Optimal. Leaf size=27 \[ \frac {3+x^2}{10+\log \left ((1-x) \left (e^3-x\right ) x^4\right )} \]
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Rubi [F]
time = 5.45, 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 {-15 x+18 x^2+15 x^3-14 x^4+e^3 \left (12-15 x-16 x^2+15 x^3\right )+\left (2 x^3-2 x^4+e^3 \left (-2 x^2+2 x^3\right )\right ) \log \left (-x^5+x^6+e^3 \left (x^4-x^5\right )\right )}{100 x^2-100 x^3+e^3 \left (-100 x+100 x^2\right )+\left (20 x^2-20 x^3+e^3 \left (-20 x+20 x^2\right )\right ) \log \left (-x^5+x^6+e^3 \left (x^4-x^5\right )\right )+\left (x^2-x^3+e^3 \left (-x+x^2\right )\right ) \log ^2\left (-x^5+x^6+e^3 \left (x^4-x^5\right )\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 {-x \left (-15+18 x+15 x^2-14 x^3\right )-e^3 \left (12-15 x-16 x^2+15 x^3\right )-2 \left (e^3-x\right ) (-1+x) x^2 \log \left ((-1+x) x^4 \left (-e^3+x\right )\right )}{(1-x) \left (e^3-x\right ) x \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2} \, dx\\ &=\int \left (\frac {\left (-4 e^3+5 \left (1+e^3\right ) x-6 x^2\right ) \left (3+x^2\right )}{(1-x) \left (e^3-x\right ) x \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2}+\frac {2 x}{10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )}\right ) \, dx\\ &=2 \int \frac {x}{10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )} \, dx+\int \frac {\left (-4 e^3+5 \left (1+e^3\right ) x-6 x^2\right ) \left (3+x^2\right )}{(1-x) \left (e^3-x\right ) x \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2} \, dx\\ &=2 \int \frac {x}{10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )} \, dx+\int \left (-\frac {1+e^3}{\left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2}+\frac {3+e^6}{\left (e^3-x\right ) \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2}-\frac {4}{(-1+x) \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2}-\frac {12}{x \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2}-\frac {6 x}{\left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2}\right ) \, dx\\ &=2 \int \frac {x}{10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )} \, dx-4 \int \frac {1}{(-1+x) \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2} \, dx-6 \int \frac {x}{\left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2} \, dx-12 \int \frac {1}{x \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2} \, dx+\left (-1-e^3\right ) \int \frac {1}{\left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2} \, dx+\left (3+e^6\right ) \int \frac {1}{\left (e^3-x\right ) \left (10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.06, size = 25, normalized size = 0.93 \begin {gather*} \frac {3+x^2}{10+\log \left ((-1+x) x^4 \left (-e^3+x\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 7.22, size = 33, normalized size = 1.22
method | result | size |
norman | \(\frac {x^{2}+3}{\ln \left (\left (-x^{5}+x^{4}\right ) {\mathrm e}^{3}+x^{6}-x^{5}\right )+10}\) | \(33\) |
risch | \(\frac {x^{2}+3}{\ln \left (\left (-x^{5}+x^{4}\right ) {\mathrm e}^{3}+x^{6}-x^{5}\right )+10}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.32, size = 25, normalized size = 0.93 \begin {gather*} \frac {x^{2} + 3}{\log \left (x - e^{3}\right ) + \log \left (x - 1\right ) + 4 \, \log \left (x\right ) + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.44, size = 33, normalized size = 1.22 \begin {gather*} \frac {x^{2} + 3}{\log \left (x^{6} - x^{5} - {\left (x^{5} - x^{4}\right )} e^{3}\right ) + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.11, size = 24, normalized size = 0.89 \begin {gather*} \frac {x^{2} + 3}{\log {\left (x^{6} - x^{5} + \left (- x^{5} + x^{4}\right ) e^{3} \right )} + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.57, size = 33, normalized size = 1.22 \begin {gather*} \frac {x^{2} + 3}{\log \left (x^{6} - x^{5} e^{3} - x^{5} + x^{4} e^{3}\right ) + 10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 11.35, size = 213, normalized size = 7.89 \begin {gather*} \frac {\frac {10\,{\mathrm {e}}^3}{3}-\frac {4\,{\mathrm {e}}^6}{3}+\frac {10\,{\mathrm {e}}^9}{3}+x\,\left (\frac {3\,{\mathrm {e}}^3}{2}+\frac {3\,{\mathrm {e}}^6}{2}-\frac {25\,{\mathrm {e}}^9}{6}-\frac {25}{6}\right )}{108\,x^2+\left (-90\,{\mathrm {e}}^3-90\right )\,x+72\,{\mathrm {e}}^3}+\frac {\frac {15\,x-12\,{\mathrm {e}}^3+15\,x\,{\mathrm {e}}^3+16\,x^2\,{\mathrm {e}}^3-15\,x^3\,{\mathrm {e}}^3-18\,x^2-15\,x^3+14\,x^4}{5\,x-4\,{\mathrm {e}}^3+5\,x\,{\mathrm {e}}^3-6\,x^2}+\frac {2\,x^2\,\ln \left ({\mathrm {e}}^3\,\left (x^4-x^5\right )-x^5+x^6\right )\,\left (x-{\mathrm {e}}^3\right )\,\left (x-1\right )}{5\,x-4\,{\mathrm {e}}^3+5\,x\,{\mathrm {e}}^3-6\,x^2}}{\ln \left ({\mathrm {e}}^3\,\left (x^4-x^5\right )-x^5+x^6\right )+10}+\frac {x^2}{3}-x\,\left (\frac {{\mathrm {e}}^3}{18}+\frac {1}{18}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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