Optimal. Leaf size=21 \[ x \left (1+\frac {4}{e^4}+x-\frac {3}{1+x}+\log (4)\right )^2 \]
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Rubi [B] time = 0.24, antiderivative size = 82, normalized size of antiderivative = 3.90, number of steps used = 3, number of rules used = 2, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {12, 2074} \begin {gather*} x^3+\frac {2 x^2 \left (4+e^4 (1+\log (4))\right )}{e^4}-\frac {9}{(x+1)^2}+\frac {x \left (16-e^8 \left (5-\log ^2(4)-\log (16)\right )+8 e^4 (1+\log (4))\right )}{e^8}+\frac {3 \left (8+e^4 (3+\log (16))\right )}{e^4 (x+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2074
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {16+48 x+48 x^2+16 x^3+e^4 \left (-16+16 x+72 x^2+56 x^3+16 x^4\right )+e^8 \left (4-20 x+16 x^3+13 x^4+3 x^5\right )+\left (e^4 \left (8+24 x+24 x^2+8 x^3\right )+e^8 \left (-4+4 x+18 x^2+14 x^3+4 x^4\right )\right ) \log (4)+e^8 \left (1+3 x+3 x^2+x^3\right ) \log ^2(4)}{1+3 x+3 x^2+x^3} \, dx}{e^8}\\ &=\frac {\int \left (3 e^8 x^2+\frac {18 e^8}{(1+x)^3}+4 e^4 x \left (4+e^4 (1+\log (4))\right )+\frac {3 e^4 \left (-8-e^4 (3+\log (16))\right )}{(1+x)^2}+16 \left (1+\frac {1}{2} e^4 \left (1+\log (4)+\frac {1}{8} e^4 \left (-5+\log ^2(4)+\log (16)\right )\right )\right )\right ) \, dx}{e^8}\\ &=x^3-\frac {9}{(1+x)^2}+\frac {2 x^2 \left (4+e^4 (1+\log (4))\right )}{e^4}+\frac {x \left (16+8 e^4 (1+\log (4))-e^8 \left (5-\log ^2(4)-\log (16)\right )\right )}{e^8}+\frac {3 \left (8+e^4 (3+\log (16))\right )}{e^4 (1+x)}\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.15, size = 99, normalized size = 4.71 \begin {gather*} \frac {16 x}{e^8}-\frac {9}{(1+x)^2}+\frac {24}{e^4 (1+x)}+\frac {8 (1+x)^2}{e^4}+(1+x)^3+\frac {8 (1+x) (-1+\log (4))}{e^4}+(1+x) \left (-6+14 \log (4)+\log ^2(4)-4 \log (256)\right )-\frac {3 (-3+2 \log (4)-\log (256))}{1+x}+\frac {1}{2} (1+x)^2 (-2+\log (256)) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.57, size = 127, normalized size = 6.05 \begin {gather*} \frac {{\left (4 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{8} \log \relax (2)^{2} + 16 \, x^{3} + 32 \, x^{2} + {\left (x^{5} + 4 \, x^{4} - 8 \, x^{2} + 4 \, x\right )} e^{8} + 8 \, {\left (x^{4} + 3 \, x^{3} + 3 \, x^{2} + 4 \, x + 3\right )} e^{4} + 4 \, {\left ({\left (x^{4} + 3 \, x^{3} + 3 \, x^{2} + 4 \, x + 3\right )} e^{8} + 4 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{4}\right )} \log \relax (2) + 16 \, x\right )} e^{\left (-8\right )}}{x^{2} + 2 \, x + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 104, normalized size = 4.95 \begin {gather*} {\left (x^{3} e^{8} + 4 \, x^{2} e^{8} \log \relax (2) + 4 \, x e^{8} \log \relax (2)^{2} + 2 \, x^{2} e^{8} + 8 \, x^{2} e^{4} + 4 \, x e^{8} \log \relax (2) + 16 \, x e^{4} \log \relax (2) - 5 \, x e^{8} + 8 \, x e^{4} + 16 \, x + \frac {3 \, {\left (4 \, x e^{8} \log \relax (2) + 3 \, x e^{8} + 8 \, x e^{4} + 4 \, e^{8} \log \relax (2) + 8 \, e^{4}\right )}}{{\left (x + 1\right )}^{2}}\right )} e^{\left (-8\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.14, size = 97, normalized size = 4.62
method | result | size |
risch | \(4 x \ln \relax (2)^{2}+4 x^{2} \ln \relax (2)+x^{3}+4 x \ln \relax (2)+2 x^{2}+16 \,{\mathrm e}^{-4} x \ln \relax (2)-5 x +8 \,{\mathrm e}^{-4} x^{2}+8 x \,{\mathrm e}^{-4}+16 x \,{\mathrm e}^{-8}+\frac {{\mathrm e}^{-8} \left (\left (9 \,{\mathrm e}^{8}+12 \,{\mathrm e}^{8} \ln \relax (2)+24 \,{\mathrm e}^{4}\right ) x +12 \,{\mathrm e}^{8} \ln \relax (2)+24 \,{\mathrm e}^{4}\right )}{x^{2}+2 x +1}\) | \(97\) |
default | \({\mathrm e}^{-8} \left (4 x \,{\mathrm e}^{8} \ln \relax (2)^{2}+4 x^{2} {\mathrm e}^{8} \ln \relax (2)+x^{3} {\mathrm e}^{8}+16 x \,{\mathrm e}^{4} \ln \relax (2)+4 \,{\mathrm e}^{8} \ln \relax (2) x +8 x^{2} {\mathrm e}^{4}+2 x^{2} {\mathrm e}^{8}+8 x \,{\mathrm e}^{4}-5 x \,{\mathrm e}^{8}+16 x -\frac {9 \,{\mathrm e}^{8}}{\left (x +1\right )^{2}}-\frac {-9 \,{\mathrm e}^{8}-12 \,{\mathrm e}^{8} \ln \relax (2)-24 \,{\mathrm e}^{4}}{x +1}\right )\) | \(103\) |
norman | \(\frac {\left (x^{5} {\mathrm e}^{4}-4 \left (3 \,{\mathrm e}^{8} \ln \relax (2)^{2}+2 \,{\mathrm e}^{8} \ln \relax (2)+12 \,{\mathrm e}^{4} \ln \relax (2)-5 \,{\mathrm e}^{8}+4 \,{\mathrm e}^{4}+12\right ) {\mathrm e}^{-4} x +\left (4 \,{\mathrm e}^{4} \ln \relax (2)+4 \,{\mathrm e}^{4}+8\right ) x^{4}+4 \left ({\mathrm e}^{8} \ln \relax (2)^{2}+3 \,{\mathrm e}^{8} \ln \relax (2)+4 \,{\mathrm e}^{4} \ln \relax (2)+6 \,{\mathrm e}^{4}+4\right ) {\mathrm e}^{-4} x^{3}-8 \left ({\mathrm e}^{8} \ln \relax (2)^{2}+4 \,{\mathrm e}^{4} \ln \relax (2)-{\mathrm e}^{8}+4\right ) {\mathrm e}^{-4}\right ) {\mathrm e}^{-4}}{\left (x +1\right )^{2}}\) | \(144\) |
gosper | \(\frac {\left (4 \ln \relax (2)^{2} {\mathrm e}^{8} x^{3}+4 \ln \relax (2) {\mathrm e}^{8} x^{4}+{\mathrm e}^{8} x^{5}+12 \ln \relax (2) {\mathrm e}^{8} x^{3}+4 x^{4} {\mathrm e}^{8}-12 x \,{\mathrm e}^{8} \ln \relax (2)^{2}+16 x^{3} {\mathrm e}^{4} \ln \relax (2)+8 x^{4} {\mathrm e}^{4}-8 \,{\mathrm e}^{8} \ln \relax (2)^{2}-8 \,{\mathrm e}^{8} \ln \relax (2) x +24 x^{3} {\mathrm e}^{4}-48 x \,{\mathrm e}^{4} \ln \relax (2)+20 x \,{\mathrm e}^{8}+16 x^{3}-32 \,{\mathrm e}^{4} \ln \relax (2)+8 \,{\mathrm e}^{8}-16 x \,{\mathrm e}^{4}-48 x -32\right ) {\mathrm e}^{-8}}{x^{2}+2 x +1}\) | \(162\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 99, normalized size = 4.71 \begin {gather*} {\left (x^{3} e^{8} + 2 \, {\left (2 \, e^{8} \log \relax (2) + e^{8} + 4 \, e^{4}\right )} x^{2} + {\left (4 \, e^{8} \log \relax (2)^{2} + 4 \, {\left (e^{8} + 4 \, e^{4}\right )} \log \relax (2) - 5 \, e^{8} + 8 \, e^{4} + 16\right )} x + \frac {3 \, {\left ({\left (4 \, e^{8} \log \relax (2) + 3 \, e^{8} + 8 \, e^{4}\right )} x + 4 \, e^{8} \log \relax (2) + 8 \, e^{4}\right )}}{x^{2} + 2 \, x + 1}\right )} e^{\left (-8\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.34, size = 124, normalized size = 5.90 \begin {gather*} \frac {x\,\left (9\,{\mathrm {e}}^4+12\,{\mathrm {e}}^4\,\ln \relax (2)+24\right )+12\,{\mathrm {e}}^4\,\ln \relax (2)+24}{{\mathrm {e}}^4\,x^2+2\,{\mathrm {e}}^4\,x+{\mathrm {e}}^4}+x^2\,\left (\frac {{\mathrm {e}}^{-8}\,\left (16\,{\mathrm {e}}^4+13\,{\mathrm {e}}^8+8\,{\mathrm {e}}^8\,\ln \relax (2)\right )}{2}-\frac {9}{2}\right )+x\,\left ({\mathrm {e}}^{-8}\,\left (56\,{\mathrm {e}}^4+16\,{\mathrm {e}}^8+4\,{\mathrm {e}}^8\,{\ln \relax (2)}^2+4\,{\mathrm {e}}^4\,\ln \relax (2)\,\left (7\,{\mathrm {e}}^4+4\right )+16\right )-3\,{\mathrm {e}}^{-8}\,\left (16\,{\mathrm {e}}^4+13\,{\mathrm {e}}^8+8\,{\mathrm {e}}^8\,\ln \relax (2)\right )+18\right )+x^3 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.68, size = 95, normalized size = 4.52 \begin {gather*} x^{3} + x^{2} \left (\frac {8}{e^{4}} + 2 + 4 \log {\relax (2 )}\right ) + x \left (-5 + \frac {16}{e^{8}} + \frac {8}{e^{4}} + \frac {16 \log {\relax (2 )}}{e^{4}} + 4 \log {\relax (2 )}^{2} + 4 \log {\relax (2 )}\right ) + \frac {x \left (24 + 12 e^{4} \log {\relax (2 )} + 9 e^{4}\right ) + 24 + 12 e^{4} \log {\relax (2 )}}{x^{2} e^{4} + 2 x e^{4} + e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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