Optimal. Leaf size=28 \[ \frac {2 x^3 \log (x)}{\left (4+\frac {e^{5-x}}{4 x}-x\right )^2} \]
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Rubi [F] time = 20.75, 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 {512 x^2-128 x^3+\left (1536 x^2-128 x^3\right ) \log (x)+\frac {e^{5-x} \left (32 x^2+\left (160 x^2+64 x^3\right ) \log (x)\right )}{x}}{4096+\frac {e^{15-3 x}}{x^3}+\frac {e^{10-2 x} (48-12 x)}{x^2}-3072 x+768 x^2-64 x^3+\frac {e^{5-x} \left (768-384 x+48 x^2\right )}{x}} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 e^{2 x} x^4 \left (e^5-4 e^x (-4+x) x-4 e^x (-12+x) x \log (x)+e^5 (5+2 x) \log (x)\right )}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx\\ &=32 \int \frac {e^{2 x} x^4 \left (e^5-4 e^x (-4+x) x-4 e^x (-12+x) x \log (x)+e^5 (5+2 x) \log (x)\right )}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx\\ &=32 \int \left (\frac {2 e^{5+2 x} x^4 \left (-4-2 x+x^2\right ) \log (x)}{(-4+x) \left (e^5+16 e^x x-4 e^x x^2\right )^3}+\frac {e^{2 x} x^4 (-4+x-12 \log (x)+x \log (x))}{(-4+x) \left (-e^5-16 e^x x+4 e^x x^2\right )^2}\right ) \, dx\\ &=32 \int \frac {e^{2 x} x^4 (-4+x-12 \log (x)+x \log (x))}{(-4+x) \left (-e^5-16 e^x x+4 e^x x^2\right )^2} \, dx+64 \int \frac {e^{5+2 x} x^4 \left (-4-2 x+x^2\right ) \log (x)}{(-4+x) \left (e^5+16 e^x x-4 e^x x^2\right )^3} \, dx\\ &=32 \int \left (\frac {64 e^{2 x} (-4+x-12 \log (x)+x \log (x))}{\left (e^5+16 e^x x-4 e^x x^2\right )^2}+\frac {256 e^{2 x} (-4+x-12 \log (x)+x \log (x))}{(-4+x) \left (-e^5-16 e^x x+4 e^x x^2\right )^2}+\frac {16 e^{2 x} x (-4+x-12 \log (x)+x \log (x))}{\left (-e^5-16 e^x x+4 e^x x^2\right )^2}+\frac {4 e^{2 x} x^2 (-4+x-12 \log (x)+x \log (x))}{\left (-e^5-16 e^x x+4 e^x x^2\right )^2}+\frac {e^{2 x} x^3 (-4+x-12 \log (x)+x \log (x))}{\left (-e^5-16 e^x x+4 e^x x^2\right )^2}\right ) \, dx-64 \int \frac {256 \int \frac {e^{5+2 x}}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx-64 \int -\frac {e^{5+2 x} x}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx-16 \int -\frac {e^{5+2 x} x^2}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx-4 \int -\frac {e^{5+2 x} x^3}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx-2 \int -\frac {e^{5+2 x} x^4}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx-\int -\frac {e^{5+2 x} x^5}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx-1024 \int \frac {e^{5+2 x}}{(-4+x) \left (-e^5+4 e^x (-4+x) x\right )^3} \, dx}{x} \, dx-(64 \log (x)) \int \frac {e^{5+2 x} x^5}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(128 \log (x)) \int \frac {e^{5+2 x} x^4}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(256 \log (x)) \int \frac {e^{5+2 x} x^3}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(1024 \log (x)) \int \frac {e^{5+2 x} x^2}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(4096 \log (x)) \int \frac {e^{5+2 x} x}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx+(16384 \log (x)) \int \frac {e^{5+2 x}}{\left (e^5+16 e^x x-4 e^x x^2\right )^3} \, dx-(65536 \log (x)) \int \frac {e^{5+2 x}}{(-4+x) \left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx\\ &=32 \int \frac {e^{2 x} x^3 (-4+x-12 \log (x)+x \log (x))}{\left (-e^5-16 e^x x+4 e^x x^2\right )^2} \, dx-64 \int \left (\frac {256 \int \frac {e^{5+2 x}}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx+64 \int \frac {e^{5+2 x} x}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx+16 \int \frac {e^{5+2 x} x^2}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx+4 \int \frac {e^{5+2 x} x^3}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx+2 \int \frac {e^{5+2 x} x^4}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx+\int \frac {e^{5+2 x} x^5}{\left (e^5-4 e^x (-4+x) x\right )^3} \, dx}{x}-\frac {1024 \int \frac {e^{5+2 x}}{(-4+x) \left (-e^5+4 e^x (-4+x) x\right )^3} \, dx}{x}\right ) \, dx+128 \int \frac {e^{2 x} x^2 (-4+x-12 \log (x)+x \log (x))}{\left (-e^5-16 e^x x+4 e^x x^2\right )^2} \, dx+512 \int \frac {e^{2 x} x (-4+x-12 \log (x)+x \log (x))}{\left (-e^5-16 e^x x+4 e^x x^2\right )^2} \, dx+2048 \int \frac {e^{2 x} (-4+x-12 \log (x)+x \log (x))}{\left (e^5+16 e^x x-4 e^x x^2\right )^2} \, dx+8192 \int \frac {e^{2 x} (-4+x-12 \log (x)+x \log (x))}{(-4+x) \left (-e^5-16 e^x x+4 e^x x^2\right )^2} \, dx-(64 \log (x)) \int \frac {e^{5+2 x} x^5}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(128 \log (x)) \int \frac {e^{5+2 x} x^4}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(256 \log (x)) \int \frac {e^{5+2 x} x^3}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(1024 \log (x)) \int \frac {e^{5+2 x} x^2}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx-(4096 \log (x)) \int \frac {e^{5+2 x} x}{\left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx+(16384 \log (x)) \int \frac {e^{5+2 x}}{\left (e^5+16 e^x x-4 e^x x^2\right )^3} \, dx-(65536 \log (x)) \int \frac {e^{5+2 x}}{(-4+x) \left (-e^5-16 e^x x+4 e^x x^2\right )^3} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 4.65, size = 27, normalized size = 0.96 \begin {gather*} \frac {32 e^{2 x} x^5 \log (x)}{\left (e^5-4 e^x (-4+x) x\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 44, normalized size = 1.57 \begin {gather*} \frac {32 \, x^{3} \log \relax (x)}{16 \, x^{2} - 8 \, {\left (x - 4\right )} e^{\left (-x - \log \relax (x) + 5\right )} - 128 \, x + e^{\left (-2 \, x - 2 \, \log \relax (x) + 10\right )} + 256} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.31, size = 1012, normalized size = 36.14 result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 26, normalized size = 0.93
method | result | size |
risch | \(\frac {32 x^{3} \ln \relax (x )}{\left (4 x -\frac {{\mathrm e}^{5-x}}{x}-16\right )^{2}}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 52, normalized size = 1.86 \begin {gather*} \frac {32 \, x^{5} e^{\left (2 \, x\right )} \log \relax (x)}{16 \, {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} e^{\left (2 \, x\right )} - 8 \, {\left (x^{2} e^{5} - 4 \, x e^{5}\right )} e^{x} + e^{10}} \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 = -0.04 \begin {gather*} \int \frac {\ln \relax (x)\,\left (1536\,x^2-128\,x^3\right )+512\,x^2-128\,x^3+{\mathrm {e}}^{5-\ln \relax (x)-x}\,\left (\ln \relax (x)\,\left (64\,x^3+160\,x^2\right )+32\,x^2\right )}{{\mathrm {e}}^{15-3\,\ln \relax (x)-3\,x}-3072\,x-{\mathrm {e}}^{10-2\,\ln \relax (x)-2\,x}\,\left (12\,x-48\right )+{\mathrm {e}}^{5-\ln \relax (x)-x}\,\left (48\,x^2-384\,x+768\right )+768\,x^2-64\,x^3+4096} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.40, size = 42, normalized size = 1.50 \begin {gather*} \frac {32 x^{5} \log {\relax (x )}}{16 x^{4} - 128 x^{3} + 256 x^{2} + \left (- 8 x^{2} + 32 x\right ) e^{5 - x} + e^{10 - 2 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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