Optimal. Leaf size=31 \[ -\frac {e^3}{x}-x+\frac {e^{-1+x} \left (\frac {4}{x}+\log (5)\right )^2}{x} \]
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Rubi [C] time = 0.27, antiderivative size = 136, normalized size of antiderivative = 4.39, number of steps used = 17, number of rules used = 5, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 14, 2199, 2177, 2178} \begin {gather*} -\frac {8 \text {Ei}(x)}{e}+\frac {\log ^2(5) \text {Ei}(x)}{e}+\frac {(8-\log (5)) \log (5) \text {Ei}(x)}{e}+\frac {8 (1-\log (5)) \text {Ei}(x)}{e}+\frac {16 e^{x-1}}{x^3}+\frac {8 e^{x-1}}{x^2}-\frac {8 e^{x-1} (1-\log (5))}{x^2}-x+\frac {8 e^{x-1}}{x}-\frac {e^3}{x}-\frac {e^{x-1} (8-\log (5)) \log (5)}{x}-\frac {8 e^{x-1} (1-\log (5))}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2177
Rule 2178
Rule 2199
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^4 x^2-e x^4+e^x \left (-48+16 x+\left (-16 x+8 x^2\right ) \log (5)+\left (-x^2+x^3\right ) \log ^2(5)\right )}{x^4} \, dx}{e}\\ &=\frac {\int \left (\frac {e^4-e x^2}{x^2}+\frac {e^x (4+x \log (5)) \left (-12+x (4-\log (5))+x^2 \log (5)\right )}{x^4}\right ) \, dx}{e}\\ &=\frac {\int \frac {e^4-e x^2}{x^2} \, dx}{e}+\frac {\int \frac {e^x (4+x \log (5)) \left (-12+x (4-\log (5))+x^2 \log (5)\right )}{x^4} \, dx}{e}\\ &=\frac {\int \left (-e+\frac {e^4}{x^2}\right ) \, dx}{e}+\frac {\int \left (-\frac {48 e^x}{x^4}-\frac {16 e^x (-1+\log (5))}{x^3}-\frac {e^x (-8+\log (5)) \log (5)}{x^2}+\frac {e^x \log ^2(5)}{x}\right ) \, dx}{e}\\ &=-\frac {e^3}{x}-x-\frac {48 \int \frac {e^x}{x^4} \, dx}{e}+\frac {(16 (1-\log (5))) \int \frac {e^x}{x^3} \, dx}{e}+\frac {((8-\log (5)) \log (5)) \int \frac {e^x}{x^2} \, dx}{e}+\frac {\log ^2(5) \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}-\frac {e^3}{x}-x-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) \log ^2(5)}{e}-\frac {16 \int \frac {e^x}{x^3} \, dx}{e}+\frac {(8 (1-\log (5))) \int \frac {e^x}{x^2} \, dx}{e}+\frac {((8-\log (5)) \log (5)) \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}+\frac {8 e^{-1+x}}{x^2}-\frac {e^3}{x}-x-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {8 e^{-1+x} (1-\log (5))}{x}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) (8-\log (5)) \log (5)}{e}+\frac {\text {Ei}(x) \log ^2(5)}{e}-\frac {8 \int \frac {e^x}{x^2} \, dx}{e}+\frac {(8 (1-\log (5))) \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}+\frac {8 e^{-1+x}}{x^2}-\frac {e^3}{x}+\frac {8 e^{-1+x}}{x}-x-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {8 e^{-1+x} (1-\log (5))}{x}+\frac {8 \text {Ei}(x) (1-\log (5))}{e}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) (8-\log (5)) \log (5)}{e}+\frac {\text {Ei}(x) \log ^2(5)}{e}-\frac {8 \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}+\frac {8 e^{-1+x}}{x^2}-\frac {e^3}{x}+\frac {8 e^{-1+x}}{x}-x-\frac {8 \text {Ei}(x)}{e}-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {8 e^{-1+x} (1-\log (5))}{x}+\frac {8 \text {Ei}(x) (1-\log (5))}{e}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) (8-\log (5)) \log (5)}{e}+\frac {\text {Ei}(x) \log ^2(5)}{e}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.17, size = 47, normalized size = 1.52 \begin {gather*} \frac {16 e^{-1+x}}{x^3}-\frac {e^3}{x}-x+\frac {8 e^{-1+x} \log (5)}{x^2}+\frac {e^{-1+x} \log ^2(5)}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.90, size = 39, normalized size = 1.26 \begin {gather*} -\frac {{\left (x^{4} e + x^{2} e^{4} - {\left (x^{2} \log \relax (5)^{2} + 8 \, x \log \relax (5) + 16\right )} e^{x}\right )} e^{\left (-1\right )}}{x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 42, normalized size = 1.35 \begin {gather*} -\frac {{\left (x^{4} e - x^{2} e^{x} \log \relax (5)^{2} + x^{2} e^{4} - 8 \, x e^{x} \log \relax (5) - 16 \, e^{x}\right )} e^{\left (-1\right )}}{x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 35, normalized size = 1.13
method | result | size |
risch | \(-x -\frac {{\mathrm e}^{3}}{x}+\frac {\left (x^{2} \ln \relax (5)^{2}+8 x \ln \relax (5)+16\right ) {\mathrm e}^{x -1}}{x^{3}}\) | \(35\) |
default | \({\mathrm e}^{-1} \left (-\frac {{\mathrm e} \,{\mathrm e}^{3}}{x}+\frac {16 \,{\mathrm e}^{x}}{x^{3}}+\frac {8 \,{\mathrm e}^{x} \ln \relax (5)}{x^{2}}+\frac {\ln \relax (5)^{2} {\mathrm e}^{x}}{x}-x \,{\mathrm e}\right )\) | \(47\) |
norman | \(\frac {{\mathrm e}^{-1} \ln \relax (5)^{2} x^{2} {\mathrm e}^{x}-x^{4}-x^{2} {\mathrm e}^{3}+16 \,{\mathrm e}^{x} {\mathrm e}^{-1}+8 \,{\mathrm e}^{-1} \ln \relax (5) x \,{\mathrm e}^{x}}{x^{3}}\) | \(51\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.39, size = 66, normalized size = 2.13 \begin {gather*} {\left ({\rm Ei}\relax (x) \log \relax (5)^{2} - \Gamma \left (-1, -x\right ) \log \relax (5)^{2} - x e + 8 \, \Gamma \left (-1, -x\right ) \log \relax (5) + 16 \, \Gamma \left (-2, -x\right ) \log \relax (5) - \frac {e^{4}}{x} - 16 \, \Gamma \left (-2, -x\right ) - 48 \, \Gamma \left (-3, -x\right )\right )} e^{\left (-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.25, size = 42, normalized size = 1.35 \begin {gather*} \frac {16\,{\mathrm {e}}^{x-1}-x^2\,\left ({\mathrm {e}}^3-{\mathrm {e}}^{x-1}\,{\ln \relax (5)}^2\right )+8\,x\,{\mathrm {e}}^{x-1}\,\ln \relax (5)}{x^3}-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 32, normalized size = 1.03 \begin {gather*} - x - \frac {e^{3}}{x} + \frac {\left (x^{2} \log {\relax (5 )}^{2} + 8 x \log {\relax (5 )} + 16\right ) e^{x}}{e x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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