Optimal. Leaf size=35 \[ \frac {e^2 x}{2 \left (5-e^x-x^2\right )}+\frac {1}{3} \log \left (-3+\frac {5}{x}\right ) \]
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Rubi [F]
time = 1.05, 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 {250+10 e^{2 x}-100 x^2+10 x^4+e^2 \left (-75 x+45 x^2-15 x^3+9 x^4\right )+e^x \left (-100+20 x^2+e^2 \left (15 x-24 x^2+9 x^3\right )\right )}{-750 x+450 x^2+300 x^3-180 x^4-30 x^5+18 x^6+e^{2 x} \left (-30 x+18 x^2\right )+e^x \left (300 x-180 x^2-60 x^3+36 x^4\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 {-10 e^{2 x}-20 e^x \left (-5+x^2\right )-10 \left (-5+x^2\right )^2-3 e^{2+x} x \left (5-8 x+3 x^2\right )-3 e^2 x \left (-25+15 x-5 x^2+3 x^3\right )}{6 (5-3 x) x \left (5-e^x-x^2\right )^2} \, dx\\ &=\frac {1}{6} \int \frac {-10 e^{2 x}-20 e^x \left (-5+x^2\right )-10 \left (-5+x^2\right )^2-3 e^{2+x} x \left (5-8 x+3 x^2\right )-3 e^2 x \left (-25+15 x-5 x^2+3 x^3\right )}{(5-3 x) x \left (5-e^x-x^2\right )^2} \, dx\\ &=\frac {1}{6} \int \left (\frac {10}{x (-5+3 x)}+\frac {3 e^2 (-1+x)}{-5+e^x+x^2}-\frac {3 e^2 x \left (-5-2 x+x^2\right )}{\left (-5+e^x+x^2\right )^2}\right ) \, dx\\ &=\frac {5}{3} \int \frac {1}{x (-5+3 x)} \, dx+\frac {1}{2} e^2 \int \frac {-1+x}{-5+e^x+x^2} \, dx-\frac {1}{2} e^2 \int \frac {x \left (-5-2 x+x^2\right )}{\left (-5+e^x+x^2\right )^2} \, dx\\ &=-\left (\frac {1}{3} \int \frac {1}{x} \, dx\right )-\frac {1}{2} e^2 \int \left (-\frac {5 x}{\left (-5+e^x+x^2\right )^2}-\frac {2 x^2}{\left (-5+e^x+x^2\right )^2}+\frac {x^3}{\left (-5+e^x+x^2\right )^2}\right ) \, dx+\frac {1}{2} e^2 \int \left (-\frac {1}{-5+e^x+x^2}+\frac {x}{-5+e^x+x^2}\right ) \, dx+\int \frac {1}{-5+3 x} \, dx\\ &=\frac {1}{3} \log (5-3 x)-\frac {\log (x)}{3}-\frac {1}{2} e^2 \int \frac {x^3}{\left (-5+e^x+x^2\right )^2} \, dx-\frac {1}{2} e^2 \int \frac {1}{-5+e^x+x^2} \, dx+\frac {1}{2} e^2 \int \frac {x}{-5+e^x+x^2} \, dx+e^2 \int \frac {x^2}{\left (-5+e^x+x^2\right )^2} \, dx+\frac {1}{2} \left (5 e^2\right ) \int \frac {x}{\left (-5+e^x+x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 3.03, size = 33, normalized size = 0.94 \begin {gather*} \frac {1}{6} \left (-\frac {3 e^2 x}{-5+e^x+x^2}+2 \log (5-3 x)-2 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.20, size = 28, normalized size = 0.80
method | result | size |
norman | \(-\frac {{\mathrm e}^{2} x}{2 \left (x^{2}+{\mathrm e}^{x}-5\right )}-\frac {\ln \left (x \right )}{3}+\frac {\ln \left (3 x -5\right )}{3}\) | \(28\) |
risch | \(-\frac {{\mathrm e}^{2} x}{2 \left (x^{2}+{\mathrm e}^{x}-5\right )}-\frac {\ln \left (x \right )}{3}+\frac {\ln \left (3 x -5\right )}{3}\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.33, size = 27, normalized size = 0.77 \begin {gather*} -\frac {x e^{2}}{2 \, {\left (x^{2} + e^{x} - 5\right )}} + \frac {1}{3} \, \log \left (3 \, x - 5\right ) - \frac {1}{3} \, \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 43, normalized size = 1.23 \begin {gather*} -\frac {3 \, x e^{2} - 2 \, {\left (x^{2} + e^{x} - 5\right )} \log \left (3 \, x - 5\right ) + 2 \, {\left (x^{2} + e^{x} - 5\right )} \log \left (x\right )}{6 \, {\left (x^{2} + e^{x} - 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.09, size = 29, normalized size = 0.83 \begin {gather*} - \frac {x e^{2}}{2 x^{2} + 2 e^{x} - 10} - \frac {\log {\left (x \right )}}{3} + \frac {\log {\left (x - \frac {5}{3} \right )}}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 63 vs.
\(2 (25) = 50\).
time = 0.43, size = 63, normalized size = 1.80 \begin {gather*} \frac {2 \, x^{2} \log \left (3 \, x - 5\right ) - 2 \, x^{2} \log \left (x\right ) - 3 \, x e^{2} + 2 \, e^{x} \log \left (3 \, x - 5\right ) - 2 \, e^{x} \log \left (x\right ) - 10 \, \log \left (3 \, x - 5\right ) + 10 \, \log \left (x\right )}{6 \, {\left (x^{2} + e^{x} - 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.37, size = 52, normalized size = 1.49 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\frac {6\,x}{5}-1\right )}{3}-\frac {-{\mathrm {e}}^2\,x^3+2\,{\mathrm {e}}^2\,x^2+5\,{\mathrm {e}}^2\,x}{2\,\left ({\mathrm {e}}^x+x^2-5\right )\,\left (-x^2+2\,x+5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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