Optimal. Leaf size=26 \[ 2 \left (x-\left (e^x+x\right )^2 \left (2-2 x+\frac {25}{\log (x)}+\log (x)\right )\right ) \]
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Rubi [F] time = 1.54, 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 {50 e^{2 x}+100 e^x x+50 x^2+\left (-100 e^{2 x} x-100 x^2+e^x \left (-100 x-100 x^2\right )\right ) \log (x)+\left (2 x-10 x^2+12 x^3+e^{2 x} \left (-2-4 x+8 x^2\right )+e^x \left (-12 x+8 x^2+8 x^3\right )\right ) \log ^2(x)+\left (-4 e^{2 x} x-4 x^2+e^x \left (-4 x-4 x^2\right )\right ) \log ^3(x)}{x \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int 2 \left (1-5 x+6 x^2+e^{2 x} \left (-2-\frac {1}{x}+4 x\right )+e^x \left (-6+4 x+4 x^2\right )+\frac {25 \left (e^x+x\right )^2}{x \log ^2(x)}-\frac {50 \left (1+e^x\right ) \left (e^x+x\right )}{\log (x)}-2 \left (1+e^x\right ) \left (e^x+x\right ) \log (x)\right ) \, dx\\ &=2 \int \left (1-5 x+6 x^2+e^{2 x} \left (-2-\frac {1}{x}+4 x\right )+e^x \left (-6+4 x+4 x^2\right )+\frac {25 \left (e^x+x\right )^2}{x \log ^2(x)}-\frac {50 \left (1+e^x\right ) \left (e^x+x\right )}{\log (x)}-2 \left (1+e^x\right ) \left (e^x+x\right ) \log (x)\right ) \, dx\\ &=2 x-5 x^2+4 x^3+2 \int e^{2 x} \left (-2-\frac {1}{x}+4 x\right ) \, dx+2 \int e^x \left (-6+4 x+4 x^2\right ) \, dx-4 \int \left (1+e^x\right ) \left (e^x+x\right ) \log (x) \, dx+50 \int \frac {\left (e^x+x\right )^2}{x \log ^2(x)} \, dx-100 \int \frac {\left (1+e^x\right ) \left (e^x+x\right )}{\log (x)} \, dx\\ &=2 x-5 x^2+4 x^3-2 \left (e^x+x\right )^2 \log (x)+2 \int \left (-2 e^{2 x}-\frac {e^{2 x}}{x}+4 e^{2 x} x\right ) \, dx+2 \int \left (-6 e^x+4 e^x x+4 e^x x^2\right ) \, dx+4 \int \frac {\left (e^x+x\right )^2}{2 x} \, dx+50 \int \left (\frac {2 e^x}{\log ^2(x)}+\frac {e^{2 x}}{x \log ^2(x)}+\frac {x}{\log ^2(x)}\right ) \, dx-100 \int \left (\frac {e^{2 x}}{\log (x)}+\frac {x}{\log (x)}+\frac {e^x (1+x)}{\log (x)}\right ) \, dx\\ &=2 x-5 x^2+4 x^3-2 \left (e^x+x\right )^2 \log (x)-2 \int \frac {e^{2 x}}{x} \, dx+2 \int \frac {\left (e^x+x\right )^2}{x} \, dx-4 \int e^{2 x} \, dx+8 \int e^x x \, dx+8 \int e^{2 x} x \, dx+8 \int e^x x^2 \, dx-12 \int e^x \, dx+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx+50 \int \frac {x}{\log ^2(x)} \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx-100 \int \frac {x}{\log (x)} \, dx-100 \int \frac {e^x (1+x)}{\log (x)} \, dx\\ &=-12 e^x-2 e^{2 x}+2 x+8 e^x x+4 e^{2 x} x-5 x^2+8 e^x x^2+4 x^3-2 \text {Ei}(2 x)-\frac {50 x^2}{\log (x)}-2 \left (e^x+x\right )^2 \log (x)+2 \int \left (2 e^x+\frac {e^{2 x}}{x}+x\right ) \, dx-4 \int e^{2 x} \, dx-8 \int e^x \, dx-16 \int e^x x \, dx+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx-100 \int \left (\frac {e^x}{\log (x)}+\frac {e^x x}{\log (x)}\right ) \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx+100 \int \frac {x}{\log (x)} \, dx-100 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-20 e^x-4 e^{2 x}+2 x-8 e^x x+4 e^{2 x} x-4 x^2+8 e^x x^2+4 x^3-2 \text {Ei}(2 x)-100 \text {Ei}(2 \log (x))-\frac {50 x^2}{\log (x)}-2 \left (e^x+x\right )^2 \log (x)+2 \int \frac {e^{2 x}}{x} \, dx+4 \int e^x \, dx+16 \int e^x \, dx+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^x}{\log (x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx-100 \int \frac {e^x x}{\log (x)} \, dx+100 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-4 e^{2 x}+2 x-8 e^x x+4 e^{2 x} x-4 x^2+8 e^x x^2+4 x^3-\frac {50 x^2}{\log (x)}-2 \left (e^x+x\right )^2 \log (x)+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^x}{\log (x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx-100 \int \frac {e^x x}{\log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.30, size = 57, normalized size = 2.19 \begin {gather*} 2 \left (2 e^{2 x} (-1+x)+x+4 e^x (-1+x) x-2 x^2+2 x^3-\frac {25 \left (e^x+x\right )^2}{\log (x)}-\left (e^x+x\right )^2 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.62, size = 77, normalized size = 2.96 \begin {gather*} -\frac {2 \, {\left ({\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} \log \relax (x)^{2} + 25 \, x^{2} + 50 \, x e^{x} - {\left (2 \, x^{3} - 2 \, x^{2} + 2 \, {\left (x - 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} - x\right )} e^{x} + x\right )} \log \relax (x) + 25 \, e^{\left (2 \, x\right )}\right )}}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.13, size = 102, normalized size = 3.92 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} \log \relax (x) + 4 \, x^{2} e^{x} \log \relax (x) - x^{2} \log \relax (x)^{2} - 2 \, x e^{x} \log \relax (x)^{2} - 2 \, x^{2} \log \relax (x) + 2 \, x e^{\left (2 \, x\right )} \log \relax (x) - 4 \, x e^{x} \log \relax (x) - e^{\left (2 \, x\right )} \log \relax (x)^{2} - 25 \, x^{2} - 50 \, x e^{x} + x \log \relax (x) - 2 \, e^{\left (2 \, x\right )} \log \relax (x) - 25 \, e^{\left (2 \, x\right )}\right )}}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 79, normalized size = 3.04
method | result | size |
risch | \(\left (-2 x^{2}-4 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{2 x}\right ) \ln \relax (x )+4 x^{3}+8 \,{\mathrm e}^{x} x^{2}+4 x \,{\mathrm e}^{2 x}-4 x^{2}-8 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{2 x}+2 x -\frac {50 \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}\right )}{\ln \relax (x )}\) | \(79\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, x^{3} - 5 \, x^{2} + 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + 8 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 8 \, {\left (x - 1\right )} e^{x} + 2 \, x - \frac {2 \, x^{2} \log \relax (x)^{2} - x^{2} \log \relax (x) + 2 \, e^{\left (2 \, x\right )} \log \relax (x)^{2} + 4 \, {\left (x \log \relax (x)^{2} + 25 \, x - \log \relax (x)\right )} e^{x}}{\log \relax (x)} - 2 \, {\rm Ei}\left (2 \, x\right ) - 2 \, e^{\left (2 \, x\right )} - 12 \, e^{x} + 100 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 100 \, \int \frac {x}{\log \relax (x)}\,{d x} + 2 \, \int -\frac {{\left (50 \, x \log \relax (x) - \log \relax (x)^{2} - 25\right )} e^{\left (2 \, x\right )}}{x \log \relax (x)^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.60, size = 87, normalized size = 3.35 \begin {gather*} 2\,x-\frac {50\,{\left (x+{\mathrm {e}}^x\right )}^2-100\,x\,\ln \relax (x)\,\left ({\mathrm {e}}^x+1\right )\,\left (x+{\mathrm {e}}^x\right )}{\ln \relax (x)}-\ln \relax (x)\,\left (2\,{\mathrm {e}}^{2\,x}+4\,x\,{\mathrm {e}}^x+2\,x^2\right )-{\mathrm {e}}^x\,\left (92\,x^2+108\,x\right )-{\mathrm {e}}^{2\,x}\,\left (96\,x+4\right )-104\,x^2+4\,x^3 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.43, size = 104, normalized size = 4.00 \begin {gather*} 4 x^{3} - 2 x^{2} \log {\relax (x )} - 4 x^{2} - \frac {50 x^{2}}{\log {\relax (x )}} + 2 x + \frac {\left (4 x \log {\relax (x )}^{2} - 2 \log {\relax (x )}^{3} - 4 \log {\relax (x )}^{2} - 50 \log {\relax (x )}\right ) e^{2 x} + \left (8 x^{2} \log {\relax (x )}^{2} - 4 x \log {\relax (x )}^{3} - 8 x \log {\relax (x )}^{2} - 100 x \log {\relax (x )}\right ) e^{x}}{\log {\relax (x )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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