Optimal. Leaf size=28 \[ \frac {4 \log \left (\frac {1}{x^3}\right )}{\left (e^{x^2}+x\right ) \left (-x+\frac {4+x}{2}\right )} \]
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Rubi [F] time = 7.25, 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 {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 \left (3 (-4+x) \left (e^{x^2}+x\right )+x \left (2 (-2+x)+e^{x^2} \left (1-8 x+2 x^2\right )\right ) \log \left (\frac {1}{x^3}\right )\right )}{(4-x)^2 x \left (e^{x^2}+x\right )^2} \, dx\\ &=8 \int \frac {3 (-4+x) \left (e^{x^2}+x\right )+x \left (2 (-2+x)+e^{x^2} \left (1-8 x+2 x^2\right )\right ) \log \left (\frac {1}{x^3}\right )}{(4-x)^2 x \left (e^{x^2}+x\right )^2} \, dx\\ &=8 \int \left (-\frac {\left (-1+2 x^2\right ) \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )^2}+\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 x \left (e^{x^2}+x\right )}\right ) \, dx\\ &=-\left (8 \int \frac {\left (-1+2 x^2\right ) \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\right )+8 \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 x \left (e^{x^2}+x\right )} \, dx\\ &=8 \int \left (\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{4 (-4+x)^2 \left (e^{x^2}+x\right )}-\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{16 (-4+x) \left (e^{x^2}+x\right )}+\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{16 x \left (e^{x^2}+x\right )}\right ) \, dx+8 \int -\frac {3 \left (8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx+2 \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx\right )}{x} \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\\ &=-\left (\frac {1}{2} \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx\right )+\frac {1}{2} \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{x \left (e^{x^2}+x\right )} \, dx+2 \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-24 \int \frac {8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx+2 \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx}{x} \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {3}{e^{x^2}+x}-\frac {12}{x \left (e^{x^2}+x\right )}+\frac {\log \left (\frac {1}{x^3}\right )}{e^{x^2}+x}-\frac {8 x \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x}+\frac {2 x^2 \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x}\right ) \, dx-\frac {1}{2} \int \left (-\frac {12}{(-4+x) \left (e^{x^2}+x\right )}+\frac {3 x}{(-4+x) \left (e^{x^2}+x\right )}+\frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )}-\frac {8 x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )}+\frac {2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )}\right ) \, dx+2 \int \left (-\frac {12}{(-4+x)^2 \left (e^{x^2}+x\right )}+\frac {3 x}{(-4+x)^2 \left (e^{x^2}+x\right )}+\frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )}-\frac {8 x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )}+\frac {2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )}\right ) \, dx-24 \int \left (\frac {8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx}{x}+\frac {2 \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx}{x}\right ) \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {\log \left (\frac {1}{x^3}\right )}{e^{x^2}+x} \, dx-\frac {1}{2} \int \frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx+\frac {3}{2} \int \frac {1}{e^{x^2}+x} \, dx-\frac {3}{2} \int \frac {x}{(-4+x) \left (e^{x^2}+x\right )} \, dx+2 \int \frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-4 \int \frac {x \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x} \, dx+4 \int \frac {x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx+4 \int \frac {x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx+6 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )} \, dx-6 \int \frac {1}{x \left (e^{x^2}+x\right )} \, dx+6 \int \frac {x}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-16 \int \frac {x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-24 \int \frac {1}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-24 \int \frac {8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx}{x} \, dx-48 \int \frac {\int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx}{x} \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx+\int \frac {x^2 \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x} \, dx-\int \frac {x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.87, size = 20, normalized size = 0.71 \begin {gather*} -\frac {8 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.80, size = 23, normalized size = 0.82 \begin {gather*} -\frac {8 \, \log \left (\frac {1}{x^{3}}\right )}{x^{2} + {\left (x - 4\right )} e^{\left (x^{2}\right )} - 4 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.37, size = 25, normalized size = 0.89 \begin {gather*} \frac {24 \, \log \relax (x)}{x^{2} + x e^{\left (x^{2}\right )} - 4 \, x - 4 \, e^{\left (x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.16, size = 140, normalized size = 5.00
method | result | size |
risch | \(\frac {24 \ln \relax (x )}{\left (x -4\right ) \left ({\mathrm e}^{x^{2}}+x \right )}-\frac {4 i \pi \left (\mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x^{2}\right )^{3}-\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x^{3}\right )^{3}\right )}{\left (x -4\right ) \left ({\mathrm e}^{x^{2}}+x \right )}\) | \(140\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.00, size = 21, normalized size = 0.75 \begin {gather*} \frac {24 \, \log \relax (x)}{x^{2} + {\left (x - 4\right )} e^{\left (x^{2}\right )} - 4 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.10, size = 19, normalized size = 0.68 \begin {gather*} -\frac {8\,\ln \left (\frac {1}{x^3}\right )}{\left (x+{\mathrm {e}}^{x^2}\right )\,\left (x-4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.30, size = 24, normalized size = 0.86 \begin {gather*} - \frac {8 \log {\left (\frac {1}{x^{3}} \right )}}{x^{2} - 4 x + \left (x - 4\right ) e^{x^{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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