Optimal. Leaf size=33 \[ x+x^2-\frac {9 x^2 \left (-e^4+x\right )^2}{25 \left (x+\frac {1}{x \log (4)}\right )^2} \]
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Rubi [B] time = 0.21, antiderivative size = 108, normalized size of antiderivative = 3.27, number of steps used = 8, number of rules used = 4, integrand size = 105, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2073, 203, 639, 199} \begin {gather*} \frac {16 x^2}{25}+\frac {27 e^4 x}{25 \left (x^2 \log (4)+1\right )}-\frac {9 \left (7 e^4 x \log (4)+3-e^8 \log (16)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )}+\frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}+\frac {1}{25} \left (25+18 e^4\right ) x \end {gather*}
Antiderivative was successfully verified.
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Rule 199
Rule 203
Rule 639
Rule 2073
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{25} \left (25+18 e^4\right )+\frac {32 x}{25}+\frac {36 e^4}{25 \left (1+x^2 \log (4)\right )}+\frac {36 \left (2 e^4-x \left (1-e^8 \log (4)\right )\right )}{25 \left (1+x^2 \log (4)\right )^3}+\frac {18 \left (-7 e^4+x \left (3-e^8 \log (16)\right )\right )}{25 \left (1+x^2 \log (4)\right )^2}\right ) \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}+\frac {18}{25} \int \frac {-7 e^4+x \left (3-e^8 \log (16)\right )}{\left (1+x^2 \log (4)\right )^2} \, dx+\frac {36}{25} \int \frac {2 e^4-x \left (1-e^8 \log (4)\right )}{\left (1+x^2 \log (4)\right )^3} \, dx+\frac {1}{25} \left (36 e^4\right ) \int \frac {1}{1+x^2 \log (4)} \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}+\frac {36 e^4 \tan ^{-1}\left (x \sqrt {\log (4)}\right )}{25 \sqrt {\log (4)}}+\frac {9 \left (1-e^8 \log (4)+2 e^4 x \log (4)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )^2}-\frac {9 \left (3+7 e^4 x \log (4)-e^8 \log (16)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )}+\frac {1}{25} \left (54 e^4\right ) \int \frac {1}{\left (1+x^2 \log (4)\right )^2} \, dx-\frac {1}{25} \left (63 e^4\right ) \int \frac {1}{1+x^2 \log (4)} \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}-\frac {27 e^4 \tan ^{-1}\left (x \sqrt {\log (4)}\right )}{25 \sqrt {\log (4)}}+\frac {9 \left (1-e^8 \log (4)+2 e^4 x \log (4)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )^2}+\frac {27 e^4 x}{25 \left (1+x^2 \log (4)\right )}-\frac {9 \left (3+7 e^4 x \log (4)-e^8 \log (16)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )}+\frac {1}{25} \left (27 e^4\right ) \int \frac {1}{1+x^2 \log (4)} \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}+\frac {9 \left (1-e^8 \log (4)+2 e^4 x \log (4)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )^2}+\frac {27 e^4 x}{25 \left (1+x^2 \log (4)\right )}-\frac {9 \left (3+7 e^4 x \log (4)-e^8 \log (16)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )}\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.07, size = 84, normalized size = 2.55 \begin {gather*} \frac {1}{25} \left (\left (25+18 e^4\right ) x+16 x^2-\frac {9 \left (-1+e^8 \log (4)-2 e^4 x \log (4)\right )}{\log (4) \left (1+x^2 \log (4)\right )^2}+\frac {9 \left (-3+2 e^8 \log (4)-4 e^4 x \log (4)\right )}{\log (4) \left (1+x^2 \log (4)\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.91, size = 92, normalized size = 2.79 \begin {gather*} \frac {4 \, {\left (16 \, x^{6} + 18 \, x^{5} e^{4} + 25 \, x^{5}\right )} \log \relax (2)^{3} + 4 \, {\left (16 \, x^{4} + 25 \, x^{3} + 9 \, x^{2} e^{8}\right )} \log \relax (2)^{2} - {\left (11 \, x^{2} - 25 \, x - 9 \, e^{8}\right )} \log \relax (2) - 9}{25 \, {\left (4 \, x^{4} \log \relax (2)^{3} + 4 \, x^{2} \log \relax (2)^{2} + \log \relax (2)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.21, size = 94, normalized size = 2.85 \begin {gather*} -\frac {9 \, {\left (8 \, x^{3} e^{4} \log \relax (2)^{2} - 4 \, x^{2} e^{8} \log \relax (2)^{2} + 3 \, x^{2} \log \relax (2) + 2 \, x e^{4} \log \relax (2) - e^{8} \log \relax (2) + 1\right )}}{25 \, {\left (2 \, x^{2} \log \relax (2) + 1\right )}^{2} \log \relax (2)} + \frac {16 \, x^{2} \log \relax (2)^{6} + 18 \, x e^{4} \log \relax (2)^{6} + 25 \, x \log \relax (2)^{6}}{25 \, \log \relax (2)^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.14, size = 66, normalized size = 2.00
method | result | size |
default | \(\frac {16 x^{2}}{25}+\frac {18 x \,{\mathrm e}^{4}}{25}+x +\frac {-\frac {72 x^{3} {\mathrm e}^{4} \ln \relax (2)}{25}+\frac {72 \left (\frac {{\mathrm e}^{8} \ln \relax (2)}{2}-\frac {3}{8}\right ) x^{2}}{25}-\frac {18 x \,{\mathrm e}^{4}}{25}+\frac {9 \left ({\mathrm e}^{8} \ln \relax (2)-1\right )}{25 \ln \relax (2)}}{\left (2 x^{2} \ln \relax (2)+1\right )^{2}}\) | \(66\) |
norman | \(\frac {x +x^{2}+\left (4 \ln \relax (2)^{2}+\frac {72 \,{\mathrm e}^{4} \ln \relax (2)^{2}}{25}\right ) x^{5}+\left (-\frac {36 \,{\mathrm e}^{8} \ln \relax (2)^{2}}{25}+4 \ln \relax (2)\right ) x^{4}+4 x^{3} \ln \relax (2)+\frac {64 x^{6} \ln \relax (2)^{2}}{25}}{\left (2 x^{2} \ln \relax (2)+1\right )^{2}}\) | \(72\) |
risch | \(\frac {18 x \,{\mathrm e}^{4}}{25}+\frac {16 x^{2}}{25}+x +\frac {-\frac {18 x^{3} {\mathrm e}^{4} \ln \relax (2)}{25}+\frac {\left (9 \,{\mathrm e}^{8} \ln \relax (2)-\frac {27}{4}\right ) x^{2}}{25}-\frac {9 x \,{\mathrm e}^{4}}{50}+\frac {\frac {9 \,{\mathrm e}^{8} \ln \relax (2)}{100}-\frac {9}{100}}{\ln \relax (2)}}{x^{4} \ln \relax (2)^{2}+x^{2} \ln \relax (2)+\frac {1}{4}}\) | \(73\) |
gosper | \(-\frac {x \left (36 \ln \relax (2)^{2} {\mathrm e}^{8} x^{3}-72 \ln \relax (2)^{2} {\mathrm e}^{4} x^{4}-64 x^{5} \ln \relax (2)^{2}-100 x^{4} \ln \relax (2)^{2}-100 x^{3} \ln \relax (2)-100 x^{2} \ln \relax (2)-25 x -25\right )}{25 \left (4 x^{4} \ln \relax (2)^{2}+4 x^{2} \ln \relax (2)+1\right )}\) | \(85\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.40, size = 84, normalized size = 2.55 \begin {gather*} \frac {16}{25} \, x^{2} + \frac {1}{25} \, x {\left (18 \, e^{4} + 25\right )} - \frac {9 \, {\left (8 \, x^{3} e^{4} \log \relax (2)^{2} - {\left (4 \, e^{8} \log \relax (2)^{2} - 3 \, \log \relax (2)\right )} x^{2} + 2 \, x e^{4} \log \relax (2) - e^{8} \log \relax (2) + 1\right )}}{25 \, {\left (4 \, x^{4} \log \relax (2)^{3} + 4 \, x^{2} \log \relax (2)^{2} + \log \relax (2)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.01, size = 77, normalized size = 2.33 \begin {gather*} \frac {16\,x^2}{25}-\frac {72\,{\mathrm {e}}^4\,\ln \relax (2)\,x^3+\left (27-36\,{\mathrm {e}}^8\,\ln \relax (2)\right )\,x^2+18\,{\mathrm {e}}^4\,x-\frac {9\,\left ({\mathrm {e}}^8\,\ln \relax (2)-1\right )}{\ln \relax (2)}}{100\,{\ln \relax (2)}^2\,x^4+100\,\ln \relax (2)\,x^2+25}+x\,\left (\frac {18\,{\mathrm {e}}^4}{25}+1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 1.55, size = 94, normalized size = 2.85 \begin {gather*} \frac {16 x^{2}}{25} + x \left (1 + \frac {18 e^{4}}{25}\right ) + \frac {- 72 x^{3} e^{4} \log {\relax (2 )}^{2} + x^{2} \left (- 27 \log {\relax (2 )} + 36 e^{8} \log {\relax (2 )}^{2}\right ) - 18 x e^{4} \log {\relax (2 )} - 9 + 9 e^{8} \log {\relax (2 )}}{100 x^{4} \log {\relax (2 )}^{3} + 100 x^{2} \log {\relax (2 )}^{2} + 25 \log {\relax (2 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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