Optimal. Leaf size=17 \[ \frac {2 x \log (4)}{\left (2+2 x+x^{x^2}\right )^4} \]
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Rubi [F] time = 1.33, 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 {(4-12 x) \log (4)+x^{x^2} \left (\left (2-8 x^2\right ) \log (4)-16 x^2 \log (4) \log (x)\right )}{32+160 x+320 x^2+320 x^3+160 x^4+32 x^5+x^{5 x^2}+x^{4 x^2} (10+10 x)+x^{3 x^2} \left (40+80 x+40 x^2\right )+x^{2 x^2} \left (80+240 x+240 x^2+80 x^3\right )+x^{x^2} \left (80+320 x+480 x^2+320 x^3+80 x^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 {2 \log (4) \left (2-6 x+x^{x^2}-4 x^{2+x^2}-8 x^{2+x^2} \log (x)\right )}{\left (2+2 x+x^{x^2}\right )^5} \, dx\\ &=(2 \log (4)) \int \frac {2-6 x+x^{x^2}-4 x^{2+x^2}-8 x^{2+x^2} \log (x)}{\left (2+2 x+x^{x^2}\right )^5} \, dx\\ &=(2 \log (4)) \int \left (\frac {8 x \left (-1+x+x^2+2 x \log (x)+2 x^2 \log (x)\right )}{\left (2+2 x+x^{x^2}\right )^5}-\frac {-1+4 x^2+8 x^2 \log (x)}{\left (2+2 x+x^{x^2}\right )^4}\right ) \, dx\\ &=-\left ((2 \log (4)) \int \frac {-1+4 x^2+8 x^2 \log (x)}{\left (2+2 x+x^{x^2}\right )^4} \, dx\right )+(16 \log (4)) \int \frac {x \left (-1+x+x^2+2 x \log (x)+2 x^2 \log (x)\right )}{\left (2+2 x+x^{x^2}\right )^5} \, dx\\ &=-\left ((2 \log (4)) \int \left (-\frac {1}{\left (2+2 x+x^{x^2}\right )^4}+\frac {4 x^2}{\left (2+2 x+x^{x^2}\right )^4}+\frac {8 x^2 \log (x)}{\left (2+2 x+x^{x^2}\right )^4}\right ) \, dx\right )+(16 \log (4)) \int \left (-\frac {x}{\left (2+2 x+x^{x^2}\right )^5}+\frac {x^2}{\left (2+2 x+x^{x^2}\right )^5}+\frac {x^3}{\left (2+2 x+x^{x^2}\right )^5}+\frac {2 x^2 \log (x)}{\left (2+2 x+x^{x^2}\right )^5}+\frac {2 x^3 \log (x)}{\left (2+2 x+x^{x^2}\right )^5}\right ) \, dx\\ &=(2 \log (4)) \int \frac {1}{\left (2+2 x+x^{x^2}\right )^4} \, dx-(8 \log (4)) \int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^4} \, dx-(16 \log (4)) \int \frac {x}{\left (2+2 x+x^{x^2}\right )^5} \, dx+(16 \log (4)) \int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^5} \, dx+(16 \log (4)) \int \frac {x^3}{\left (2+2 x+x^{x^2}\right )^5} \, dx-(16 \log (4)) \int \frac {x^2 \log (x)}{\left (2+2 x+x^{x^2}\right )^4} \, dx+(32 \log (4)) \int \frac {x^2 \log (x)}{\left (2+2 x+x^{x^2}\right )^5} \, dx+(32 \log (4)) \int \frac {x^3 \log (x)}{\left (2+2 x+x^{x^2}\right )^5} \, dx\\ &=(2 \log (4)) \int \frac {1}{\left (2+2 x+x^{x^2}\right )^4} \, dx-(8 \log (4)) \int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^4} \, dx-(16 \log (4)) \int \frac {x}{\left (2+2 x+x^{x^2}\right )^5} \, dx+(16 \log (4)) \int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^5} \, dx+(16 \log (4)) \int \frac {x^3}{\left (2+2 x+x^{x^2}\right )^5} \, dx+(16 \log (4)) \int \frac {\int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^4} \, dx}{x} \, dx-(32 \log (4)) \int \frac {\int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^5} \, dx}{x} \, dx-(32 \log (4)) \int \frac {\int \frac {x^3}{\left (2+2 x+x^{x^2}\right )^5} \, dx}{x} \, dx-(16 \log (4) \log (x)) \int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^4} \, dx+(32 \log (4) \log (x)) \int \frac {x^2}{\left (2+2 x+x^{x^2}\right )^5} \, dx+(32 \log (4) \log (x)) \int \frac {x^3}{\left (2+2 x+x^{x^2}\right )^5} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.24, size = 17, normalized size = 1.00 \begin {gather*} \frac {2 x \log (4)}{\left (2+2 x+x^{x^2}\right )^4} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.86, size = 83, normalized size = 4.88 \begin {gather*} \frac {4 \, x \log \relax (2)}{16 \, x^{4} + 64 \, x^{3} + 8 \, {\left (x + 1\right )} x^{3 \, x^{2}} + 24 \, {\left (x^{2} + 2 \, x + 1\right )} x^{2 \, x^{2}} + 32 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} x^{\left (x^{2}\right )} + 96 \, x^{2} + 64 \, x + x^{4 \, x^{2}} + 16} \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 = 0.06, size = 18, normalized size = 1.06
method | result | size |
risch | \(\frac {4 x \ln \relax (2)}{\left (2 x +2+x^{x^{2}}\right )^{4}}\) | \(18\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.33, size = 83, normalized size = 4.88 \begin {gather*} \frac {4 \, x \log \relax (2)}{16 \, x^{4} + 64 \, x^{3} + 8 \, {\left (x + 1\right )} x^{3 \, x^{2}} + 24 \, {\left (x^{2} + 2 \, x + 1\right )} x^{2 \, x^{2}} + 32 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} x^{\left (x^{2}\right )} + 96 \, x^{2} + 64 \, x + x^{4 \, x^{2}} + 16} \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 = -0.06 \begin {gather*} \int -\frac {2\,\ln \relax (2)\,\left (12\,x-4\right )+{\mathrm {e}}^{x^2\,\ln \relax (x)}\,\left (2\,\ln \relax (2)\,\left (8\,x^2-2\right )+32\,x^2\,\ln \relax (2)\,\ln \relax (x)\right )}{160\,x+{\mathrm {e}}^{5\,x^2\,\ln \relax (x)}+{\mathrm {e}}^{4\,x^2\,\ln \relax (x)}\,\left (10\,x+10\right )+{\mathrm {e}}^{3\,x^2\,\ln \relax (x)}\,\left (40\,x^2+80\,x+40\right )+{\mathrm {e}}^{2\,x^2\,\ln \relax (x)}\,\left (80\,x^3+240\,x^2+240\,x+80\right )+{\mathrm {e}}^{x^2\,\ln \relax (x)}\,\left (80\,x^4+320\,x^3+480\,x^2+320\,x+80\right )+320\,x^2+320\,x^3+160\,x^4+32\,x^5+32} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.48, size = 95, normalized size = 5.59 \begin {gather*} \frac {4 x \log {\relax (2 )}}{16 x^{4} + 64 x^{3} + 96 x^{2} + 64 x + \left (8 x + 8\right ) e^{3 x^{2} \log {\relax (x )}} + \left (24 x^{2} + 48 x + 24\right ) e^{2 x^{2} \log {\relax (x )}} + \left (32 x^{3} + 96 x^{2} + 96 x + 32\right ) e^{x^{2} \log {\relax (x )}} + e^{4 x^{2} \log {\relax (x )}} + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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