Optimal. Leaf size=24 \[ \frac {8 x^2 \left (e^{x/5}+\log (x)\right )}{\frac {4}{5}-x} \]
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Rubi [B] time = 0.45, antiderivative size = 57, normalized size of antiderivative = 2.38, number of steps used = 20, number of rules used = 12, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {27, 6742, 2199, 2194, 2176, 2177, 2178, 43, 2357, 2295, 2314, 31} \begin {gather*} -8 e^{x/5} x-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}+\frac {32 x \log (x)}{4-5 x}-8 x \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 31
Rule 43
Rule 2176
Rule 2177
Rule 2178
Rule 2194
Rule 2199
Rule 2295
Rule 2314
Rule 2357
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{(-4+5 x)^2} \, dx\\ &=\int \left (-\frac {8 e^{x/5} x \left (-40+21 x+5 x^2\right )}{(-4+5 x)^2}-\frac {40 x (-4+5 x-8 \log (x)+5 x \log (x))}{(-4+5 x)^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{x/5} x \left (-40+21 x+5 x^2\right )}{(-4+5 x)^2} \, dx\right )-40 \int \frac {x (-4+5 x-8 \log (x)+5 x \log (x))}{(-4+5 x)^2} \, dx\\ &=-\left (8 \int \left (\frac {29 e^{x/5}}{25}+\frac {1}{5} e^{x/5} x-\frac {16 e^{x/5}}{(-4+5 x)^2}+\frac {16 e^{x/5}}{25 (-4+5 x)}\right ) \, dx\right )-40 \int \left (\frac {x}{-4+5 x}+\frac {x (-8+5 x) \log (x)}{(-4+5 x)^2}\right ) \, dx\\ &=-\left (\frac {8}{5} \int e^{x/5} x \, dx\right )-\frac {128}{25} \int \frac {e^{x/5}}{-4+5 x} \, dx-\frac {232}{25} \int e^{x/5} \, dx-40 \int \frac {x}{-4+5 x} \, dx-40 \int \frac {x (-8+5 x) \log (x)}{(-4+5 x)^2} \, dx+128 \int \frac {e^{x/5}}{(-4+5 x)^2} \, dx\\ &=-\frac {232 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-\frac {128}{125} e^{4/25} \text {Ei}\left (\frac {1}{25} (-4+5 x)\right )+\frac {128}{25} \int \frac {e^{x/5}}{-4+5 x} \, dx+8 \int e^{x/5} \, dx-40 \int \left (\frac {1}{5}+\frac {4}{5 (-4+5 x)}\right ) \, dx-40 \int \left (\frac {\log (x)}{5}-\frac {16 \log (x)}{5 (-4+5 x)^2}\right ) \, dx\\ &=-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 x-8 e^{x/5} x-\frac {32}{5} \log (4-5 x)-8 \int \log (x) \, dx+128 \int \frac {\log (x)}{(-4+5 x)^2} \, dx\\ &=-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-\frac {32}{5} \log (4-5 x)-8 x \log (x)+\frac {32 x \log (x)}{4-5 x}+32 \int \frac {1}{-4+5 x} \, dx\\ &=-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-8 x \log (x)+\frac {32 x \log (x)}{4-5 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.19, size = 22, normalized size = 0.92 \begin {gather*} -\frac {40 x^2 \left (e^{x/5}+\log (x)\right )}{-4+5 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.02, size = 24, normalized size = 1.00 \begin {gather*} -\frac {40 \, {\left (x^{2} e^{\left (\frac {1}{5} \, x\right )} + x^{2} \log \relax (x)\right )}}{5 \, x - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 24, normalized size = 1.00 \begin {gather*} -\frac {40 \, {\left (x^{2} e^{\left (\frac {1}{5} \, x\right )} + x^{2} \log \relax (x)\right )}}{5 \, x - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.29, size = 26, normalized size = 1.08
method | result | size |
norman | \(\frac {-40 x^{2} \ln \relax (x )-40 \,{\mathrm e}^{\frac {x}{5}} x^{2}}{5 x -4}\) | \(26\) |
default | \(-8 x \ln \relax (x )-\frac {32 \ln \relax (x ) x}{5 x -4}-\frac {128 \,{\mathrm e}^{\frac {x}{5}}}{125 \left (\frac {x}{5}-\frac {4}{25}\right )}-\frac {32 \,{\mathrm e}^{\frac {x}{5}}}{5}-8 x \,{\mathrm e}^{\frac {x}{5}}\) | \(45\) |
risch | \(-\frac {8 \left (25 x^{2}-20 x +16\right ) \ln \relax (x )}{5 \left (5 x -4\right )}-\frac {8 \left (25 \,{\mathrm e}^{\frac {x}{5}} x^{2}+20 x \ln \relax (x )-16 \ln \relax (x )\right )}{5 \left (5 x -4\right )}\) | \(51\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.40, size = 48, normalized size = 2.00 \begin {gather*} -8 \, x - \frac {8 \, {\left (25 \, x^{2} e^{\left (\frac {1}{5} \, x\right )} - 25 \, x^{2} + {\left (25 \, x^{2} - 20 \, x + 16\right )} \log \relax (x) + 20 \, x\right )}}{5 \, {\left (5 \, x - 4\right )}} - \frac {32}{5} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.40, size = 19, normalized size = 0.79 \begin {gather*} -\frac {40\,x^2\,\left ({\mathrm {e}}^{x/5}+\ln \relax (x)\right )}{5\,x-4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.42, size = 39, normalized size = 1.62 \begin {gather*} - \frac {40 x^{2} e^{\frac {x}{5}}}{5 x - 4} - \frac {32 \log {\relax (x )}}{5} + \frac {\left (- 200 x^{2} + 160 x - 128\right ) \log {\relax (x )}}{25 x - 20} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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