Optimal. Leaf size=26 \[ \left (-1-x+\frac {e^{e^{2 x}}}{5+x}-\log ^4(3)\right )^2 \]
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Rubi [F] time = 2.30, 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 {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, 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 \left (e^{e^{2 x}}-2 e^{e^{2 x}+2 x} (5+x)+(5+x)^2\right ) \left (-e^{e^{2 x}}+(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^3} \, dx\\ &=2 \int \frac {\left (e^{e^{2 x}}-2 e^{e^{2 x}+2 x} (5+x)+(5+x)^2\right ) \left (-e^{e^{2 x}}+(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^3} \, dx\\ &=2 \int \left (\frac {2 e^{e^{2 x}+2 x} \left (e^{e^{2 x}}-x^2-6 x \left (1+\frac {\log ^4(3)}{6}\right )-5 \left (1+\log ^4(3)\right )\right )}{(5+x)^2}+\frac {\left (25+e^{e^{2 x}}+10 x+x^2\right ) \left (-e^{e^{2 x}}+x^2+6 x \left (1+\frac {\log ^4(3)}{6}\right )+5 \left (1+\log ^4(3)\right )\right )}{(5+x)^3}\right ) \, dx\\ &=2 \int \frac {\left (25+e^{e^{2 x}}+10 x+x^2\right ) \left (-e^{e^{2 x}}+x^2+6 x \left (1+\frac {\log ^4(3)}{6}\right )+5 \left (1+\log ^4(3)\right )\right )}{(5+x)^3} \, dx+4 \int \frac {e^{e^{2 x}+2 x} \left (e^{e^{2 x}}-x^2-6 x \left (1+\frac {\log ^4(3)}{6}\right )-5 \left (1+\log ^4(3)\right )\right )}{(5+x)^2} \, dx\\ &=2 \int \frac {\left (e^{e^{2 x}}+(5+x)^2\right ) \left (-e^{e^{2 x}}+(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^3} \, dx+4 \int \frac {e^{e^{2 x}+2 x} \left (e^{e^{2 x}}-(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^2} \, dx\\ &=2 \int \left (1+x-\frac {e^{2 e^{2 x}}}{(5+x)^3}+\log ^4(3)+\frac {e^{e^{2 x}} \left (-4+\log ^4(3)\right )}{(5+x)^2}\right ) \, dx+4 \int \left (\frac {e^{2 e^{2 x}+2 x}}{(5+x)^2}+\frac {e^{e^{2 x}+2 x} \left (-1-x-\log ^4(3)\right )}{5+x}\right ) \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx+4 \int \frac {e^{2 e^{2 x}+2 x}}{(5+x)^2} \, dx+4 \int \frac {e^{e^{2 x}+2 x} \left (-1-x-\log ^4(3)\right )}{5+x} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx+4 \int \left (-e^{e^{2 x}+2 x}+\frac {e^{e^{2 x}+2 x} \left (4-\log ^4(3)\right )}{5+x}\right ) \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx-4 \int e^{e^{2 x}+2 x} \, dx+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx+\left (4 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}+2 x}}{5+x} \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx-2 \operatorname {Subst}\left (\int e^x \, dx,x,e^{2 x}\right )+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx+\left (4 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}+2 x}}{5+x} \, dx\\ &=-2 e^{e^{2 x}}+x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx+\left (4 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}+2 x}}{5+x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.34, size = 44, normalized size = 1.69 \begin {gather*} \frac {\left (-e^{e^{2 x}}+x (5+x)\right ) \left (-e^{e^{2 x}}+(5+x) \left (2+x+2 \log ^4(3)\right )\right )}{(5+x)^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.69, size = 76, normalized size = 2.92 \begin {gather*} \frac {2 \, {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} \log \relax (3)^{4} + x^{4} + 12 \, x^{3} + 45 \, x^{2} - 2 \, {\left ({\left (x + 5\right )} \log \relax (3)^{4} + x^{2} + 6 \, x + 5\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 50 \, x + e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 108, normalized size = 4.15 \begin {gather*} \frac {2 \, x^{3} \log \relax (3)^{4} + 20 \, x^{2} \log \relax (3)^{4} - 2 \, x e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (3)^{4} + 50 \, x \log \relax (3)^{4} - 10 \, e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (3)^{4} + x^{4} + 12 \, x^{3} - 2 \, x^{2} e^{\left (e^{\left (2 \, x\right )}\right )} + 45 \, x^{2} - 12 \, x e^{\left (e^{\left (2 \, x\right )}\right )} + 50 \, x + e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - 10 \, e^{\left (e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.10, size = 52, normalized size = 2.00
method | result | size |
risch | \(2 x \ln \relax (3)^{4}+x^{2}+2 x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{2 x}}}{x^{2}+10 x +25}-\frac {2 \left (\ln \relax (3)^{4}+x +1\right ) {\mathrm e}^{{\mathrm e}^{2 x}}}{5+x}\) | \(52\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 234, normalized size = 9.00 \begin {gather*} {\left (2 \, x - \frac {25 \, {\left (6 \, x + 25\right )}}{x^{2} + 10 \, x + 25} - 30 \, \log \left (x + 5\right )\right )} \log \relax (3)^{4} + 15 \, {\left (\frac {5 \, {\left (4 \, x + 15\right )}}{x^{2} + 10 \, x + 25} + 2 \, \log \left (x + 5\right )\right )} \log \relax (3)^{4} - \frac {75 \, {\left (2 \, x + 5\right )} \log \relax (3)^{4}}{x^{2} + 10 \, x + 25} - \frac {125 \, \log \relax (3)^{4}}{x^{2} + 10 \, x + 25} + x^{2} + 2 \, x - \frac {2 \, {\left (5 \, \log \relax (3)^{4} + {\left (\log \relax (3)^{4} + 6\right )} x + x^{2} + 5\right )} e^{\left (e^{\left (2 \, x\right )}\right )} - e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} + \frac {125 \, {\left (8 \, x + 35\right )}}{x^{2} + 10 \, x + 25} - \frac {400 \, {\left (6 \, x + 25\right )}}{x^{2} + 10 \, x + 25} + \frac {450 \, {\left (4 \, x + 15\right )}}{x^{2} + 10 \, x + 25} - \frac {200 \, {\left (2 \, x + 5\right )}}{x^{2} + 10 \, x + 25} - \frac {125}{x^{2} + 10 \, x + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.70, size = 55, normalized size = 2.12 \begin {gather*} x\,\left (2\,{\ln \relax (3)}^4+2\right )+\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}}{x^2+10\,x+25}+x^2-\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\left (2\,x+2\,{\ln \relax (3)}^4+2\right )}{x+5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.29, size = 87, normalized size = 3.35 \begin {gather*} x^{2} + x \left (2 + 2 \log {\relax (3 )}^{4}\right ) + \frac {\left (x + 5\right ) e^{2 e^{2 x}} + \left (- 2 x^{3} - 22 x^{2} - 2 x^{2} \log {\relax (3 )}^{4} - 70 x - 20 x \log {\relax (3 )}^{4} - 50 \log {\relax (3 )}^{4} - 50\right ) e^{e^{2 x}}}{x^{3} + 15 x^{2} + 75 x + 125} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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