Optimal. Leaf size=26 \[ \frac {1}{x+x^2+\frac {(-4+x) \log (4)}{20 \left (-2+e^x\right ) x}} \]
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Rubi [F] time = 13.31, 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 {-1600 x^2-3200 x^3+e^{2 x} \left (-400 x^2-800 x^3\right )+160 \log (4)+e^x \left (1600 x^2+3200 x^3+\left (-80-80 x+20 x^2\right ) \log (4)\right )}{1600 x^4+3200 x^5+1600 x^6+e^{2 x} \left (400 x^4+800 x^5+400 x^6\right )+\left (320 x^2+240 x^3-80 x^4\right ) \log (4)+\left (16-8 x+x^2\right ) \log ^2(4)+e^x \left (-1600 x^4-3200 x^5-1600 x^6+\left (-160 x^2-120 x^3+40 x^4\right ) \log (4)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20 \left (-40 \left (-2+e^x\right )^2 x^3-4 \left (-2+e^x\right ) \log (4)-4 e^x x \log (4)-x^2 \left (80+20 e^{2 x}-e^x (80+\log (4))\right )\right )}{\left (20 \left (-2+e^x\right ) x^2+20 \left (-2+e^x\right ) x^3-4 \log (4)+x \log (4)\right )^2} \, dx\\ &=20 \int \frac {-40 \left (-2+e^x\right )^2 x^3-4 \left (-2+e^x\right ) \log (4)-4 e^x x \log (4)-x^2 \left (80+20 e^{2 x}-e^x (80+\log (4))\right )}{\left (20 \left (-2+e^x\right ) x^2+20 \left (-2+e^x\right ) x^3-4 \log (4)+x \log (4)\right )^2} \, dx\\ &=20 \int \left (\frac {-1-2 x}{20 x^2 (1+x)^2}+\frac {\left (-12-22 x+x^2+x^3\right ) \log (4)}{20 x^2 (1+x)^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )}+\frac {(4-x) \log (4) \left (-80 x^4-40 x^5-40 x^3 \left (1-\frac {\log (2)}{20}\right )-8 \log (4)-15 x \log (4)-x^2 \log (4)\right )}{20 x^2 (1+x)^2 \left (40 x^2-20 e^x x^2+40 x^3-20 e^x x^3+4 \log (4)-x \log (4)\right )^2}\right ) \, dx\\ &=\log (4) \int \frac {-12-22 x+x^2+x^3}{x^2 (1+x)^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )} \, dx+\log (4) \int \frac {(4-x) \left (-80 x^4-40 x^5-40 x^3 \left (1-\frac {\log (2)}{20}\right )-8 \log (4)-15 x \log (4)-x^2 \log (4)\right )}{x^2 (1+x)^2 \left (40 x^2-20 e^x x^2+40 x^3-20 e^x x^3+4 \log (4)-x \log (4)\right )^2} \, dx+\int \frac {-1-2 x}{x^2 (1+x)^2} \, dx\\ &=\frac {1}{x (1+x)}+\log (4) \int \left (-\frac {12}{x^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )}+\frac {2}{x \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )}+\frac {10}{(1+x)^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )}-\frac {1}{(1+x) \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )}\right ) \, dx+\log (4) \int \left (-\frac {160 x}{\left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2}+\frac {40 x^2}{\left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2}-\frac {\log (4)}{\left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2}-\frac {32 \log (4)}{x^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2}+\frac {12 \log (4)}{x \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2}+\frac {10 \log (32)}{(1+x)^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2}-\frac {\log (1024)}{(1+x) \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2}\right ) \, dx\\ &=\frac {1}{x (1+x)}-\log (4) \int \frac {1}{(1+x) \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )} \, dx+(2 \log (4)) \int \frac {1}{x \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )} \, dx+(10 \log (4)) \int \frac {1}{(1+x)^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )} \, dx-(12 \log (4)) \int \frac {1}{x^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )} \, dx+(40 \log (4)) \int \frac {x^2}{\left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2} \, dx-(160 \log (4)) \int \frac {x}{\left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2} \, dx-\log ^2(4) \int \frac {1}{\left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2} \, dx+\left (12 \log ^2(4)\right ) \int \frac {1}{x \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2} \, dx-\left (32 \log ^2(4)\right ) \int \frac {1}{x^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2} \, dx+(10 \log (4) \log (32)) \int \frac {1}{(1+x)^2 \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2} \, dx-(\log (4) \log (1024)) \int \frac {1}{(1+x) \left (-40 x^2+20 e^x x^2-40 x^3+20 e^x x^3-4 \log (4)+x \log (4)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 39, normalized size = 1.50 \begin {gather*} \frac {20 \left (-2+e^x\right ) x}{20 \left (-2+e^x\right ) x^2+20 \left (-2+e^x\right ) x^3-4 \log (4)+x \log (4)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 41, normalized size = 1.58 \begin {gather*} -\frac {10 \, {\left (x e^{x} - 2 \, x\right )}}{20 \, x^{3} + 20 \, x^{2} - 10 \, {\left (x^{3} + x^{2}\right )} e^{x} - {\left (x - 4\right )} \log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.38, size = 46, normalized size = 1.77
method | result | size |
norman | \(\frac {-20 x +10 \,{\mathrm e}^{x} x}{10 \,{\mathrm e}^{x} x^{3}+10 \,{\mathrm e}^{x} x^{2}-20 x^{3}+x \ln \relax (2)-20 x^{2}-4 \ln \relax (2)}\) | \(46\) |
risch | \(\frac {1}{x \left (x +1\right )}-\frac {\left (x -4\right ) \ln \relax (2)}{\left (x +1\right ) x \left (10 \,{\mathrm e}^{x} x^{3}+10 \,{\mathrm e}^{x} x^{2}-20 x^{3}+x \ln \relax (2)-20 x^{2}-4 \ln \relax (2)\right )}\) | \(61\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.56, size = 43, normalized size = 1.65 \begin {gather*} -\frac {10 \, {\left (x e^{x} - 2 \, x\right )}}{20 \, x^{3} + 20 \, x^{2} - 10 \, {\left (x^{3} + x^{2}\right )} e^{x} - x \log \relax (2) + 4 \, \log \relax (2)} \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 = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{2\,x}\,\left (800\,x^3+400\,x^2\right )-{\mathrm {e}}^x\,\left (1600\,x^2-2\,\ln \relax (2)\,\left (-20\,x^2+80\,x+80\right )+3200\,x^3\right )-320\,\ln \relax (2)+1600\,x^2+3200\,x^3}{4\,{\ln \relax (2)}^2\,\left (x^2-8\,x+16\right )-{\mathrm {e}}^x\,\left (2\,\ln \relax (2)\,\left (-40\,x^4+120\,x^3+160\,x^2\right )+1600\,x^4+3200\,x^5+1600\,x^6\right )+{\mathrm {e}}^{2\,x}\,\left (400\,x^6+800\,x^5+400\,x^4\right )+2\,\ln \relax (2)\,\left (-80\,x^4+240\,x^3+320\,x^2\right )+1600\,x^4+3200\,x^5+1600\,x^6} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.46, size = 71, normalized size = 2.73 \begin {gather*} \frac {- x \log {\relax (2 )} + 4 \log {\relax (2 )}}{- 20 x^{5} - 40 x^{4} - 20 x^{3} + x^{3} \log {\relax (2 )} - 3 x^{2} \log {\relax (2 )} - 4 x \log {\relax (2 )} + \left (10 x^{5} + 20 x^{4} + 10 x^{3}\right ) e^{x}} + \frac {1}{x^{2} + x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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