Optimal. Leaf size=28 \[ 2 e^5 \left (3-x-20 \left (x+\frac {x}{\frac {2}{5}+\log (2 x)}\right )\right )^2 \]
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Rubi [B] time = 0.45, antiderivative size = 70, normalized size of antiderivative = 2.50, number of steps used = 23, number of rules used = 9, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.112, Rules used = {6688, 12, 6742, 2306, 2309, 2178, 2320, 2330, 2299} \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2178
Rule 2299
Rule 2306
Rule 2309
Rule 2320
Rule 2330
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \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 ) \operatorname {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} \text {Ei}\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 ) \operatorname {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 ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right )+\left (4200 e^5\right ) \operatorname {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} \text {Ei}\left (\frac {1}{5} (2+5 \log (2 x))\right )+40 e^{21/5} \text {Ei}\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 ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right )+\left (1200 e^5\right ) \operatorname {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 {aligned} \end {gather*}
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Mathematica [A] time = 0.30, size = 49, normalized size = 1.75 \begin {gather*} 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 ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.56, size = 70, normalized size = 2.50 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.37, size = 165, normalized size = 5.89 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 45, normalized size = 1.61
method | result | size |
risch | \(126 \,{\mathrm e}^{5} x \left (7 x -2\right )+\frac {400 x \,{\mathrm e}^{5} \left (105 x \ln \left (2 x \right )+92 x -15 \ln \left (2 x \right )-6\right )}{\left (5 \ln \left (2 x \right )+2\right )^{2}}\) | \(45\) |
norman | \(\frac {-3408 x \,{\mathrm e}^{5}+40328 x^{2} {\mathrm e}^{5}-11040 x \,{\mathrm e}^{5} \ln \left (2 x \right )-6300 x \,{\mathrm e}^{5} \ln \left (2 x \right )^{2}+59640 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )+22050 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )^{2}}{\left (5 \ln \left (2 x \right )+2\right )^{2}}\) | \(69\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2 \, {\left ({\left (275625 \, \log \relax (2)^{2} + 438780 \, \log \relax (2) + 228052\right )} x^{2} e^{5} - 6 \, {\left (13125 \, \log \relax (2)^{2} + 21920 \, \log \relax (2) + 5588\right )} x e^{5} + 39375 \, {\left (7 \, x^{2} e^{5} - 2 \, x e^{5}\right )} \log \relax (x)^{2} + 30 \, {\left (x^{2} {\left (18375 \, \log \relax (2) + 14626\right )} e^{5} - 2 \, x {\left (2625 \, \log \relax (2) + 2192\right )} e^{5}\right )} \log \relax (x)\right )}}{25 \, {\left (25 \, \log \relax (2)^{2} + 10 \, {\left (5 \, \log \relax (2) + 2\right )} \log \relax (x) + 25 \, \log \relax (x)^{2} + 20 \, \log \relax (2) + 4\right )}} - \frac {2592 \, e^{\frac {23}{5}} E_{3}\left (-\log \left (2 \, x\right ) - \frac {2}{5}\right )}{5 \, {\left (5 \, \log \left (2 \, x\right ) + 2\right )}^{2}} + \frac {30672 \, e^{\frac {21}{5}} E_{3}\left (-2 \, \log \left (2 \, x\right ) - \frac {4}{5}\right )}{5 \, {\left (5 \, \log \left (2 \, x\right ) + 2\right )}^{2}} + 4 \, \int \frac {216 \, {\left (284 \, x e^{5} - 3 \, e^{5}\right )}}{25 \, {\left (5 \, \log \relax (2) + 5 \, \log \relax (x) + 2\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.64, size = 41, normalized size = 1.46 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.20, size = 68, normalized size = 2.43 \begin {gather*} 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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