Optimal. Leaf size=33 \[ \log \left (-1+x+\frac {1}{5} \left (-e^x+\frac {3}{2 x+\frac {\log ^2(x)}{5 x}}\right )\right ) \]
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Rubi [F] time = 9.44, 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 {150 x^2-500 x^4+100 e^x x^4+30 \log (x)+\left (-15-100 x^2+20 e^x x^2\right ) \log ^2(x)+\left (-5+e^x\right ) \log ^4(x)}{-150 x^3+500 x^4+100 e^x x^4-500 x^5+\left (-15 x+100 x^2+20 e^x x^2-100 x^3\right ) \log ^2(x)+\left (5+e^x-5 x\right ) \log ^4(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-150 x^2+500 x^4-100 e^x x^4-30 \log (x)-\left (-15-100 x^2+20 e^x x^2\right ) \log ^2(x)-\left (-5+e^x\right ) \log ^4(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx\\ &=\int \left (1-\frac {5 \left (30 x^2+30 x^3-200 x^4+100 x^5+6 \log (x)-3 \log ^2(x)+3 x \log ^2(x)-40 x^2 \log ^2(x)+20 x^3 \log ^2(x)-2 \log ^4(x)+x \log ^4(x)\right )}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}\right ) \, dx\\ &=x-5 \int \frac {30 x^2+30 x^3-200 x^4+100 x^5+6 \log (x)-3 \log ^2(x)+3 x \log ^2(x)-40 x^2 \log ^2(x)+20 x^3 \log ^2(x)-2 \log ^4(x)+x \log ^4(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx\\ &=x-5 \int \left (\frac {30 x^2}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}+\frac {30 x^3}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}-\frac {200 x^4}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}+\frac {100 x^5}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}+\frac {6 \log (x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}-\frac {3 \log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}+\frac {3 x \log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}-\frac {40 x^2 \log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}+\frac {20 x^3 \log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}-\frac {2 \log ^4(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}+\frac {x \log ^4(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )}\right ) \, dx\\ &=x-5 \int \frac {x \log ^4(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx+10 \int \frac {\log ^4(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx+15 \int \frac {\log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx-15 \int \frac {x \log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx-30 \int \frac {\log (x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx-100 \int \frac {x^3 \log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx-150 \int \frac {x^2}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx-150 \int \frac {x^3}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx+200 \int \frac {x^2 \log ^2(x)}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx-500 \int \frac {x^5}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx+1000 \int \frac {x^4}{\left (10 x^2+\log ^2(x)\right ) \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.17, size = 59, normalized size = 1.79 \begin {gather*} -\log \left (10 x^2+\log ^2(x)\right )+\log \left (15 x-50 x^2-10 e^x x^2+50 x^3-5 \log ^2(x)-e^x \log ^2(x)+5 x \log ^2(x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.85, size = 70, normalized size = 2.12 \begin {gather*} -\log \left (10 \, x^{2} + \log \relax (x)^{2}\right ) + \log \left (-5 \, x + e^{x} + 5\right ) + \log \left (\frac {50 \, x^{3} - 10 \, x^{2} e^{x} + {\left (5 \, x - e^{x} - 5\right )} \log \relax (x)^{2} - 50 \, x^{2} + 15 \, x}{5 \, x - e^{x} - 5}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 9.02, size = 56, normalized size = 1.70 \begin {gather*} \log \left (-50 \, x^{3} + 10 \, x^{2} e^{x} - 5 \, x \log \relax (x)^{2} + e^{x} \log \relax (x)^{2} + 50 \, x^{2} + 5 \, \log \relax (x)^{2} - 15 \, x\right ) - \log \left (10 \, x^{2} + \log \relax (x)^{2}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 58, normalized size = 1.76
method | result | size |
risch | \(\ln \left ({\mathrm e}^{x}-5 x +5\right )+\ln \left (\ln \relax (x )^{2}+\frac {5 x \left (10 x^{2}-2 \,{\mathrm e}^{x} x -10 x +3\right )}{5 x -{\mathrm e}^{x}-5}\right )-\ln \left (10 x^{2}+\ln \relax (x )^{2}\right )\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 52, normalized size = 1.58 \begin {gather*} \log \left (-\frac {50 \, x^{3} + 5 \, {\left (x - 1\right )} \log \relax (x)^{2} - 50 \, x^{2} - {\left (10 \, x^{2} + \log \relax (x)^{2}\right )} e^{x} + 15 \, x}{10 \, x^{2} + \log \relax (x)^{2}}\right ) \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 = -0.03 \begin {gather*} \int \frac {30\,\ln \relax (x)+100\,x^4\,{\mathrm {e}}^x-{\ln \relax (x)}^2\,\left (100\,x^2-20\,x^2\,{\mathrm {e}}^x+15\right )+150\,x^2-500\,x^4+{\ln \relax (x)}^4\,\left ({\mathrm {e}}^x-5\right )}{100\,x^4\,{\mathrm {e}}^x-{\ln \relax (x)}^2\,\left (15\,x-20\,x^2\,{\mathrm {e}}^x-100\,x^2+100\,x^3\right )+{\ln \relax (x)}^4\,\left ({\mathrm {e}}^x-5\,x+5\right )-150\,x^3+500\,x^4-500\,x^5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 1.07, size = 42, normalized size = 1.27 \begin {gather*} \log {\left (e^{x} + \frac {- 50 x^{3} + 50 x^{2} - 5 x \log {\relax (x )}^{2} - 15 x + 5 \log {\relax (x )}^{2}}{10 x^{2} + \log {\relax (x )}^{2}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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