} - \frac {32}{5} \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=-\frac {8 \, {\left (25 \, x^{2} e^{\left (\frac {1}{5} \, x\right )} + 25 \, x^{2} \log \left (x\right ) + 20 \, x \log \left (\frac {1}{5} \, x\right ) - 20 \, x \log \left (x\right ) - 16 \, \log \left (\frac {1}{5} \, x\right ) + 16 \, \log \left (x\right )\right )}}{5 \, {\left (5 \, x - 4\right )}} \]
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Time = 13.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{16-40 x+25 x^2} \, dx=-\frac {40\,x^2\,\left ({\mathrm {e}}^{x/5}+\ln \left (x\right )\right )}{5\,x-4} \]
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