Optimal. Leaf size=32 \[ (3+2 x)^2 \left (-x+\frac {-\frac {e^5}{x}+x}{5 \log (x)}\right )^2 \]
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Rubi [F] time = 1.07, 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 {-18 x^4-24 x^5-8 x^6+e^{10} \left (-18-24 x-8 x^2\right )+e^5 \left (36 x^2+48 x^3+16 x^4\right )+\left (e^{10} (-18-12 x)+108 x^4+156 x^5+56 x^6+e^5 \left (-90 x^2-144 x^3-56 x^4\right )\right ) \log (x)+\left (-180 x^4-360 x^5-160 x^6+e^5 \left (120 x^3+80 x^4\right )\right ) \log ^2(x)+\left (450 x^4+900 x^5+400 x^6\right ) \log ^3(x)}{25 x^3 \log ^3(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {-18 x^4-24 x^5-8 x^6+e^{10} \left (-18-24 x-8 x^2\right )+e^5 \left (36 x^2+48 x^3+16 x^4\right )+\left (e^{10} (-18-12 x)+108 x^4+156 x^5+56 x^6+e^5 \left (-90 x^2-144 x^3-56 x^4\right )\right ) \log (x)+\left (-180 x^4-360 x^5-160 x^6+e^5 \left (120 x^3+80 x^4\right )\right ) \log ^2(x)+\left (450 x^4+900 x^5+400 x^6\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx\\ &=\frac {1}{25} \int \frac {2 (3+2 x) \left (-\left ((3+2 x) \left (e^5-x^2\right )^2\right )-\left (3 e^{10}-2 x^4 (9+7 x)+e^5 x^2 (15+14 x)\right ) \log (x)-10 x^3 \left (-2 e^5+x (3+4 x)\right ) \log ^2(x)+25 x^4 (3+4 x) \log ^3(x)\right )}{x^3 \log ^3(x)} \, dx\\ &=\frac {2}{25} \int \frac {(3+2 x) \left (-\left ((3+2 x) \left (e^5-x^2\right )^2\right )-\left (3 e^{10}-2 x^4 (9+7 x)+e^5 x^2 (15+14 x)\right ) \log (x)-10 x^3 \left (-2 e^5+x (3+4 x)\right ) \log ^2(x)+25 x^4 (3+4 x) \log ^3(x)\right )}{x^3 \log ^3(x)} \, dx\\ &=\frac {2}{25} \int \left (25 x (3+2 x) (3+4 x)-\frac {(3+2 x)^2 \left (-e^5+x^2\right )^2}{x^3 \log ^3(x)}+\frac {(3+2 x) \left (-e^5+x^2\right ) \left (3 e^5+18 x^2+14 x^3\right )}{x^3 \log ^2(x)}-\frac {10 (3+2 x) \left (-2 e^5+3 x+4 x^2\right )}{\log (x)}\right ) \, dx\\ &=-\left (\frac {2}{25} \int \frac {(3+2 x)^2 \left (-e^5+x^2\right )^2}{x^3 \log ^3(x)} \, dx\right )+\frac {2}{25} \int \frac {(3+2 x) \left (-e^5+x^2\right ) \left (3 e^5+18 x^2+14 x^3\right )}{x^3 \log ^2(x)} \, dx-\frac {4}{5} \int \frac {(3+2 x) \left (-2 e^5+3 x+4 x^2\right )}{\log (x)} \, dx+2 \int x (3+2 x) (3+4 x) \, dx\\ &=x^2 (3+2 x)^2-\frac {2}{25} \int \frac {(3+2 x)^2 \left (-e^5+x^2\right )^2}{x^3 \log ^3(x)} \, dx+\frac {2}{25} \int \frac {(3+2 x) \left (-e^5+x^2\right ) \left (3 e^5+18 x^2+14 x^3\right )}{x^3 \log ^2(x)} \, dx-\frac {4}{5} \int \left (-\frac {6 e^5}{\log (x)}+\frac {\left (9-4 e^5\right ) x}{\log (x)}+\frac {18 x^2}{\log (x)}+\frac {8 x^3}{\log (x)}\right ) \, dx\\ &=x^2 (3+2 x)^2-\frac {2}{25} \int \frac {(3+2 x)^2 \left (-e^5+x^2\right )^2}{x^3 \log ^3(x)} \, dx+\frac {2}{25} \int \frac {(3+2 x) \left (-e^5+x^2\right ) \left (3 e^5+18 x^2+14 x^3\right )}{x^3 \log ^2(x)} \, dx-\frac {32}{5} \int \frac {x^3}{\log (x)} \, dx-\frac {72}{5} \int \frac {x^2}{\log (x)} \, dx+\frac {1}{5} \left (24 e^5\right ) \int \frac {1}{\log (x)} \, dx-\frac {1}{5} \left (4 \left (9-4 e^5\right )\right ) \int \frac {x}{\log (x)} \, dx\\ &=x^2 (3+2 x)^2+\frac {24 e^5 \text {li}(x)}{5}-\frac {2}{25} \int \frac {(3+2 x)^2 \left (-e^5+x^2\right )^2}{x^3 \log ^3(x)} \, dx+\frac {2}{25} \int \frac {(3+2 x) \left (-e^5+x^2\right ) \left (3 e^5+18 x^2+14 x^3\right )}{x^3 \log ^2(x)} \, dx-\frac {32}{5} \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right )-\frac {72}{5} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-\frac {1}{5} \left (4 \left (9-4 e^5\right )\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=x^2 (3+2 x)^2-\frac {4}{5} \left (9-4 e^5\right ) \text {Ei}(2 \log (x))-\frac {72}{5} \text {Ei}(3 \log (x))-\frac {32}{5} \text {Ei}(4 \log (x))+\frac {24 e^5 \text {li}(x)}{5}-\frac {2}{25} \int \frac {(3+2 x)^2 \left (-e^5+x^2\right )^2}{x^3 \log ^3(x)} \, dx+\frac {2}{25} \int \frac {(3+2 x) \left (-e^5+x^2\right ) \left (3 e^5+18 x^2+14 x^3\right )}{x^3 \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 36, normalized size = 1.12 \begin {gather*} \frac {(3+2 x)^2 \left (e^5-x^2+5 x^2 \log (x)\right )^2}{25 x^2 \log ^2(x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.14, size = 120, normalized size = 3.75 \begin {gather*} \frac {4 \, x^{6} + 12 \, x^{5} + 9 \, x^{4} + 25 \, {\left (4 \, x^{6} + 12 \, x^{5} + 9 \, x^{4}\right )} \log \relax (x)^{2} + {\left (4 \, x^{2} + 12 \, x + 9\right )} e^{10} - 2 \, {\left (4 \, x^{4} + 12 \, x^{3} + 9 \, x^{2}\right )} e^{5} - 10 \, {\left (4 \, x^{6} + 12 \, x^{5} + 9 \, x^{4} - {\left (4 \, x^{4} + 12 \, x^{3} + 9 \, x^{2}\right )} e^{5}\right )} \log \relax (x)}{25 \, x^{2} \log \relax (x)^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.46, size = 137, normalized size = 4.28 \begin {gather*} \frac {100 \, x^{6} \log \relax (x)^{2} - 40 \, x^{6} \log \relax (x) + 300 \, x^{5} \log \relax (x)^{2} + 4 \, x^{6} - 120 \, x^{5} \log \relax (x) + 40 \, x^{4} e^{5} \log \relax (x) + 225 \, x^{4} \log \relax (x)^{2} + 12 \, x^{5} - 8 \, x^{4} e^{5} - 90 \, x^{4} \log \relax (x) + 120 \, x^{3} e^{5} \log \relax (x) + 9 \, x^{4} - 24 \, x^{3} e^{5} + 90 \, x^{2} e^{5} \log \relax (x) + 4 \, x^{2} e^{10} - 18 \, x^{2} e^{5} + 12 \, x e^{10} + 9 \, e^{10}}{25 \, x^{2} \log \relax (x)^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.04, size = 127, normalized size = 3.97
method | result | size |
risch | \(4 x^{4}+12 x^{3}+9 x^{2}+\frac {-40 x^{6} \ln \relax (x )+40 \ln \relax (x ) {\mathrm e}^{5} x^{4}+4 x^{6}-120 x^{5} \ln \relax (x )-8 x^{4} {\mathrm e}^{5}+120 x^{3} {\mathrm e}^{5} \ln \relax (x )+12 x^{5}-90 x^{4} \ln \relax (x )+4 \,{\mathrm e}^{10} x^{2}-24 x^{3} {\mathrm e}^{5}+90 x^{2} {\mathrm e}^{5} \ln \relax (x )+9 x^{4}+12 x \,{\mathrm e}^{10}-18 x^{2} {\mathrm e}^{5}+9 \,{\mathrm e}^{10}}{25 x^{2} \ln \relax (x )^{2}}\) | \(127\) |
default | \(-\frac {18 x^{2}}{5 \ln \relax (x )}+4 x^{4}+12 x^{3}+9 x^{2}+\frac {9 x^{2}}{25 \ln \relax (x )^{2}}-\frac {8 x^{4}}{5 \ln \relax (x )}-\frac {24 x^{3}}{5 \ln \relax (x )}+\frac {48 \,{\mathrm e}^{5} \left (-\frac {x}{2 \ln \relax (x )^{2}}-\frac {x}{2 \ln \relax (x )}-\frac {\expIntegralEi \left (1, -\ln \relax (x )\right )}{2}\right )}{25}-\frac {18 \,{\mathrm e}^{10} \left (-\frac {1}{x^{2} \ln \relax (x )}+2 \expIntegralEi \left (1, 2 \ln \relax (x )\right )\right )}{25}-\frac {16 \,{\mathrm e}^{5} \expIntegralEi \left (1, -2 \ln \relax (x )\right )}{5}-\frac {24 \,{\mathrm e}^{5} \expIntegralEi \left (1, -\ln \relax (x )\right )}{5}-\frac {56 \,{\mathrm e}^{5} \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )}{25}-\frac {18 \,{\mathrm e}^{10} \left (-\frac {1}{2 x^{2} \ln \relax (x )^{2}}+\frac {1}{x^{2} \ln \relax (x )}-2 \expIntegralEi \left (1, 2 \ln \relax (x )\right )\right )}{25}+\frac {16 \,{\mathrm e}^{5} \left (-\frac {x^{2}}{2 \ln \relax (x )^{2}}-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )}{25}+\frac {4 \,{\mathrm e}^{10}}{25 \ln \relax (x )^{2}}-\frac {24 \,{\mathrm e}^{10} \left (-\frac {1}{2 x \ln \relax (x )^{2}}+\frac {1}{2 x \ln \relax (x )}-\frac {\expIntegralEi \left (1, \ln \relax (x )\right )}{2}\right )}{25}-\frac {144 \,{\mathrm e}^{5} \left (-\frac {x}{\ln \relax (x )}-\expIntegralEi \left (1, -\ln \relax (x )\right )\right )}{25}-\frac {12 \,{\mathrm e}^{10} \left (-\frac {1}{x \ln \relax (x )}+\expIntegralEi \left (1, \ln \relax (x )\right )\right )}{25}+\frac {4 x^{4}}{25 \ln \relax (x )^{2}}+\frac {12 x^{3}}{25 \ln \relax (x )^{2}}-\frac {18 \,{\mathrm e}^{5}}{25 \ln \relax (x )^{2}}+\frac {18 \,{\mathrm e}^{5}}{5 \ln \relax (x )}\) | \(324\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.75, size = 201, normalized size = 6.28 \begin {gather*} 4 \, x^{4} + 12 \, x^{3} + 9 \, x^{2} + \frac {16}{5} \, {\rm Ei}\left (2 \, \log \relax (x)\right ) e^{5} + \frac {24}{5} \, {\rm Ei}\left (\log \relax (x)\right ) e^{5} + \frac {36}{25} \, e^{10} \Gamma \left (-1, 2 \, \log \relax (x)\right ) - \frac {144}{25} \, e^{5} \Gamma \left (-1, -\log \relax (x)\right ) - \frac {112}{25} \, e^{5} \Gamma \left (-1, -2 \, \log \relax (x)\right ) + \frac {12}{25} \, e^{10} \Gamma \left (-1, \log \relax (x)\right ) + \frac {72}{25} \, e^{10} \Gamma \left (-2, 2 \, \log \relax (x)\right ) - \frac {48}{25} \, e^{5} \Gamma \left (-2, -\log \relax (x)\right ) - \frac {64}{25} \, e^{5} \Gamma \left (-2, -2 \, \log \relax (x)\right ) + \frac {24}{25} \, e^{10} \Gamma \left (-2, \log \relax (x)\right ) + \frac {18 \, e^{5}}{5 \, \log \relax (x)} + \frac {4 \, e^{10}}{25 \, \log \relax (x)^{2}} - \frac {18 \, e^{5}}{25 \, \log \relax (x)^{2}} - \frac {32}{5} \, {\rm Ei}\left (4 \, \log \relax (x)\right ) - \frac {72}{5} \, {\rm Ei}\left (3 \, \log \relax (x)\right ) - \frac {36}{5} \, {\rm Ei}\left (2 \, \log \relax (x)\right ) + \frac {216}{25} \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) + \frac {468}{25} \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) + \frac {224}{25} \, \Gamma \left (-1, -4 \, \log \relax (x)\right ) + \frac {72}{25} \, \Gamma \left (-2, -2 \, \log \relax (x)\right ) + \frac {216}{25} \, \Gamma \left (-2, -3 \, \log \relax (x)\right ) + \frac {128}{25} \, \Gamma \left (-2, -4 \, \log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.50, size = 33, normalized size = 1.03 \begin {gather*} \frac {{\left (2\,x+3\right )}^2\,{\left ({\mathrm {e}}^5+5\,x^2\,\ln \relax (x)-x^2\right )}^2}{25\,x^2\,{\ln \relax (x)}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.22, size = 129, normalized size = 4.03 \begin {gather*} 4 x^{4} + 12 x^{3} + 9 x^{2} + \frac {4 x^{6} + 12 x^{5} - 8 x^{4} e^{5} + 9 x^{4} - 24 x^{3} e^{5} - 18 x^{2} e^{5} + 4 x^{2} e^{10} + 12 x e^{10} + \left (- 40 x^{6} - 120 x^{5} - 90 x^{4} + 40 x^{4} e^{5} + 120 x^{3} e^{5} + 90 x^{2} e^{5}\right ) \log {\relax (x )} + 9 e^{10}}{25 x^{2} \log {\relax (x )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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