Optimal. Leaf size=29 \[ \left (-x+\frac {10}{\left (x-e^{-x} x\right ) \log \left (\frac {5}{\log (625)}\right )}\right )^2 \]
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Rubi [F] time = 1.79, 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 {-200 e^{3 x}+e^{2 x} (200-200 x)+\left (-20 e^x x^3+20 e^{2 x} x^3\right ) \log \left (\frac {5}{\log (625)}\right )+\left (-2 x^4+6 e^x x^4-6 e^{2 x} x^4+2 e^{3 x} x^4\right ) \log ^2\left (\frac {5}{\log (625)}\right )}{\left (-x^3+3 e^x x^3-3 e^{2 x} x^3+e^{3 x} x^3\right ) \log ^2\left (\frac {5}{\log (625)}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-200 e^{3 x}+e^{2 x} (200-200 x)+\left (-20 e^x x^3+20 e^{2 x} x^3\right ) \log \left (\frac {5}{\log (625)}\right )+\left (-2 x^4+6 e^x x^4-6 e^{2 x} x^4+2 e^{3 x} x^4\right ) \log ^2\left (\frac {5}{\log (625)}\right )}{-x^3+3 e^x x^3-3 e^{2 x} x^3+e^{3 x} x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}\\ &=\frac {\int \frac {200 e^{3 x}-e^{2 x} (200-200 x)-\left (-20 e^x x^3+20 e^{2 x} x^3\right ) \log \left (\frac {5}{\log (625)}\right )-\left (-2 x^4+6 e^x x^4-6 e^{2 x} x^4+2 e^{3 x} x^4\right ) \log ^2\left (\frac {5}{\log (625)}\right )}{\left (1-e^x\right )^3 x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}\\ &=\frac {\int \frac {200 e^{3 x}+200 e^{2 x} (-1+x)-20 e^x \left (-1+e^x\right ) x^3 \log \left (\frac {5}{\log (625)}\right )-2 \left (-1+e^x\right )^3 x^4 \log ^2\left (\frac {5}{\log (625)}\right )}{\left (x-e^x x\right )^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}\\ &=\frac {\int \left (-\frac {200}{\left (-1+e^x\right )^3 x^2}+\frac {20 \left (-10-20 x+x^3 \log \left (\frac {5}{\log (625)}\right )\right )}{\left (-1+e^x\right )^2 x^3}+\frac {20 \left (-20-10 x+x^3 \log \left (\frac {5}{\log (625)}\right )\right )}{\left (-1+e^x\right ) x^3}+\frac {2 \left (-100+x^4 \log ^2\left (\frac {5}{\log (625)}\right )\right )}{x^3}\right ) \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}\\ &=\frac {2 \int \frac {-100+x^4 \log ^2\left (\frac {5}{\log (625)}\right )}{x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20 \int \frac {-10-20 x+x^3 \log \left (\frac {5}{\log (625)}\right )}{\left (-1+e^x\right )^2 x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20 \int \frac {-20-10 x+x^3 \log \left (\frac {5}{\log (625)}\right )}{\left (-1+e^x\right ) x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^3 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}\\ &=\frac {2 \int \left (-\frac {100}{x^3}+x \log ^2\left (\frac {5}{\log (625)}\right )\right ) \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20 \int \left (-\frac {10}{\left (-1+e^x\right )^2 x^3}-\frac {20}{\left (-1+e^x\right )^2 x^2}+\frac {\log \left (\frac {5}{\log (625)}\right )}{\left (-1+e^x\right )^2}\right ) \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20 \int \left (-\frac {20}{\left (-1+e^x\right ) x^3}-\frac {10}{\left (-1+e^x\right ) x^2}+\frac {\log \left (\frac {5}{\log (625)}\right )}{-1+e^x}\right ) \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^3 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}\\ &=x^2+\frac {100}{x^2 \log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^2 x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^3 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right ) x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right ) x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right )^2 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20 \int \frac {1}{\left (-1+e^x\right )^2} \, dx}{\log \left (\frac {5}{\log (625)}\right )}+\frac {20 \int \frac {1}{-1+e^x} \, dx}{\log \left (\frac {5}{\log (625)}\right )}\\ &=x^2+\frac {100}{x^2 \log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^2 x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^3 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right ) x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right ) x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right )^2 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20 \operatorname {Subst}\left (\int \frac {1}{(-1+x)^2 x} \, dx,x,e^x\right )}{\log \left (\frac {5}{\log (625)}\right )}+\frac {20 \operatorname {Subst}\left (\int \frac {1}{(-1+x) x} \, dx,x,e^x\right )}{\log \left (\frac {5}{\log (625)}\right )}\\ &=x^2+\frac {100}{x^2 \log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^2 x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^3 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right ) x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right ) x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right )^2 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20 \operatorname {Subst}\left (\int \left (\frac {1}{1-x}+\frac {1}{(-1+x)^2}+\frac {1}{x}\right ) \, dx,x,e^x\right )}{\log \left (\frac {5}{\log (625)}\right )}+\frac {20 \operatorname {Subst}\left (\int \frac {1}{-1+x} \, dx,x,e^x\right )}{\log \left (\frac {5}{\log (625)}\right )}-\frac {20 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{\log \left (\frac {5}{\log (625)}\right )}\\ &=x^2+\frac {100}{x^2 \log ^2\left (\frac {5}{\log (625)}\right )}+\frac {20}{\left (1-e^x\right ) \log \left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^2 x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right )^3 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {200 \int \frac {1}{\left (-1+e^x\right ) x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right ) x^3} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}-\frac {400 \int \frac {1}{\left (-1+e^x\right )^2 x^2} \, dx}{\log ^2\left (\frac {5}{\log (625)}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 57, normalized size = 1.97 \begin {gather*} \frac {\frac {100 e^{2 x}}{\left (-1+e^x\right )^2 x^2}+\log \left (\frac {5}{\log (625)}\right ) \left (-\frac {20}{-1+e^x}+x^2 \log \left (\frac {5}{\log (625)}\right )\right )}{\log ^2\left (\frac {5}{\log (625)}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.60, size = 88, normalized size = 3.03 \begin {gather*} \frac {{\left (x^{4} e^{\left (2 \, x\right )} - 2 \, x^{4} e^{x} + x^{4}\right )} \log \left (\frac {5}{4 \, \log \relax (5)}\right )^{2} - 20 \, {\left (x^{2} e^{x} - x^{2}\right )} \log \left (\frac {5}{4 \, \log \relax (5)}\right ) + 100 \, e^{\left (2 \, x\right )}}{{\left (x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} + x^{2}\right )} \log \left (\frac {5}{4 \, \log \relax (5)}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.41, size = 291, normalized size = 10.03 \begin {gather*} \frac {x^{4} e^{\left (2 \, x\right )} \log \relax (5)^{2} - 2 \, x^{4} e^{x} \log \relax (5)^{2} - 4 \, x^{4} e^{\left (2 \, x\right )} \log \relax (5) \log \relax (2) + 8 \, x^{4} e^{x} \log \relax (5) \log \relax (2) + 4 \, x^{4} e^{\left (2 \, x\right )} \log \relax (2)^{2} - 8 \, x^{4} e^{x} \log \relax (2)^{2} - 2 \, x^{4} e^{\left (2 \, x\right )} \log \relax (5) \log \left (\log \relax (5)\right ) + 4 \, x^{4} e^{x} \log \relax (5) \log \left (\log \relax (5)\right ) + 4 \, x^{4} e^{\left (2 \, x\right )} \log \relax (2) \log \left (\log \relax (5)\right ) - 8 \, x^{4} e^{x} \log \relax (2) \log \left (\log \relax (5)\right ) + x^{4} e^{\left (2 \, x\right )} \log \left (\log \relax (5)\right )^{2} - 2 \, x^{4} e^{x} \log \left (\log \relax (5)\right )^{2} + x^{4} \log \relax (5)^{2} - 4 \, x^{4} \log \relax (5) \log \relax (2) + 4 \, x^{4} \log \relax (2)^{2} - 2 \, x^{4} \log \relax (5) \log \left (\log \relax (5)\right ) + 4 \, x^{4} \log \relax (2) \log \left (\log \relax (5)\right ) + x^{4} \log \left (\log \relax (5)\right )^{2} - 20 \, x^{2} e^{x} \log \relax (5) + 40 \, x^{2} e^{x} \log \relax (2) + 20 \, x^{2} e^{x} \log \left (\log \relax (5)\right ) + 20 \, x^{2} \log \relax (5) - 40 \, x^{2} \log \relax (2) - 20 \, x^{2} \log \left (\log \relax (5)\right ) + 100 \, e^{\left (2 \, x\right )}}{{\left (x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} + x^{2}\right )} \log \left (\frac {5}{4 \, \log \relax (5)}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.39, size = 109, normalized size = 3.76
method | result | size |
norman | \(\frac {20 x^{2}+\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right ) x^{4}-20 \,{\mathrm e}^{x} x^{2}+\left (4 \ln \relax (2)-2 \ln \relax (5)+2 \ln \left (\ln \relax (5)\right )\right ) x^{4} {\mathrm e}^{x}+\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right ) x^{4} {\mathrm e}^{2 x}-\frac {100 \,{\mathrm e}^{2 x}}{2 \ln \relax (2)-\ln \relax (5)+\ln \left (\ln \relax (5)\right )}}{x^{2} \left ({\mathrm e}^{x}-1\right )^{2} \ln \left (\frac {5}{4 \ln \relax (5)}\right )}\) | \(109\) |
risch | \(\frac {4 x^{2} \ln \relax (2)^{2}}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2}}-\frac {4 x^{2} \ln \relax (2) \ln \relax (5)}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2}}+\frac {4 x^{2} \ln \relax (2) \ln \left (\ln \relax (5)\right )}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2}}+\frac {x^{2} \ln \relax (5)^{2}}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2}}-\frac {2 x^{2} \ln \relax (5) \ln \left (\ln \relax (5)\right )}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2}}+\frac {x^{2} \ln \left (\ln \relax (5)\right )^{2}}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2}}+\frac {100}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2} x^{2}}+\frac {40 x^{2} \ln \relax (2) {\mathrm e}^{x}-20 x^{2} \ln \relax (5) {\mathrm e}^{x}+20 \ln \left (\ln \relax (5)\right ) x^{2} {\mathrm e}^{x}-40 x^{2} \ln \relax (2)+20 x^{2} \ln \relax (5)-20 \ln \left (\ln \relax (5)\right ) x^{2}+200 \,{\mathrm e}^{x}-100}{\left (\ln \relax (5)-2 \ln \relax (2)-\ln \left (\ln \relax (5)\right )\right )^{2} x^{2} \left ({\mathrm e}^{x}-1\right )^{2}}\) | \(239\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.50, size = 195, normalized size = 6.72 \begin {gather*} \frac {{\left (\log \relax (5)^{2} - 4 \, {\left (\log \relax (5) - \log \left (\log \relax (5)\right )\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, \log \relax (5) \log \left (\log \relax (5)\right ) + \log \left (\log \relax (5)\right )^{2}\right )} x^{4} + 20 \, x^{2} {\left (\log \relax (5) - 2 \, \log \relax (2) - \log \left (\log \relax (5)\right )\right )} + {\left ({\left (\log \relax (5)^{2} - 4 \, {\left (\log \relax (5) - \log \left (\log \relax (5)\right )\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, \log \relax (5) \log \left (\log \relax (5)\right ) + \log \left (\log \relax (5)\right )^{2}\right )} x^{4} + 100\right )} e^{\left (2 \, x\right )} - 2 \, {\left ({\left (\log \relax (5)^{2} - 4 \, {\left (\log \relax (5) - \log \left (\log \relax (5)\right )\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, \log \relax (5) \log \left (\log \relax (5)\right ) + \log \left (\log \relax (5)\right )^{2}\right )} x^{4} + 10 \, x^{2} {\left (\log \relax (5) - 2 \, \log \relax (2) - \log \left (\log \relax (5)\right )\right )}\right )} e^{x}}{{\left (x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} + x^{2}\right )} \log \left (\frac {5}{4 \, \log \relax (5)}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.41, size = 107, normalized size = 3.69 \begin {gather*} \frac {100}{x^2\,{\ln \left (\frac {5}{4\,\ln \relax (5)}\right )}^2}+x^2-\frac {x^3\,\ln \left (\frac {1099511627776\,{\ln \relax (5)}^{20}}{95367431640625}\right )-200\,x+20\,x^3\,\ln \left (\frac {5}{4\,\ln \relax (5)}\right )}{2\,x^3\,{\ln \left (\frac {5}{4\,\ln \relax (5)}\right )}^2\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )}+\frac {20\,\left (10\,x-x^3\,\ln \left (\frac {5}{4\,\ln \relax (5)}\right )\right )}{x^3\,{\ln \left (\frac {5}{4\,\ln \relax (5)}\right )}^2\,\left ({\mathrm {e}}^x-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.90, size = 379, normalized size = 13.07 \begin {gather*} \frac {x^{2} \left (- 4 \log {\relax (2 )} \log {\relax (5 )} - 2 \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + \log {\left (\log {\relax (5 )} \right )}^{2} + 4 \log {\relax (2 )} \log {\left (\log {\relax (5 )} \right )} + 4 \log {\relax (2 )}^{2} + \log {\relax (5 )}^{2}\right ) + \frac {100}{x^{2}}}{- 4 \log {\relax (2 )} \log {\relax (5 )} - 2 \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + \log {\left (\log {\relax (5 )} \right )}^{2} + 4 \log {\relax (2 )} \log {\left (\log {\relax (5 )} \right )} + 4 \log {\relax (2 )}^{2} + \log {\relax (5 )}^{2}} + \frac {- 40 x^{2} \log {\relax (2 )} - 20 x^{2} \log {\left (\log {\relax (5 )} \right )} + 20 x^{2} \log {\relax (5 )} + \left (- 20 x^{2} \log {\relax (5 )} + 20 x^{2} \log {\left (\log {\relax (5 )} \right )} + 40 x^{2} \log {\relax (2 )} + 200\right ) e^{x} - 100}{- 4 x^{2} \log {\relax (2 )} \log {\relax (5 )} - 2 x^{2} \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + x^{2} \log {\left (\log {\relax (5 )} \right )}^{2} + 4 x^{2} \log {\relax (2 )} \log {\left (\log {\relax (5 )} \right )} + 4 x^{2} \log {\relax (2 )}^{2} + x^{2} \log {\relax (5 )}^{2} + \left (- 2 x^{2} \log {\relax (5 )}^{2} - 8 x^{2} \log {\relax (2 )}^{2} - 8 x^{2} \log {\relax (2 )} \log {\left (\log {\relax (5 )} \right )} - 2 x^{2} \log {\left (\log {\relax (5 )} \right )}^{2} + 4 x^{2} \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + 8 x^{2} \log {\relax (2 )} \log {\relax (5 )}\right ) e^{x} + \left (- 4 x^{2} \log {\relax (2 )} \log {\relax (5 )} - 2 x^{2} \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + x^{2} \log {\left (\log {\relax (5 )} \right )}^{2} + 4 x^{2} \log {\relax (2 )} \log {\left (\log {\relax (5 )} \right )} + 4 x^{2} \log {\relax (2 )}^{2} + x^{2} \log {\relax (5 )}^{2}\right ) e^{2 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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