Integrand size = 80, antiderivative size = 28 \[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=2 e^5 \left (3-x-20 \left (x+\frac {x}{\frac {2}{5}+\log (2 x)}\right )\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(28)=56\).
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.50, number of steps used = 23, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.112, Rules used = {6820, 12, 6874, 2343, 2346, 2209, 2357, 2367, 2336} \[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=\frac {8000 e^5 x^2}{5 \log (2 x)+2}+\frac {20000 e^5 x^2}{(5 \log (2 x)+2)^2}+18 e^5 (1-7 x)^2-\frac {400 e^5 (3-x) x}{5 \log (2 x)+2} \]
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Rule 12
Rule 2209
Rule 2336
Rule 2343
Rule 2346
Rule 2357
Rule 2367
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e^5 (6-142 x-15 (-1+7 x) \log (2 x)) \left (216-920 \log (2 x)-525 \log ^2(2 x)\right )}{(2+5 \log (2 x))^3} \, dx \\ & = \left (4 e^5\right ) \int \frac {(6-142 x-15 (-1+7 x) \log (2 x)) \left (216-920 \log (2 x)-525 \log ^2(2 x)\right )}{(2+5 \log (2 x))^3} \, dx \\ & = \left (4 e^5\right ) \int \left (63 (-1+7 x)-\frac {50000 x}{(2+5 \log (2 x))^3}-\frac {500 (-3+x)}{(2+5 \log (2 x))^2}+\frac {300 (-1+14 x)}{2+5 \log (2 x)}\right ) \, dx \\ & = 18 e^5 (1-7 x)^2+\left (1200 e^5\right ) \int \frac {-1+14 x}{2+5 \log (2 x)} \, dx-\left (2000 e^5\right ) \int \frac {-3+x}{(2+5 \log (2 x))^2} \, dx-\left (200000 e^5\right ) \int \frac {x}{(2+5 \log (2 x))^3} \, dx \\ & = 18 e^5 (1-7 x)^2+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}-\left (800 e^5\right ) \int \frac {-3+x}{2+5 \log (2 x)} \, dx-\left (1200 e^5\right ) \int \frac {1}{2+5 \log (2 x)} \, dx+\left (1200 e^5\right ) \int \left (-\frac {1}{2+5 \log (2 x)}+\frac {14 x}{2+5 \log (2 x)}\right ) \, dx-\left (40000 e^5\right ) \int \frac {x}{(2+5 \log (2 x))^2} \, dx \\ & = 18 e^5 (1-7 x)^2+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)}-\left (600 e^5\right ) \text {Subst}\left (\int \frac {e^x}{2+5 x} \, dx,x,\log (2 x)\right )-\left (800 e^5\right ) \int \left (-\frac {3}{2+5 \log (2 x)}+\frac {x}{2+5 \log (2 x)}\right ) \, dx-\left (1200 e^5\right ) \int \frac {1}{2+5 \log (2 x)} \, dx-\left (16000 e^5\right ) \int \frac {x}{2+5 \log (2 x)} \, dx+\left (16800 e^5\right ) \int \frac {x}{2+5 \log (2 x)} \, dx \\ & = 18 e^5 (1-7 x)^2-120 e^{23/5} \operatorname {ExpIntegralEi}\left (\frac {1}{5} (2+5 \log (2 x))\right )+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)}-\left (600 e^5\right ) \text {Subst}\left (\int \frac {e^x}{2+5 x} \, dx,x,\log (2 x)\right )-\left (800 e^5\right ) \int \frac {x}{2+5 \log (2 x)} \, dx+\left (2400 e^5\right ) \int \frac {1}{2+5 \log (2 x)} \, dx-\left (4000 e^5\right ) \text {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right )+\left (4200 e^5\right ) \text {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right ) \\ & = 18 e^5 (1-7 x)^2-240 e^{23/5} \operatorname {ExpIntegralEi}\left (\frac {1}{5} (2+5 \log (2 x))\right )+40 e^{21/5} \operatorname {ExpIntegralEi}\left (\frac {2}{5} (2+5 \log (2 x))\right )+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)}-\left (200 e^5\right ) \text {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right )+\left (1200 e^5\right ) \text {Subst}\left (\int \frac {e^x}{2+5 x} \, dx,x,\log (2 x)\right ) \\ & = 18 e^5 (1-7 x)^2+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)} \\ \end{align*}
Time = 0.66 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=4 e^5 \left (-63 x+\frac {441 x^2}{2}+\frac {5000 x^2}{(2+5 \log (2 x))^2}+\frac {150 x (-2+14 x)}{2+5 \log (2 x)}\right ) \]
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Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61
method | result | size |
risch | \(126 \,{\mathrm e}^{5} x \left (7 x -2\right )+\frac {400 x \,{\mathrm e}^{5} \left (105 x \ln \left (2 x \right )+92 x -15 \ln \left (2 x \right )-6\right )}{\left (5 \ln \left (2 x \right )+2\right )^{2}}\) | \(45\) |
derivativedivides | \({\mathrm e}^{5} \left (-\frac {240 x}{\frac {2}{5}+\ln \left (2 x \right )}+\frac {800 x^{2}}{\left (\frac {2}{5}+\ln \left (2 x \right )\right )^{2}}+\frac {1680 x^{2}}{\frac {2}{5}+\ln \left (2 x \right )}-252 x +882 x^{2}\right )\) | \(50\) |
default | \({\mathrm e}^{5} \left (-\frac {240 x}{\frac {2}{5}+\ln \left (2 x \right )}+\frac {800 x^{2}}{\left (\frac {2}{5}+\ln \left (2 x \right )\right )^{2}}+\frac {1680 x^{2}}{\frac {2}{5}+\ln \left (2 x \right )}-252 x +882 x^{2}\right )\) | \(50\) |
norman | \(\frac {-3408 x \,{\mathrm e}^{5}+40328 x^{2} {\mathrm e}^{5}-11040 x \,{\mathrm e}^{5} \ln \left (2 x \right )-6300 x \,{\mathrm e}^{5} \ln \left (2 x \right )^{2}+59640 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )+22050 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )^{2}}{\left (5 \ln \left (2 x \right )+2\right )^{2}}\) | \(69\) |
parallelrisch | \(\frac {88200 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )^{2}+238560 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )-25200 x \,{\mathrm e}^{5} \ln \left (2 x \right )^{2}+161312 x^{2} {\mathrm e}^{5}-44160 x \,{\mathrm e}^{5} \ln \left (2 x \right )-13632 x \,{\mathrm e}^{5}}{100 \ln \left (2 x \right )^{2}+80 \ln \left (2 x \right )+16}\) | \(78\) |
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (24) = 48\).
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.50 \[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=\frac {2 \, {\left (1575 \, {\left (7 \, x^{2} - 2 \, x\right )} e^{5} \log \left (2 \, x\right )^{2} + 60 \, {\left (497 \, x^{2} - 92 \, x\right )} e^{5} \log \left (2 \, x\right ) + 284 \, {\left (71 \, x^{2} - 6 \, x\right )} e^{5}\right )}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.43 \[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=882 x^{2} e^{5} - 252 x e^{5} + \frac {1472 x^{2} e^{5} - 96 x e^{5} + \left (1680 x^{2} e^{5} - 240 x e^{5}\right ) \log {\left (2 x \right )}}{\log {\left (2 x \right )}^{2} + \frac {4 \log {\left (2 x \right )}}{5} + \frac {4}{25}} \]
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\[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=\int { \frac {4 \, {\left (7875 \, {\left (7 \, x - 1\right )} e^{5} \log \left (2 \, x\right )^{3} + 150 \, {\left (1141 \, x - 113\right )} e^{5} \log \left (2 \, x\right )^{2} + 40 \, {\left (2699 \, x - 57\right )} e^{5} \log \left (2 \, x\right ) - 432 \, {\left (71 \, x - 3\right )} e^{5}\right )}}{125 \, \log \left (2 \, x\right )^{3} + 150 \, \log \left (2 \, x\right )^{2} + 60 \, \log \left (2 \, x\right ) + 8} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (24) = 48\).
Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 5.89 \[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=\frac {22050 \, x^{2} e^{5} \log \left (2 \, x\right )^{2}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} + \frac {59640 \, x^{2} e^{5} \log \left (2 \, x\right )}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} - \frac {6300 \, x e^{5} \log \left (2 \, x\right )^{2}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} + \frac {40328 \, x^{2} e^{5}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} - \frac {11040 \, x e^{5} \log \left (2 \, x\right )}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} - \frac {3408 \, x e^{5}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} \]
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Time = 7.44 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx=\frac {2\,x\,{\mathrm {e}}^5\,\left (105\,\ln \left (2\,x\right )+142\right )\,\left (142\,x-30\,\ln \left (2\,x\right )+105\,x\,\ln \left (2\,x\right )-12\right )}{{\left (5\,\ln \left (2\,x\right )+2\right )}^2} \]
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