Optimal. Leaf size=34 \[ e^2 x \left (e^{\left (e^3-x\right )^2 x^2}+\frac {x^2 \log (x)}{5+4 x}\right ) \]
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Rubi [B] time = 1.36, antiderivative size = 105, normalized size of antiderivative = 3.09, number of steps used = 16, number of rules used = 11, integrand size = 138, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {27, 6688, 12, 6742, 2288, 43, 2357, 2295, 2304, 2314, 31} \begin {gather*} \frac {1}{4} e^2 x^2 \log (x)+\frac {e^{\left (e^3-x\right )^2 x^2+2} \left (2 x^4-3 e^3 x^3+e^6 x^2\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+\frac {25 e^2 x \log (x)}{16 (4 x+5)}-\frac {5}{16} e^2 x \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 31
Rule 43
Rule 2288
Rule 2295
Rule 2304
Rule 2314
Rule 2357
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{(5+4 x)^2} \, dx\\ &=\int \frac {e^2 \left (x^2 (5+4 x)+e^{\left (e^3-x\right )^2 x^2} (5+4 x)^2 \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right )+x^2 (15+8 x) \log (x)\right )}{(5+4 x)^2} \, dx\\ &=e^2 \int \frac {x^2 (5+4 x)+e^{\left (e^3-x\right )^2 x^2} (5+4 x)^2 \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right )+x^2 (15+8 x) \log (x)}{(5+4 x)^2} \, dx\\ &=e^2 \int \left (e^{\left (e^3-x\right )^2 x^2} \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right )+\frac {x^2 (5+4 x+15 \log (x)+8 x \log (x))}{(5+4 x)^2}\right ) \, dx\\ &=e^2 \int e^{\left (e^3-x\right )^2 x^2} \left (1+2 e^6 x^2-6 e^3 x^3+4 x^4\right ) \, dx+e^2 \int \frac {x^2 (5+4 x+15 \log (x)+8 x \log (x))}{(5+4 x)^2} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+e^2 \int \left (\frac {x^2}{5+4 x}+\frac {x^2 (15+8 x) \log (x)}{(5+4 x)^2}\right ) \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+e^2 \int \frac {x^2}{5+4 x} \, dx+e^2 \int \frac {x^2 (15+8 x) \log (x)}{(5+4 x)^2} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+e^2 \int \left (-\frac {5}{16}+\frac {x}{4}+\frac {25}{16 (5+4 x)}\right ) \, dx+e^2 \int \left (-\frac {5 \log (x)}{16}+\frac {1}{2} x \log (x)+\frac {125 \log (x)}{16 (5+4 x)^2}\right ) \, dx\\ &=-\frac {5 e^2 x}{16}+\frac {e^2 x^2}{8}+\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+\frac {25}{64} e^2 \log (5+4 x)-\frac {1}{16} \left (5 e^2\right ) \int \log (x) \, dx+\frac {1}{2} e^2 \int x \log (x) \, dx+\frac {1}{16} \left (125 e^2\right ) \int \frac {\log (x)}{(5+4 x)^2} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}-\frac {5}{16} e^2 x \log (x)+\frac {1}{4} e^2 x^2 \log (x)+\frac {25 e^2 x \log (x)}{16 (5+4 x)}+\frac {25}{64} e^2 \log (5+4 x)-\frac {1}{16} \left (25 e^2\right ) \int \frac {1}{5+4 x} \, dx\\ &=\frac {e^{2+\left (e^3-x\right )^2 x^2} \left (e^6 x^2-3 e^3 x^3+2 x^4\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}-\frac {5}{16} e^2 x \log (x)+\frac {1}{4} e^2 x^2 \log (x)+\frac {25 e^2 x \log (x)}{16 (5+4 x)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.23, size = 35, normalized size = 1.03 \begin {gather*} e^2 \left (e^{\left (e^3-x\right )^2 x^2} x+\frac {x^3 \log (x)}{5+4 x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 46, normalized size = 1.35 \begin {gather*} \frac {x^{3} e^{2} \log \relax (x) + {\left (4 \, x^{2} + 5 \, x\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6} + 2\right )}}{4 \, x + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (8 \, x^{3} + 15 \, x^{2}\right )} e^{2} \log \relax (x) + {\left (4 \, x^{3} + 5 \, x^{2}\right )} e^{2} + {\left (2 \, {\left (16 \, x^{4} + 40 \, x^{3} + 25 \, x^{2}\right )} e^{8} - 6 \, {\left (16 \, x^{5} + 40 \, x^{4} + 25 \, x^{3}\right )} e^{5} + {\left (64 \, x^{6} + 160 \, x^{5} + 100 \, x^{4} + 16 \, x^{2} + 40 \, x + 25\right )} e^{2}\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6}\right )}}{16 \, x^{2} + 40 \, x + 25}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 52, normalized size = 1.53
method | result | size |
risch | \(\frac {{\mathrm e}^{2} \left (64 x^{3}-100 x -125\right ) \ln \relax (x )}{256 x +320}+\frac {25 \,{\mathrm e}^{2} \ln \relax (x )}{64}+x \,{\mathrm e}^{-2 x^{3} {\mathrm e}^{3}+x^{4}+x^{2} {\mathrm e}^{6}+2}\) | \(52\) |
default | \({\mathrm e}^{2} x \,{\mathrm e}^{x^{2} {\mathrm e}^{6}-2 x^{3} {\mathrm e}^{3}+x^{4}}+\frac {x^{2} {\mathrm e}^{2} \ln \relax (x )}{4}-\frac {5 x \,{\mathrm e}^{2} \ln \relax (x )}{16}+\frac {25 \,{\mathrm e}^{2} \ln \relax (x ) x}{16 \left (4 x +5\right )}\) | \(56\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.60, size = 155, normalized size = 4.56 \begin {gather*} \frac {1}{64} \, {\left (8 \, x^{2} - 40 \, x + \frac {125}{4 \, x + 5} + 75 \, \log \left (4 \, x + 5\right )\right )} e^{2} + \frac {5}{64} \, {\left (4 \, x - \frac {25}{4 \, x + 5} - 10 \, \log \left (4 \, x + 5\right )\right )} e^{2} - \frac {25}{64} \, e^{2} \log \left (4 \, x + 5\right ) + \frac {25}{64} \, e^{2} \log \relax (x) - \frac {32 \, x^{3} e^{2} - 40 \, x^{2} e^{2} - 100 \, x e^{2} - 64 \, {\left (4 \, x^{2} e^{2} + 5 \, x e^{2}\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6}\right )} - {\left (64 \, x^{3} e^{2} - 100 \, x e^{2} - 125 \, e^{2}\right )} \log \relax (x)}{64 \, {\left (4 \, x + 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.37, size = 91, normalized size = 2.68 \begin {gather*} x\,{\mathrm {e}}^{x^4-2\,{\mathrm {e}}^3\,x^3+{\mathrm {e}}^6\,x^2+2}-\frac {25\,\ln \left (x+\frac {5}{4}\right )\,{\mathrm {e}}^2}{64}+\frac {{\mathrm {e}}^2\,\left (25\,\ln \left (x+\frac {5}{4}\right )-20\,x+8\,x^2\right )}{64}+\frac {\frac {25\,x^2\,{\mathrm {e}}^2}{16}+\frac {5\,x^3\,{\mathrm {e}}^2}{8}-\frac {x^4\,{\mathrm {e}}^2}{2}+x^4\,{\mathrm {e}}^2\,\ln \relax (x)}{4\,x^2+5\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.56, size = 63, normalized size = 1.85 \begin {gather*} x e^{2} e^{x^{4} - 2 x^{3} e^{3} + x^{2} e^{6}} + \frac {25 e^{2} \log {\relax (x )}}{64} + \frac {\left (64 x^{3} e^{2} - 100 x e^{2} - 125 e^{2}\right ) \log {\relax (x )}}{256 x + 320} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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