Optimal. Leaf size=35 \[ x^2+\frac {1}{25} \left (-x+\frac {e^{(4-x)^2}}{4+x}+\frac {\log (3)}{x}\right )^2 \]
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Rubi [F] time = 25.55, 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 {3328 x^4+2496 x^5+624 x^6+52 x^7+e^{32-16 x+2 x^2} \left (-66 x^3+4 x^5\right )+\left (-128-96 x-24 x^2-2 x^3\right ) \log ^2(3)+e^{16-8 x+x^2} \left (-32 x^3+248 x^4+64 x^5-16 x^6-4 x^7+\left (-32 x-280 x^2-68 x^3+16 x^4+4 x^5\right ) \log (3)\right )}{1600 x^3+1200 x^4+300 x^5+25 x^6} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-16 x} \left (2 e^{32+2 x^2} x^3 \left (-33+2 x^2\right )+2 e^{16 x} (4+x)^3 \left (26 x^4-\log ^2(3)\right )-4 e^{(4+x)^2} x (4+x) \left (2 x^2+x^5+17 x \log (3)-x^3 (16+\log (3))+\log (9)\right )\right )}{25 x^3 (4+x)^3} \, dx\\ &=\frac {1}{25} \int \frac {e^{-16 x} \left (2 e^{32+2 x^2} x^3 \left (-33+2 x^2\right )+2 e^{16 x} (4+x)^3 \left (26 x^4-\log ^2(3)\right )-4 e^{(4+x)^2} x (4+x) \left (2 x^2+x^5+17 x \log (3)-x^3 (16+\log (3))+\log (9)\right )\right )}{x^3 (4+x)^3} \, dx\\ &=\frac {1}{25} \int \left (\frac {2 e^{-16 x} \left (-33 e^{32+2 x^2} x^3+1664 e^{16 x} x^4+1248 e^{16 x} x^5+2 e^{32+2 x^2} x^5+312 e^{16 x} x^6+26 e^{16 x} x^7-64 e^{16 x} \log ^2(3)-48 e^{16 x} x \log ^2(3)-12 e^{16 x} x^2 \log ^2(3)-e^{16 x} x^3 \log ^2(3)\right )}{x^3 (4+x)^3}+\frac {4 e^{16-8 x+x^2} \left (-2 x^2-x^5+16 x^3 \left (1+\frac {\log (3)}{16}\right )-17 x \log (3)-\log (9)\right )}{x^2 (4+x)^2}\right ) \, dx\\ &=\frac {2}{25} \int \frac {e^{-16 x} \left (-33 e^{32+2 x^2} x^3+1664 e^{16 x} x^4+1248 e^{16 x} x^5+2 e^{32+2 x^2} x^5+312 e^{16 x} x^6+26 e^{16 x} x^7-64 e^{16 x} \log ^2(3)-48 e^{16 x} x \log ^2(3)-12 e^{16 x} x^2 \log ^2(3)-e^{16 x} x^3 \log ^2(3)\right )}{x^3 (4+x)^3} \, dx+\frac {4}{25} \int \frac {e^{16-8 x+x^2} \left (-2 x^2-x^5+16 x^3 \left (1+\frac {\log (3)}{16}\right )-17 x \log (3)-\log (9)\right )}{x^2 (4+x)^2} \, dx\\ &=\frac {2}{25} \int \left (26 x+\frac {e^{2 (-4+x)^2} \left (-33+2 x^2\right )}{(4+x)^3}-\frac {\log ^2(3)}{x^3}\right ) \, dx+\frac {4}{25} \int \left (8 e^{16-8 x+x^2}-e^{16-8 x+x^2} x+\frac {e^{16-8 x+x^2} (-1024+66 \log (3)-\log (9))}{32 (4+x)}+\frac {e^{16-8 x+x^2} (-32+\log (9))}{16 (4+x)^2}-\frac {e^{16-8 x+x^2} \log (9)}{16 x^2}+\frac {e^{16-8 x+x^2} (-34 \log (3)+\log (9))}{32 x}\right ) \, dx\\ &=\frac {26 x^2}{25}+\frac {\log ^2(3)}{25 x^2}+\frac {2}{25} \int \frac {e^{2 (-4+x)^2} \left (-33+2 x^2\right )}{(4+x)^3} \, dx-\frac {4}{25} \int e^{16-8 x+x^2} x \, dx+\frac {32}{25} \int e^{16-8 x+x^2} \, dx-\frac {1}{25} (4 \log (3)) \int \frac {e^{16-8 x+x^2}}{x} \, dx+\frac {1}{200} (-1024+66 \log (3)-\log (9)) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{100} (-32+\log (9)) \int \frac {e^{16-8 x+x^2}}{(4+x)^2} \, dx-\frac {1}{100} \log (9) \int \frac {e^{16-8 x+x^2}}{x^2} \, dx\\ &=\frac {26 x^2}{25}+\frac {e^{2 (4-x)^2}}{25 (4+x)^2}+\frac {\log ^2(3)}{25 x^2}+\frac {e^{16-8 x+x^2} (32-\log (9))}{100 (4+x)}+\frac {e^{16-8 x+x^2} \log (9)}{100 x}-\frac {4}{25} \int e^{(-4+x)^2} x \, dx+\frac {32}{25} \int e^{(-4+x)^2} \, dx-\frac {1}{25} (4 \log (3)) \int \frac {e^{16-8 x+x^2}}{x} \, dx+\frac {1}{25} (4 (32-\log (9))) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{200} (-1024+66 \log (3)-\log (9)) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{50} (-32+\log (9)) \int e^{16-8 x+x^2} \, dx-\frac {1}{50} \log (9) \int e^{16-8 x+x^2} \, dx+\frac {1}{25} (2 \log (9)) \int \frac {e^{16-8 x+x^2}}{x} \, dx\\ &=\frac {26 x^2}{25}+\frac {e^{2 (4-x)^2}}{25 (4+x)^2}-\frac {16}{25} \sqrt {\pi } \text {erfi}(4-x)+\frac {\log ^2(3)}{25 x^2}+\frac {e^{16-8 x+x^2} (32-\log (9))}{100 (4+x)}+\frac {e^{16-8 x+x^2} \log (9)}{100 x}-\frac {4}{25} \int \left (4 e^{(-4+x)^2}+e^{(-4+x)^2} (-4+x)\right ) \, dx-\frac {1}{25} (4 \log (3)) \int \frac {e^{16-8 x+x^2}}{x} \, dx+\frac {1}{25} (4 (32-\log (9))) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{200} (-1024+66 \log (3)-\log (9)) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{50} (-32+\log (9)) \int e^{(-4+x)^2} \, dx-\frac {1}{50} \log (9) \int e^{(-4+x)^2} \, dx+\frac {1}{25} (2 \log (9)) \int \frac {e^{16-8 x+x^2}}{x} \, dx\\ &=\frac {26 x^2}{25}+\frac {e^{2 (4-x)^2}}{25 (4+x)^2}-\frac {16}{25} \sqrt {\pi } \text {erfi}(4-x)+\frac {\log ^2(3)}{25 x^2}+\frac {e^{16-8 x+x^2} (32-\log (9))}{100 (4+x)}+\frac {1}{100} \sqrt {\pi } \text {erfi}(4-x) (32-\log (9))+\frac {e^{16-8 x+x^2} \log (9)}{100 x}+\frac {1}{100} \sqrt {\pi } \text {erfi}(4-x) \log (9)-\frac {4}{25} \int e^{(-4+x)^2} (-4+x) \, dx-\frac {16}{25} \int e^{(-4+x)^2} \, dx-\frac {1}{25} (4 \log (3)) \int \frac {e^{16-8 x+x^2}}{x} \, dx+\frac {1}{25} (4 (32-\log (9))) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{200} (-1024+66 \log (3)-\log (9)) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{25} (2 \log (9)) \int \frac {e^{16-8 x+x^2}}{x} \, dx\\ &=-\frac {2}{25} e^{(-4+x)^2}+\frac {26 x^2}{25}+\frac {e^{2 (4-x)^2}}{25 (4+x)^2}-\frac {8}{25} \sqrt {\pi } \text {erfi}(4-x)+\frac {\log ^2(3)}{25 x^2}+\frac {e^{16-8 x+x^2} (32-\log (9))}{100 (4+x)}+\frac {1}{100} \sqrt {\pi } \text {erfi}(4-x) (32-\log (9))+\frac {e^{16-8 x+x^2} \log (9)}{100 x}+\frac {1}{100} \sqrt {\pi } \text {erfi}(4-x) \log (9)-\frac {1}{25} (4 \log (3)) \int \frac {e^{16-8 x+x^2}}{x} \, dx+\frac {1}{25} (4 (32-\log (9))) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{200} (-1024+66 \log (3)-\log (9)) \int \frac {e^{16-8 x+x^2}}{4+x} \, dx+\frac {1}{25} (2 \log (9)) \int \frac {e^{16-8 x+x^2}}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.94, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3328 x^4+2496 x^5+624 x^6+52 x^7+e^{32-16 x+2 x^2} \left (-66 x^3+4 x^5\right )+\left (-128-96 x-24 x^2-2 x^3\right ) \log ^2(3)+e^{16-8 x+x^2} \left (-32 x^3+248 x^4+64 x^5-16 x^6-4 x^7+\left (-32 x-280 x^2-68 x^3+16 x^4+4 x^5\right ) \log (3)\right )}{1600 x^3+1200 x^4+300 x^5+25 x^6} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [B] time = 0.80, size = 93, normalized size = 2.66 \begin {gather*} \frac {26 \, x^{6} + 208 \, x^{5} + 416 \, x^{4} + x^{2} e^{\left (2 \, x^{2} - 16 \, x + 32\right )} + {\left (x^{2} + 8 \, x + 16\right )} \log \relax (3)^{2} - 2 \, {\left (x^{4} + 4 \, x^{3} - {\left (x^{2} + 4 \, x\right )} \log \relax (3)\right )} e^{\left (x^{2} - 8 \, x + 16\right )}}{25 \, {\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.19, size = 128, normalized size = 3.66 \begin {gather*} \frac {26 \, x^{6} + 208 \, x^{5} - 2 \, x^{4} e^{\left (x^{2} - 8 \, x + 16\right )} + 416 \, x^{4} - 8 \, x^{3} e^{\left (x^{2} - 8 \, x + 16\right )} + 2 \, x^{2} e^{\left (x^{2} - 8 \, x + 16\right )} \log \relax (3) + x^{2} \log \relax (3)^{2} + x^{2} e^{\left (2 \, x^{2} - 16 \, x + 32\right )} + 8 \, x e^{\left (x^{2} - 8 \, x + 16\right )} \log \relax (3) + 8 \, x \log \relax (3)^{2} + 16 \, \log \relax (3)^{2}}{25 \, {\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 55, normalized size = 1.57
method | result | size |
risch | \(\frac {26 x^{2}}{25}+\frac {\ln \relax (3)^{2}}{25 x^{2}}+\frac {{\mathrm e}^{2 \left (x -4\right )^{2}}}{25 \left (4+x \right )^{2}}+\frac {2 \left (\ln \relax (3)-x^{2}\right ) {\mathrm e}^{\left (x -4\right )^{2}}}{25 \left (4+x \right ) x}\) | \(55\) |
norman | \(\frac {-\frac {3328 x^{3}}{25}+\left (\frac {\ln \relax (3)^{2}}{25}-\frac {6656}{25}\right ) x^{2}+\frac {208 x^{5}}{25}+\frac {26 x^{6}}{25}+\frac {16 \ln \relax (3)^{2}}{25}+\frac {8 x \ln \relax (3)^{2}}{25}+\frac {x^{2} {\mathrm e}^{2 x^{2}-16 x +32}}{25}-\frac {8 \,{\mathrm e}^{x^{2}-8 x +16} x^{3}}{25}-\frac {2 \,{\mathrm e}^{x^{2}-8 x +16} x^{4}}{25}+\frac {8 \ln \relax (3) {\mathrm e}^{x^{2}-8 x +16} x}{25}+\frac {2 \ln \relax (3) {\mathrm e}^{x^{2}-8 x +16} x^{2}}{25}}{\left (4+x \right )^{2} x^{2}}\) | \(125\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 277, normalized size = 7.91 \begin {gather*} -\frac {1}{100} \, {\left (\frac {4 \, {\left (3 \, x^{3} + 18 \, x^{2} + 16 \, x - 16\right )}}{x^{4} + 8 \, x^{3} + 16 \, x^{2}} - 3 \, \log \left (x + 4\right ) + 3 \, \log \relax (x)\right )} \log \relax (3)^{2} + \frac {3}{200} \, {\left (\frac {4 \, {\left (3 \, x^{2} + 18 \, x + 16\right )}}{x^{3} + 8 \, x^{2} + 16 \, x} - 3 \, \log \left (x + 4\right ) + 3 \, \log \relax (x)\right )} \log \relax (3)^{2} - \frac {3}{200} \, {\left (\frac {4 \, {\left (x + 6\right )}}{x^{2} + 8 \, x + 16} - \log \left (x + 4\right ) + \log \relax (x)\right )} \log \relax (3)^{2} + \frac {26}{25} \, x^{2} + \frac {{\left (x e^{\left (2 \, x^{2} + 32\right )} - 2 \, {\left (x^{3} e^{16} + 4 \, x^{2} e^{16} - x e^{16} \log \relax (3) - 4 \, e^{16} \log \relax (3)\right )} e^{\left (x^{2} + 8 \, x\right )}\right )} e^{\left (-16 \, x\right )}}{25 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )}} + \frac {\log \relax (3)^{2}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} - \frac {9984 \, {\left (3 \, x + 10\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} + \frac {6656 \, {\left (2 \, x + 7\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} + \frac {19968 \, {\left (x + 3\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} - \frac {3328 \, {\left (x + 2\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.85, size = 96, normalized size = 2.74 \begin {gather*} \frac {{\ln \relax (3)}^2}{25\,x^2}-\frac {2\,{\mathrm {e}}^{x^2-8\,x+16}}{25}+\frac {26\,x^2}{25}+\frac {{\mathrm {e}}^{x^2-8\,x+16}\,\left (\ln \relax (9)-2\,\ln \relax (3)+{\mathrm {e}}^{x^2-8\,x+16}\right )}{25\,{\left (x+4\right )}^2}+\frac {{\mathrm {e}}^{x^2-8\,x+16}\,\ln \relax (3)}{50\,x}-\frac {{\mathrm {e}}^{x^2-8\,x+16}\,\left (\ln \relax (3)-16\right )}{50\,\left (x+4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.31, size = 100, normalized size = 2.86 \begin {gather*} \frac {26 x^{2}}{25} + \frac {\left (25 x^{2} + 100 x\right ) e^{2 x^{2} - 16 x + 32} + \left (- 50 x^{4} - 400 x^{3} - 800 x^{2} + 50 x^{2} \log {\relax (3 )} + 400 x \log {\relax (3 )} + 800 \log {\relax (3 )}\right ) e^{x^{2} - 8 x + 16}}{625 x^{4} + 7500 x^{3} + 30000 x^{2} + 40000 x} + \frac {\log {\relax (3 )}^{2}}{25 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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