Optimal. Leaf size=33 \[ 3+\frac {-e^5+x+\log \left (x \left (-e^{x^2}+\left (e^4+5 x\right )^2\right )\right )}{x} \]
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Rubi [A] time = 2.09, antiderivative size = 46, normalized size of antiderivative = 1.39, number of steps used = 15, number of rules used = 3, integrand size = 139, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6742, 14, 2551} \begin {gather*} \frac {\log \left (25 x^3+10 e^4 x^2-e^{x^2} x+e^8 x\right )}{x}-\frac {1+e^5}{x}+\frac {1}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2551
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3\right )}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}+\frac {1+e^5+2 x^2-\log \left (x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx+\int \frac {1+e^5+2 x^2-\log \left (x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )\right )}{x^2} \, dx\\ &=2 \int \left (-\frac {25 \left (1-\frac {e^8}{25}\right )}{-e^8+e^{x^2}-10 e^4 x-25 x^2}+\frac {5 e^4}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}-\frac {10 e^4 x}{e^8-e^{x^2}+10 e^4 x+25 x^2}-\frac {25 x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2}\right ) \, dx+\int \left (\frac {1+e^5+2 x^2}{x^2}-\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x^2}\right ) \, dx\\ &=-\left (50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx\right )+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx+\int \frac {1+e^5+2 x^2}{x^2} \, dx-\int \frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x^2} \, dx\\ &=\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx+\int \left (2+\frac {1+e^5}{x^2}\right ) \, dx-\int \frac {e^8+20 e^4 x+75 x^2-e^{x^2} \left (1+2 x^2\right )}{x^2 \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx\\ &=-\frac {1+e^5}{x}+2 x+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx-\int \left (\frac {1+2 x^2}{x^2}+\frac {2 \left (5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3\right )}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}\right ) \, dx\\ &=-\frac {1+e^5}{x}+2 x+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-2 \int \frac {5 e^4+\left (25-e^8\right ) x-10 e^4 x^2-25 x^3}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx-\int \frac {1+2 x^2}{x^2} \, dx\\ &=-\frac {1+e^5}{x}+2 x+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}-2 \int \left (-\frac {25 \left (1-\frac {e^8}{25}\right )}{-e^8+e^{x^2}-10 e^4 x-25 x^2}+\frac {5 e^4}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )}-\frac {10 e^4 x}{e^8-e^{x^2}+10 e^4 x+25 x^2}-\frac {25 x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2}\right ) \, dx-50 \int \frac {x^2}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx+\left (10 e^4\right ) \int \frac {1}{x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )} \, dx-\left (20 e^4\right ) \int \frac {x}{e^8-e^{x^2}+10 e^4 x+25 x^2} \, dx-\left (2 \left (25-e^8\right )\right ) \int \frac {1}{-e^8+e^{x^2}-10 e^4 x-25 x^2} \, dx-\int \left (2+\frac {1}{x^2}\right ) \, dx\\ &=\frac {1}{x}-\frac {1+e^5}{x}+\frac {\log \left (e^8 x-e^{x^2} x+10 e^4 x^2+25 x^3\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 38, normalized size = 1.15 \begin {gather*} -\frac {e^5}{x}+\frac {\log \left (x \left (e^8-e^{x^2}+10 e^4 x+25 x^2\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 35, normalized size = 1.06 \begin {gather*} -\frac {e^{5} - \log \left (25 \, x^{3} + 10 \, x^{2} e^{4} + x e^{8} - x e^{\left (x^{2}\right )}\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 37, normalized size = 1.12
method | result | size |
norman | \(\frac {-{\mathrm e}^{5}+\ln \left (-{\mathrm e}^{x^{2}} x +x \,{\mathrm e}^{8}+10 x^{2} {\mathrm e}^{4}+25 x^{3}\right )}{x}\) | \(37\) |
risch | \(\frac {\ln \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )}{x}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right ) \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right ) \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )^{2}+i \pi \mathrm {csgn}\left (i x \left (-{\mathrm e}^{x^{2}}+{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+25 x^{2}\right )\right )^{3}+2 \,{\mathrm e}^{5}-2 \ln \relax (x )}{2 x}\) | \(214\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 34, normalized size = 1.03 \begin {gather*} -\frac {e^{5} - \log \left (25 \, x^{2} + 10 \, x e^{4} + e^{8} - e^{\left (x^{2}\right )}\right ) - \log \relax (x)}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.13, size = 35, normalized size = 1.06 \begin {gather*} -\frac {{\mathrm {e}}^5-\ln \left (x\,{\mathrm {e}}^8-x\,{\mathrm {e}}^{x^2}+10\,x^2\,{\mathrm {e}}^4+25\,x^3\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.57, size = 32, normalized size = 0.97 \begin {gather*} \frac {\log {\left (25 x^{3} + 10 x^{2} e^{4} - x e^{x^{2}} + x e^{8} \right )}}{x} - \frac {e^{5}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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