Optimal. Leaf size=30 \[ \frac {e^{x/5} \left (e^4-x \left (-x+\log \left (\frac {x^2}{4096}\right )\right )\right )}{x} \]
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Rubi [C] time = 0.12, antiderivative size = 78, normalized size of antiderivative = 2.60, number of steps used = 12, number of rules used = 7, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.127, Rules used = {12, 14, 2194, 2177, 2178, 2176, 2554} \begin {gather*} -\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} e^4 \text {Ei}\left (\frac {x}{5}\right )+2 \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )+e^{x/5} x+\frac {e^{\frac {x}{5}+4}}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2176
Rule 2177
Rule 2178
Rule 2194
Rule 2554
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{x/5} \left (e^4 (-5+x)-10 x+5 x^2+x^3\right )-e^{x/5} x^2 \log \left (\frac {x^2}{4096}\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (5 e^{x/5}-\frac {5 e^{4+\frac {x}{5}}}{x^2}+\frac {\left (1-\frac {10}{e^4}\right ) e^{4+\frac {x}{5}}}{x}+e^{x/5} x-e^{x/5} \log \left (\frac {x^2}{4096}\right )\right ) \, dx\\ &=\frac {1}{5} \int e^{x/5} x \, dx-\frac {1}{5} \int e^{x/5} \log \left (\frac {x^2}{4096}\right ) \, dx+\frac {1}{5} \left (1-\frac {10}{e^4}\right ) \int \frac {e^{4+\frac {x}{5}}}{x} \, dx+\int e^{x/5} \, dx-\int \frac {e^{4+\frac {x}{5}}}{x^2} \, dx\\ &=5 e^{x/5}+\frac {e^{4+\frac {x}{5}}}{x}+e^{x/5} x-\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )-\frac {1}{5} \int \frac {e^{4+\frac {x}{5}}}{x} \, dx+\frac {1}{5} \int \frac {10 e^{x/5}}{x} \, dx-\int e^{x/5} \, dx\\ &=\frac {e^{4+\frac {x}{5}}}{x}+e^{x/5} x-\frac {1}{5} e^4 \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )+2 \int \frac {e^{x/5}}{x} \, dx\\ &=\frac {e^{4+\frac {x}{5}}}{x}+e^{x/5} x+2 \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} e^4 \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 29, normalized size = 0.97 \begin {gather*} \frac {e^{x/5} \left (e^4+x^2-x \log \left (\frac {x^2}{4096}\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.39, size = 30, normalized size = 1.00 \begin {gather*} -\frac {x e^{\left (\frac {1}{5} \, x\right )} \log \left (\frac {1}{4096} \, x^{2}\right ) - {\left (x^{2} + e^{4}\right )} e^{\left (\frac {1}{5} \, x\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 32, normalized size = 1.07 \begin {gather*} \frac {x^{2} e^{\left (\frac {1}{5} \, x\right )} - x e^{\left (\frac {1}{5} \, x\right )} \log \left (\frac {1}{4096} \, x^{2}\right ) + e^{\left (\frac {1}{5} \, x + 4\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 34, normalized size = 1.13
method | result | size |
norman | \(\frac {x^{2} {\mathrm e}^{\frac {x}{5}}+{\mathrm e}^{4} {\mathrm e}^{\frac {x}{5}}-x \,{\mathrm e}^{\frac {x}{5}} \ln \left (\frac {x^{2}}{4096}\right )}{x}\) | \(34\) |
default | \(\frac {5 x^{2} {\mathrm e}^{\frac {x}{5}}+5 \,{\mathrm e}^{4} {\mathrm e}^{\frac {x}{5}}-5 \left (\ln \left (\frac {x^{2}}{4096}\right )-2 \ln \relax (x )\right ) x \,{\mathrm e}^{\frac {x}{5}}-10 \ln \relax (x ) {\mathrm e}^{\frac {x}{5}} x}{5 x}\) | \(51\) |
risch | \(-2 \,{\mathrm e}^{\frac {x}{5}} \ln \relax (x )+\frac {\left (i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) x -2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} x +i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} x +24 x \ln \relax (2)+2 x^{2}+2 \,{\mathrm e}^{4}\right ) {\mathrm e}^{\frac {x}{5}}}{2 x}\) | \(86\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.39, size = 44, normalized size = 1.47 \begin {gather*} \frac {1}{5} \, {\rm Ei}\left (\frac {1}{5} \, x\right ) e^{4} + {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} - \frac {1}{5} \, e^{4} \Gamma \left (-1, -\frac {1}{5} \, x\right ) - e^{\left (\frac {1}{5} \, x\right )} \log \left (\frac {1}{4096} \, x^{2}\right ) + 5 \, e^{\left (\frac {1}{5} \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.75, size = 26, normalized size = 0.87 \begin {gather*} {\mathrm {e}}^{x/5}\,\left (x-\ln \left (\frac {x^2}{4096}\right )\right )+\frac {{\mathrm {e}}^{\frac {x}{5}+4}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.31, size = 20, normalized size = 0.67 \begin {gather*} \frac {\left (x^{2} - x \log {\left (\frac {x^{2}}{4096} \right )} + e^{4}\right ) e^{\frac {x}{5}}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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