Optimal. Leaf size=39 \[ \frac {1+e^x+2 x-\frac {x}{5 \left (-3+\frac {-\frac {4}{x}+\frac {\log ^2(x)}{4}}{x}\right )}}{x} \]
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Rubi [F] time = 3.53, 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 {-1280-1920 x^2+128 x^3-720 x^4+e^x \left (-1280+1280 x-1920 x^2+1920 x^3-720 x^4+720 x^5\right )+8 x^4 \log (x)+\left (160 x+120 x^3-4 x^4+e^x \left (160 x-160 x^2+120 x^3-120 x^4\right )\right ) \log ^2(x)+\left (-5 x^2+e^x \left (-5 x^2+5 x^3\right )\right ) \log ^4(x)}{1280 x^2+1920 x^4+720 x^6+\left (-160 x^3-120 x^5\right ) \log ^2(x)+5 x^4 \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 {16 \left (-80-120 x^2+8 x^3-45 x^4+5 e^x (-1+x) \left (4+3 x^2\right )^2\right )+8 x^4 \log (x)-4 x \left (-40-30 x^2+x^3+10 e^x \left (-4+4 x-3 x^2+3 x^3\right )\right ) \log ^2(x)+5 \left (-1+e^x (-1+x)\right ) x^2 \log ^4(x)}{5 x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {16 \left (-80-120 x^2+8 x^3-45 x^4+5 e^x (-1+x) \left (4+3 x^2\right )^2\right )+8 x^4 \log (x)-4 x \left (-40-30 x^2+x^3+10 e^x \left (-4+4 x-3 x^2+3 x^3\right )\right ) \log ^2(x)+5 \left (-1+e^x (-1+x)\right ) x^2 \log ^4(x)}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {5 e^x (-1+x)}{x^2}-\frac {1920}{\left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {1280}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {128 x}{\left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {720 x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {8 x^2 \log (x)}{\left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {160 \log ^2(x)}{x \left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {120 x \log ^2(x)}{\left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {4 x^2 \log ^2(x)}{\left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {5 \log ^4(x)}{\left (-16-12 x^2+x \log ^2(x)\right )^2}\right ) \, dx\\ &=-\left (\frac {4}{5} \int \frac {x^2 \log ^2(x)}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\right )+\frac {8}{5} \int \frac {x^2 \log (x)}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+24 \int \frac {x \log ^2(x)}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+\frac {128}{5} \int \frac {x}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+32 \int \frac {\log ^2(x)}{x \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-144 \int \frac {x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-256 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-384 \int \frac {1}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+\int \frac {e^x (-1+x)}{x^2} \, dx-\int \frac {\log ^4(x)}{\left (-16-12 x^2+x \log ^2(x)\right )^2} \, dx\\ &=\frac {e^x}{x}-\frac {4}{5} \int \left (\frac {4 x \left (4+3 x^2\right )}{\left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {x}{16+12 x^2-x \log ^2(x)}\right ) \, dx+\frac {8}{5} \int \frac {x^2 \log (x)}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+24 \int \left (\frac {4 \left (4+3 x^2\right )}{\left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {1}{16+12 x^2-x \log ^2(x)}\right ) \, dx+\frac {128}{5} \int \frac {x}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+32 \int \left (\frac {4 \left (4+3 x^2\right )}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )}\right ) \, dx-144 \int \frac {x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-256 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-384 \int \frac {1}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-\int \left (\frac {1}{x^2}+\frac {16 \left (4+3 x^2\right )^2}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2}-\frac {8 \left (4+3 x^2\right )}{x^2 \left (16+12 x^2-x \log ^2(x)\right )}\right ) \, dx\\ &=\frac {1}{x}+\frac {e^x}{x}+\frac {4}{5} \int \frac {x}{16+12 x^2-x \log ^2(x)} \, dx+\frac {8}{5} \int \frac {x^2 \log (x)}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-\frac {16}{5} \int \frac {x \left (4+3 x^2\right )}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+8 \int \frac {4+3 x^2}{x^2 \left (16+12 x^2-x \log ^2(x)\right )} \, dx-16 \int \frac {\left (4+3 x^2\right )^2}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-24 \int \frac {1}{16+12 x^2-x \log ^2(x)} \, dx+\frac {128}{5} \int \frac {x}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-32 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )} \, dx+96 \int \frac {4+3 x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+128 \int \frac {4+3 x^2}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-144 \int \frac {x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-256 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-384 \int \frac {1}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\\ &=\frac {1}{x}+\frac {e^x}{x}+\frac {4}{5} \int \frac {x}{16+12 x^2-x \log ^2(x)} \, dx+\frac {8}{5} \int \frac {x^2 \log (x)}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-\frac {16}{5} \int \left (\frac {4 x}{\left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {3 x^3}{\left (16+12 x^2-x \log ^2(x)\right )^2}\right ) \, dx+8 \int \left (\frac {3}{16+12 x^2-x \log ^2(x)}+\frac {4}{x^2 \left (16+12 x^2-x \log ^2(x)\right )}\right ) \, dx-16 \int \left (\frac {24}{\left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {16}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {9 x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2}\right ) \, dx-24 \int \frac {1}{16+12 x^2-x \log ^2(x)} \, dx+\frac {128}{5} \int \frac {x}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-32 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )} \, dx+96 \int \left (\frac {4}{\left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {3 x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2}\right ) \, dx+128 \int \left (\frac {3}{\left (16+12 x^2-x \log ^2(x)\right )^2}+\frac {4}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2}\right ) \, dx-144 \int \frac {x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-256 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-384 \int \frac {1}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\\ &=\frac {1}{x}+\frac {e^x}{x}+\frac {4}{5} \int \frac {x}{16+12 x^2-x \log ^2(x)} \, dx+\frac {8}{5} \int \frac {x^2 \log (x)}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-\frac {48}{5} \int \frac {x^3}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-\frac {64}{5} \int \frac {x}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+\frac {128}{5} \int \frac {x}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx-2 \left (144 \int \frac {x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\right )-2 \left (256 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\right )+288 \int \frac {x^2}{\left (16+12 x^2-x \log ^2(x)\right )^2} \, dx+512 \int \frac {1}{x^2 \left (16+12 x^2-x \log ^2(x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 38, normalized size = 0.97 \begin {gather*} \frac {1}{5} \left (\frac {5}{x}+\frac {5 e^x}{x}-\frac {4 x^2}{-16-12 x^2+x \log ^2(x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.98, size = 56, normalized size = 1.44 \begin {gather*} -\frac {4 \, x^{3} - 5 \, {\left (x e^{x} + x\right )} \log \relax (x)^{2} + 60 \, x^{2} + 20 \, {\left (3 \, x^{2} + 4\right )} e^{x} + 80}{5 \, {\left (x^{2} \log \relax (x)^{2} - 12 \, x^{3} - 16 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.38, size = 60, normalized size = 1.54 \begin {gather*} \frac {5 \, x e^{x} \log \relax (x)^{2} - 4 \, x^{3} - 60 \, x^{2} e^{x} + 5 \, x \log \relax (x)^{2} - 60 \, x^{2} - 80 \, e^{x} - 80}{5 \, {\left (x^{2} \log \relax (x)^{2} - 12 \, x^{3} - 16 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 31, normalized size = 0.79
method | result | size |
risch | \(\frac {{\mathrm e}^{x}+1}{x}+\frac {4 x^{2}}{5 \left (-x \ln \relax (x )^{2}+12 x^{2}+16\right )}\) | \(31\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.39, size = 57, normalized size = 1.46 \begin {gather*} -\frac {4 \, x^{3} - 5 \, x \log \relax (x)^{2} + 60 \, x^{2} - 5 \, {\left (x \log \relax (x)^{2} - 12 \, x^{2} - 16\right )} e^{x} + 80}{5 \, {\left (x^{2} \log \relax (x)^{2} - 12 \, x^{3} - 16 \, x\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 {8\,x^4\,\ln \relax (x)-{\ln \relax (x)}^4\,\left ({\mathrm {e}}^x\,\left (5\,x^2-5\,x^3\right )+5\,x^2\right )+{\mathrm {e}}^x\,\left (720\,x^5-720\,x^4+1920\,x^3-1920\,x^2+1280\,x-1280\right )+{\ln \relax (x)}^2\,\left (160\,x+{\mathrm {e}}^x\,\left (-120\,x^4+120\,x^3-160\,x^2+160\,x\right )+120\,x^3-4\,x^4\right )-1920\,x^2+128\,x^3-720\,x^4-1280}{5\,x^4\,{\ln \relax (x)}^4-{\ln \relax (x)}^2\,\left (120\,x^5+160\,x^3\right )+1280\,x^2+1920\,x^4+720\,x^6} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.40, size = 27, normalized size = 0.69 \begin {gather*} - \frac {4 x^{2}}{- 60 x^{2} + 5 x \log {\relax (x )}^{2} - 80} + \frac {e^{x}}{x} + \frac {1}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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