Optimal. Leaf size=30 \[ 5+x-x \left (e^{x/5}-\frac {2}{x}+x\right ) \left (1+x (x+\log (4))^2\right ) \]
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Rubi [B] time = 0.30, antiderivative size = 99, normalized size of antiderivative = 3.30, number of steps used = 34, number of rules used = 4, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 2196, 2194, 2176} \begin {gather*} -x^5-e^{x/5} x^4-2 x^4 \log (4)+2 x^3-x^3 \log ^2(4)-2 e^{x/5} x^3 \log (4)-x^2-e^{x/5} x^2 \log ^2(4)+4 x^2 \log (4)-e^{x/5} x+x+2 x \log ^2(4) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2194
Rule 2196
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (5-10 x+30 x^2-25 x^4+\left (40 x-40 x^3\right ) \log (4)+\left (10-15 x^2\right ) \log ^2(4)+e^{x/5} \left (-5-x-20 x^3-x^4+\left (-30 x^2-2 x^3\right ) \log (4)+\left (-10 x-x^2\right ) \log ^2(4)\right )\right ) \, dx\\ &=x-x^2+2 x^3-x^5+\frac {1}{5} \int e^{x/5} \left (-5-x-20 x^3-x^4+\left (-30 x^2-2 x^3\right ) \log (4)+\left (-10 x-x^2\right ) \log ^2(4)\right ) \, dx+\frac {1}{5} \log (4) \int \left (40 x-40 x^3\right ) \, dx+\frac {1}{5} \log ^2(4) \int \left (10-15 x^2\right ) \, dx\\ &=x-x^2+2 x^3-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)+\frac {1}{5} \int \left (-5 e^{x/5}-e^{x/5} x-20 e^{x/5} x^3-e^{x/5} x^4-2 e^{x/5} x^2 (15+x) \log (4)-e^{x/5} x (10+x) \log ^2(4)\right ) \, dx\\ &=x-x^2+2 x^3-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)-\frac {1}{5} \int e^{x/5} x \, dx-\frac {1}{5} \int e^{x/5} x^4 \, dx-4 \int e^{x/5} x^3 \, dx-\frac {1}{5} (2 \log (4)) \int e^{x/5} x^2 (15+x) \, dx-\frac {1}{5} \log ^2(4) \int e^{x/5} x (10+x) \, dx-\int e^{x/5} \, dx\\ &=-5 e^{x/5}+x-e^{x/5} x-x^2+2 x^3-20 e^{x/5} x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)+4 \int e^{x/5} x^3 \, dx+60 \int e^{x/5} x^2 \, dx-\frac {1}{5} (2 \log (4)) \int \left (15 e^{x/5} x^2+e^{x/5} x^3\right ) \, dx-\frac {1}{5} \log ^2(4) \int \left (10 e^{x/5} x+e^{x/5} x^2\right ) \, dx+\int e^{x/5} \, dx\\ &=x-e^{x/5} x-x^2+300 e^{x/5} x^2+2 x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)-60 \int e^{x/5} x^2 \, dx-600 \int e^{x/5} x \, dx-\frac {1}{5} (2 \log (4)) \int e^{x/5} x^3 \, dx-(6 \log (4)) \int e^{x/5} x^2 \, dx-\frac {1}{5} \log ^2(4) \int e^{x/5} x^2 \, dx-\left (2 \log ^2(4)\right ) \int e^{x/5} x \, dx\\ &=x-3001 e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-30 e^{x/5} x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-10 e^{x/5} x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)+600 \int e^{x/5} x \, dx+3000 \int e^{x/5} \, dx+(6 \log (4)) \int e^{x/5} x^2 \, dx+(60 \log (4)) \int e^{x/5} x \, dx+\left (2 \log ^2(4)\right ) \int e^{x/5} x \, dx+\left (10 \log ^2(4)\right ) \int e^{x/5} \, dx\\ &=15000 e^{x/5}+x-e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5+300 e^{x/5} x \log (4)+4 x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+50 e^{x/5} \log ^2(4)+2 x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)-3000 \int e^{x/5} \, dx-(60 \log (4)) \int e^{x/5} x \, dx-(300 \log (4)) \int e^{x/5} \, dx-\left (10 \log ^2(4)\right ) \int e^{x/5} \, dx\\ &=x-e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5-1500 e^{x/5} \log (4)+4 x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)+(300 \log (4)) \int e^{x/5} \, dx\\ &=x-e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.17, size = 76, normalized size = 2.53 \begin {gather*} x-x^5-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \left (-2+\log ^2(4)\right )-\frac {1}{5} e^{x/5} \left (5 x+5 x^4+10 x^3 \log (4)+5 x^2 \log ^2(4)\right )+x^2 (-1+\log (256)) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.85, size = 70, normalized size = 2.33 \begin {gather*} -x^{5} + 2 \, x^{3} - 4 \, {\left (x^{3} - 2 \, x\right )} \log \relax (2)^{2} - x^{2} - {\left (x^{4} + 4 \, x^{3} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} + x\right )} e^{\left (\frac {1}{5} \, x\right )} - 4 \, {\left (x^{4} - 2 \, x^{2}\right )} \log \relax (2) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.24, size = 70, normalized size = 2.33 \begin {gather*} -x^{5} + 2 \, x^{3} - 4 \, {\left (x^{3} - 2 \, x\right )} \log \relax (2)^{2} - x^{2} - {\left (x^{4} + 4 \, x^{3} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} + x\right )} e^{\left (\frac {1}{5} \, x\right )} - 4 \, {\left (x^{4} - 2 \, x^{2}\right )} \log \relax (2) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 79, normalized size = 2.63
method | result | size |
risch | \(\frac {\left (-20 x^{2} \ln \relax (2)^{2}-20 x^{3} \ln \relax (2)-5 x^{4}-5 x \right ) {\mathrm e}^{\frac {x}{5}}}{5}-4 x^{3} \ln \relax (2)^{2}+8 x \ln \relax (2)^{2}-4 x^{4} \ln \relax (2)+8 x^{2} \ln \relax (2)-x^{5}+2 x^{3}-x^{2}+x\) | \(79\) |
norman | \(\left (2-4 \ln \relax (2)^{2}\right ) x^{3}+\left (8 \ln \relax (2)^{2}+1\right ) x +\left (8 \ln \relax (2)-1\right ) x^{2}-x^{5}-x \,{\mathrm e}^{\frac {x}{5}}-x^{4} {\mathrm e}^{\frac {x}{5}}-4 x^{4} \ln \relax (2)-4 \ln \relax (2) {\mathrm e}^{\frac {x}{5}} x^{3}-4 \ln \relax (2)^{2} {\mathrm e}^{\frac {x}{5}} x^{2}\) | \(86\) |
derivativedivides | \(x -x \,{\mathrm e}^{\frac {x}{5}}-x^{4} {\mathrm e}^{\frac {x}{5}}-4 \ln \relax (2) {\mathrm e}^{\frac {x}{5}} x^{3}-4 \ln \relax (2)^{2} {\mathrm e}^{\frac {x}{5}} x^{2}-x^{2}+2 x^{3}-x^{5}-4 x^{3} \ln \relax (2)^{2}+8 x \ln \relax (2)^{2}-4 x^{4} \ln \relax (2)+8 x^{2} \ln \relax (2)\) | \(88\) |
default | \(x -x \,{\mathrm e}^{\frac {x}{5}}-x^{4} {\mathrm e}^{\frac {x}{5}}-4 \ln \relax (2) {\mathrm e}^{\frac {x}{5}} x^{3}-4 \ln \relax (2)^{2} {\mathrm e}^{\frac {x}{5}} x^{2}-x^{2}+2 x^{3}-x^{5}-4 x^{3} \ln \relax (2)^{2}+8 x \ln \relax (2)^{2}-4 x^{4} \ln \relax (2)+8 x^{2} \ln \relax (2)\) | \(88\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.45, size = 70, normalized size = 2.33 \begin {gather*} -x^{5} + 2 \, x^{3} - 4 \, {\left (x^{3} - 2 \, x\right )} \log \relax (2)^{2} - x^{2} - {\left (x^{4} + 4 \, x^{3} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} + x\right )} e^{\left (\frac {1}{5} \, x\right )} - 4 \, {\left (x^{4} - 2 \, x^{2}\right )} \log \relax (2) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.21, size = 86, normalized size = 2.87 \begin {gather*} x\,\left (8\,{\ln \relax (2)}^2+1\right )-x\,{\mathrm {e}}^{x/5}+x^2\,\left (8\,\ln \relax (2)-1\right )-x^4\,{\mathrm {e}}^{x/5}-4\,x^4\,\ln \relax (2)-x^3\,\left (4\,{\ln \relax (2)}^2-2\right )-x^5-4\,x^2\,{\mathrm {e}}^{x/5}\,{\ln \relax (2)}^2-4\,x^3\,{\mathrm {e}}^{x/5}\,\ln \relax (2) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.18, size = 73, normalized size = 2.43 \begin {gather*} - x^{5} - 4 x^{4} \log {\relax (2 )} + x^{3} \left (2 - 4 \log {\relax (2 )}^{2}\right ) + x^{2} \left (-1 + 8 \log {\relax (2 )}\right ) + x \left (1 + 8 \log {\relax (2 )}^{2}\right ) + \left (- x^{4} - 4 x^{3} \log {\relax (2 )} - 4 x^{2} \log {\relax (2 )}^{2} - x\right ) e^{\frac {x}{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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