Optimal. Leaf size=21 \[ -5+\frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \]
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Rubi [F] time = 3.85, 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 {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (e^{3+x^2}+x\right ) (5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=\int \frac {-6 x-e^{3+x^2} \left (1+10 x^2\right )-x \left (1+2 e^{3+x^2} x\right ) \log (x)+\left (e^{3+x^2}+x\right ) (5+\log (x)) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{\left (e^{3+x^2}+x\right ) (5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=\int \left (\frac {x \left (-1+2 x^2\right )}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {-1-10 x^2-2 x^2 \log (x)+5 \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )+\log (x) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx\\ &=\int \frac {x \left (-1+2 x^2\right )}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {-1-10 x^2-2 x^2 \log (x)+5 \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )+\log (x) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=\int \left (-\frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {2 x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx+\int \left (\frac {-1-10 x^2-2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx\\ &=2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {-1-10 x^2-2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \left (-\frac {1}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}-\frac {10 x^2}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}-\frac {2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-2 \int \frac {x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-10 \int \frac {x^2}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {1}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 19, normalized size = 0.90 \begin {gather*} \frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 28, normalized size = 1.33 \begin {gather*} \frac {x}{\log \left ({\left (x + e^{\left (x^{2} + 3\right )}\right )} \log \relax (x) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.73, size = 30, normalized size = 1.43 \begin {gather*} \frac {x}{\log \left (x \log \relax (x) + e^{\left (x^{2} + 3\right )} \log \relax (x) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.11, size = 148, normalized size = 7.05
method | result | size |
risch | \(\frac {2 i x}{\pi \,\mathrm {csgn}\left (i \left (5+\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )-\pi \,\mathrm {csgn}\left (i \left (5+\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )^{2}+\pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )^{3}+2 i \ln \left (5+\ln \relax (x )\right )+2 i \ln \left ({\mathrm e}^{x^{2}+3}+x \right )}\) | \(148\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.80, size = 19, normalized size = 0.90 \begin {gather*} \frac {x}{\log \left (x + e^{\left (x^{2} + 3\right )}\right ) + \log \left (\log \relax (x) + 5\right )} \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 = -0.05 \begin {gather*} \int -\frac {6\,x+{\mathrm {e}}^{x^2+3}\,\left (10\,x^2+1\right )-\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )+\ln \relax (x)\,\left (x+2\,x^2\,{\mathrm {e}}^{x^2+3}\right )}{{\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )}^2\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.01, size = 26, normalized size = 1.24 \begin {gather*} \frac {x}{\log {\left (5 x + \left (x + e^{x^{2} + 3}\right ) \log {\relax (x )} + 5 e^{x^{2} + 3} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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