Optimal. Leaf size=24 \[ \frac {e^{e^x-x+x^2} \left (x+\log \left ((4+x)^2\right )\right )}{x} \]
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Rubi [F]
time = 8.26, 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 {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int \frac {e^{e^x+(-1+x) x} \left (x \left (2+4 \left (-1+e^x\right ) x+\left (7+e^x\right ) x^2+2 x^3\right )+(4+x) \left (-1+\left (-1+e^x\right ) x+2 x^2\right ) \log \left ((4+x)^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int \left (\frac {e^{e^x+x+(-1+x) x} \left (x+\log \left ((4+x)^2\right )\right )}{x}+\frac {e^{e^x+(-1+x) x} \left (2 x-4 x^2+7 x^3+2 x^4-4 \log \left ((4+x)^2\right )-5 x \log \left ((4+x)^2\right )+7 x^2 \log \left ((4+x)^2\right )+2 x^3 \log \left ((4+x)^2\right )\right )}{x^2 (4+x)}\right ) \, dx\\ &=\int \frac {e^{e^x+x+(-1+x) x} \left (x+\log \left ((4+x)^2\right )\right )}{x} \, dx+\int \frac {e^{e^x+(-1+x) x} \left (2 x-4 x^2+7 x^3+2 x^4-4 \log \left ((4+x)^2\right )-5 x \log \left ((4+x)^2\right )+7 x^2 \log \left ((4+x)^2\right )+2 x^3 \log \left ((4+x)^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int \left (e^{e^x+x+(-1+x) x}+\frac {e^{e^x+x+(-1+x) x} \log \left ((4+x)^2\right )}{x}\right ) \, dx+\int \frac {e^{e^x+(-1+x) x} \left (x \left (2-4 x+7 x^2+2 x^3\right )+\left (-4-5 x+7 x^2+2 x^3\right ) \log \left ((4+x)^2\right )\right )}{x^2 (4+x)} \, dx\\ &=\int e^{e^x+x+(-1+x) x} \, dx+\int \frac {e^{e^x+x+(-1+x) x} \log \left ((4+x)^2\right )}{x} \, dx+\int \left (\frac {e^{e^x+(-1+x) x} \left (2-4 x+7 x^2+2 x^3\right )}{x (4+x)}+\frac {e^{e^x+(-1+x) x} (-1+x) (1+2 x) \log \left ((4+x)^2\right )}{x^2}\right ) \, dx\\ &=\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx+\int \frac {e^{e^x+(-1+x) x} \left (2-4 x+7 x^2+2 x^3\right )}{x (4+x)} \, dx+\int \frac {e^{e^x+(-1+x) x} (-1+x) (1+2 x) \log \left ((4+x)^2\right )}{x^2} \, dx-\int \frac {2 \int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx\\ &=-\left (2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx\right )-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx+\int \left (-e^{e^x+(-1+x) x}+\frac {e^{e^x+(-1+x) x}}{2 x}+2 e^{e^x+(-1+x) x} x-\frac {e^{e^x+(-1+x) x}}{2 (4+x)}\right ) \, dx-\int \frac {4 \int e^{e^x+(-1+x) x} \, dx-2 \left (\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx+\int \frac {e^{e^x+(-1+x) x}}{x} \, dx\right )}{4+x} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx-\int \left (\frac {2 \left (2 \int e^{e^x+(-1+x) x} \, dx-\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx\right )}{4+x}-\frac {2 \int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx-2 \int \frac {2 \int e^{e^x+(-1+x) x} \, dx-\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx}{4+x} \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x} \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx-2 \int \left (\frac {2 \int e^{e^x+(-1+x) x} \, dx}{4+x}-\frac {\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx}{4+x}\right ) \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x} \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{x} \, dx-\frac {1}{2} \int \frac {e^{e^x+(-1+x) x}}{4+x} \, dx+2 \int e^{e^x+(-1+x) x} x \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx}{4+x} \, dx+2 \int \frac {\int \frac {e^{e^x+(-1+x) x}}{x} \, dx}{4+x} \, dx-2 \int \frac {\int \frac {e^{e^x+x^2}}{x} \, dx}{4+x} \, dx-4 \int \frac {\int e^{e^x+(-1+x) x} \, dx}{4+x} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x^2} \, dx-\log \left ((4+x)^2\right ) \int \frac {e^{e^x+(-1+x) x}}{x} \, dx+\log \left ((4+x)^2\right ) \int \frac {e^{e^x+x+(-1+x) x}}{x} \, dx+\left (2 \log \left ((4+x)^2\right )\right ) \int e^{e^x+(-1+x) x} \, dx-\int e^{e^x+(-1+x) x} \, dx+\int e^{e^x+x+(-1+x) x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.34, size = 23, normalized size = 0.96 \begin {gather*} \frac {e^{e^x+(-1+x) x} \left (x+\log \left ((4+x)^2\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.76, size = 85, normalized size = 3.54
method | result | size |
risch | \(\frac {\left (-i \pi \mathrm {csgn}\left (i \left (4+x \right )\right )^{2} \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i \left (4+x \right )\right ) \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{3}+2 x +4 \ln \left (4+x \right )\right ) {\mathrm e}^{x^{2}+{\mathrm e}^{x}-x}}{2 x}\) | \(85\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.34, size = 22, normalized size = 0.92 \begin {gather*} \frac {{\left (x + 2 \, \log \left (x + 4\right )\right )} e^{\left (x^{2} - x + e^{x}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.47, size = 31, normalized size = 1.29 \begin {gather*} \frac {{\left (x e^{x} + e^{x} \log \left (x^{2} + 8 \, x + 16\right )\right )} e^{\left (x^{2} - 2 \, x + e^{x}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.14, size = 31, normalized size = 1.29 \begin {gather*} \frac {\left (x e^{x} + e^{x} \log {\left (x^{2} + 8 x + 16 \right )}\right ) e^{x^{2} - 2 x + e^{x}}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.46, size = 35, normalized size = 1.46 \begin {gather*} \frac {{\left (x e^{\left (x^{2} + e^{x}\right )} + e^{\left (x^{2} + e^{x}\right )} \log \left (x^{2} + 8 \, x + 16\right )\right )} e^{\left (-x\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{{\mathrm {e}}^x-2\,x+x^2}\,\left ({\mathrm {e}}^{2\,x}\,\left (x^3+4\,x^2\right )+{\mathrm {e}}^x\,\left (2\,x^4+7\,x^3-4\,x^2+2\,x\right )+\ln \left (x^2+8\,x+16\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (x^2+4\,x\right )-{\mathrm {e}}^x\,\left (-2\,x^3-7\,x^2+5\,x+4\right )\right )\right )}{x^3+4\,x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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