Optimal. Leaf size=28 \[ \frac {1}{x+\left (5-2 \left (4-3 e^{e^{e^x}}+\frac {4 x}{5}+\log (3)\right )\right )^2} \]
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Rubi [F] time = 9.41, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-6625-45000 e^{2 e^{e^x}+e^x+x}-3200 x-4000 \log (3)+e^{e^{e^x}} \left (12000+e^{e^x+x} (22500+12000 x+15000 \log (3))\right )}{50625+810000 e^{4 e^{e^x}}+119250 x+99025 x^2+33920 x^3+4096 x^4+e^{3 e^{e^x}} (-1620000-864000 x-1080000 \log (3))+\left (135000+231000 x+123200 x^2+20480 x^3\right ) \log (3)+\left (135000+149000 x+38400 x^2\right ) \log ^2(3)+(60000+32000 x) \log ^3(3)+10000 \log ^4(3)+e^{2 e^{e^x}} \left (1215000+1341000 x+345600 x^2+(1620000+864000 x) \log (3)+540000 \log ^2(3)\right )+e^{e^{e^x}} \left (-405000-693000 x-369600 x^2-61440 x^3+\left (-810000-894000 x-230400 x^2\right ) \log (3)+(-540000-288000 x) \log ^2(3)-120000 \log ^3(3)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 \left (480 e^{e^{e^x}}-1800 e^{2 e^{e^x}+e^x+x}-128 x-265 \left (1+\frac {32 \log (3)}{53}\right )+60 e^{e^{e^x}+e^x+x} (8 x+5 (3+\log (9)))\right )}{\left (900 e^{2 e^{e^x}}+64 x^2+5 x (53+32 \log (3))+25 (3+\log (9))^2-60 e^{e^{e^x}} (8 x+5 (3+\log (9)))\right )^2} \, dx\\ &=25 \int \frac {480 e^{e^{e^x}}-1800 e^{2 e^{e^x}+e^x+x}-128 x-265 \left (1+\frac {32 \log (3)}{53}\right )+60 e^{e^{e^x}+e^x+x} (8 x+5 (3+\log (9)))}{\left (900 e^{2 e^{e^x}}+64 x^2+5 x (53+32 \log (3))+25 (3+\log (9))^2-60 e^{e^{e^x}} (8 x+5 (3+\log (9)))\right )^2} \, dx\\ &=25 \int \left (\frac {480 e^{e^{e^x}}}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2}-\frac {128 x}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2}+\frac {60 e^{e^{e^x}+e^x+x} \left (-30 e^{e^{e^x}}+8 x+15 \left (1+\frac {2 \log (3)}{3}\right )\right )}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2}+\frac {5 (-53-32 \log (3))}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2}\right ) \, dx\\ &=1500 \int \frac {e^{e^{e^x}+e^x+x} \left (-30 e^{e^{e^x}}+8 x+15 \left (1+\frac {2 \log (3)}{3}\right )\right )}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx-3200 \int \frac {x}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx+12000 \int \frac {e^{e^{e^x}}}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx-(125 (53+32 \log (3))) \int \frac {1}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx\\ &=1500 \int \left (-\frac {30 e^{2 e^{e^x}+e^x+x}}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2}+\frac {8 e^{e^{e^x}+e^x+x} x}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2}+\frac {5 e^{e^{e^x}+e^x+x} (3+\log (9))}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2}\right ) \, dx-3200 \int \frac {x}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx+12000 \int \frac {e^{e^{e^x}}}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx-(125 (53+32 \log (3))) \int \frac {1}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx\\ &=-\left (3200 \int \frac {x}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx\right )+12000 \int \frac {e^{e^{e^x}}}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx+12000 \int \frac {e^{e^{e^x}+e^x+x} x}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx-45000 \int \frac {e^{2 e^{e^x}+e^x+x}}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx-(125 (53+32 \log (3))) \int \frac {1}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx+(7500 (3+\log (9))) \int \frac {e^{e^{e^x}+e^x+x}}{\left (900 e^{2 e^{e^x}}-480 e^{e^{e^x}} x+64 x^2+265 x \left (1+\frac {32 \log (3)}{53}\right )-900 e^{e^{e^x}} \left (1+\frac {2 \log (3)}{3}\right )+225 \left (1+\frac {2}{9} (3+\log (3)) \log (9)\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.33, size = 57, normalized size = 2.04 \begin {gather*} \frac {25}{900 e^{2 e^{e^x}}+64 x^2+5 x (53+32 \log (3))+25 (3+\log (9))^2-60 e^{e^{e^x}} (8 x+5 (3+\log (9)))} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.57, size = 52, normalized size = 1.86 \begin {gather*} \frac {25}{64 \, x^{2} - 60 \, {\left (8 \, x + 10 \, \log \relax (3) + 15\right )} e^{\left (e^{\left (e^{x}\right )}\right )} + 20 \, {\left (8 \, x + 15\right )} \log \relax (3) + 100 \, \log \relax (3)^{2} + 265 \, x + 900 \, e^{\left (2 \, e^{\left (e^{x}\right )}\right )} + 225} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {25 \, {\left (60 \, {\left ({\left (8 \, x + 10 \, \log \relax (3) + 15\right )} e^{\left (x + e^{x}\right )} + 8\right )} e^{\left (e^{\left (e^{x}\right )}\right )} - 128 \, x - 1800 \, e^{\left (x + e^{x} + 2 \, e^{\left (e^{x}\right )}\right )} - 160 \, \log \relax (3) - 265\right )}}{4096 \, x^{4} + 4000 \, {\left (8 \, x + 15\right )} \log \relax (3)^{3} + 10000 \, \log \relax (3)^{4} + 33920 \, x^{3} + 200 \, {\left (192 \, x^{2} + 745 \, x + 675\right )} \log \relax (3)^{2} + 99025 \, x^{2} - 108000 \, {\left (8 \, x + 10 \, \log \relax (3) + 15\right )} e^{\left (3 \, e^{\left (e^{x}\right )}\right )} + 1800 \, {\left (192 \, x^{2} + 60 \, {\left (8 \, x + 15\right )} \log \relax (3) + 300 \, \log \relax (3)^{2} + 745 \, x + 675\right )} e^{\left (2 \, e^{\left (e^{x}\right )}\right )} - 120 \, {\left (512 \, x^{3} + 300 \, {\left (8 \, x + 15\right )} \log \relax (3)^{2} + 1000 \, \log \relax (3)^{3} + 3080 \, x^{2} + 10 \, {\left (192 \, x^{2} + 745 \, x + 675\right )} \log \relax (3) + 5775 \, x + 3375\right )} e^{\left (e^{\left (e^{x}\right )}\right )} + 40 \, {\left (512 \, x^{3} + 3080 \, x^{2} + 5775 \, x + 3375\right )} \log \relax (3) + 119250 \, x + 810000 \, e^{\left (4 \, e^{\left (e^{x}\right )}\right )} + 50625}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 59, normalized size = 2.11
method | result | size |
risch | \(\frac {25}{100 \ln \relax (3)^{2}-600 \ln \relax (3) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}}+160 x \ln \relax (3)+900 \,{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{x}}}-480 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}} x +64 x^{2}+300 \ln \relax (3)-900 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}}+265 x +225}\) | \(59\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.82, size = 53, normalized size = 1.89 \begin {gather*} \frac {25}{64 \, x^{2} + 5 \, x {\left (32 \, \log \relax (3) + 53\right )} - 60 \, {\left (8 \, x + 10 \, \log \relax (3) + 15\right )} e^{\left (e^{\left (e^{x}\right )}\right )} + 100 \, \log \relax (3)^{2} + 900 \, e^{\left (2 \, e^{\left (e^{x}\right )}\right )} + 300 \, \log \relax (3) + 225} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {3200\,x+4000\,\ln \relax (3)-{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\left (12000\,x+15000\,\ln \relax (3)+22500\right )+12000\right )+45000\,{\mathrm {e}}^{2\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x+6625}{119250\,x+810000\,{\mathrm {e}}^{4\,{\mathrm {e}}^{{\mathrm {e}}^x}}+{\mathrm {e}}^{2\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,\left (1341000\,x+\ln \relax (3)\,\left (864000\,x+1620000\right )+540000\,{\ln \relax (3)}^2+345600\,x^2+1215000\right )+{\ln \relax (3)}^3\,\left (32000\,x+60000\right )-{\mathrm {e}}^{3\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,\left (864000\,x+1080000\,\ln \relax (3)+1620000\right )+\ln \relax (3)\,\left (20480\,x^3+123200\,x^2+231000\,x+135000\right )-{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}\,\left (693000\,x+\ln \relax (3)\,\left (230400\,x^2+894000\,x+810000\right )+{\ln \relax (3)}^2\,\left (288000\,x+540000\right )+120000\,{\ln \relax (3)}^3+369600\,x^2+61440\,x^3+405000\right )+{\ln \relax (3)}^2\,\left (38400\,x^2+149000\,x+135000\right )+10000\,{\ln \relax (3)}^4+99025\,x^2+33920\,x^3+4096\,x^4+50625} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.51, size = 58, normalized size = 2.07 \begin {gather*} \frac {25}{64 x^{2} + 160 x \log {\relax (3 )} + 265 x + \left (- 480 x - 900 - 600 \log {\relax (3 )}\right ) e^{e^{e^{x}}} + 900 e^{2 e^{e^{x}}} + 100 \log {\relax (3 )}^{2} + 225 + 300 \log {\relax (3 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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