Optimal. Leaf size=27 \[ e^{\frac {2}{5}+\frac {\log (5)}{5-e^{-2+2 x} x^3+\log (3)}} \]
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Rubi [F]
time = 19.92, 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 {\exp \left (-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\log (5) \int \frac {\exp \left (-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) \left (3 x^2+2 x^3\right )}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx\\ &=\log (5) \int \frac {\exp \left (-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2 (3+2 x)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx\\ &=\log (5) \int \frac {\exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2 (3+2 x)}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2} \, dx\\ &=\log (5) \int \left (\frac {3 \exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}+\frac {2 \exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^3}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}\right ) \, dx\\ &=(2 \log (5)) \int \frac {\exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^3}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2} \, dx+(3 \log (5)) \int \frac {\exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.49, size = 33, normalized size = 1.22 \begin {gather*} 5^{\frac {e^2}{-e^{2 x} x^3+e^2 (5+\log (3))}} e^{2/5} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.74, size = 42, normalized size = 1.56
method | result | size |
risch | \({\mathrm e}^{\frac {-2 x^{3} {\mathrm e}^{2 x -2}+5 \ln \left (5\right )+2 \ln \left (3\right )+10}{-5 x^{3} {\mathrm e}^{2 x -2}+5 \ln \left (3\right )+25}}\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.69, size = 30, normalized size = 1.11 \begin {gather*} e^{\left (-\frac {e^{2} \log \left (5\right )}{x^{3} e^{\left (2 \, x\right )} - e^{2} \log \left (3\right ) - 5 \, e^{2}} + \frac {2}{5}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 63 vs.
\(2 (25) = 50\).
time = 0.50, size = 63, normalized size = 2.33 \begin {gather*} e^{\left (-2 \, x + \frac {2 \, {\left (5 \, x^{4} - 4 \, x^{3}\right )} e^{\left (2 \, x - 2\right )} - 2 \, {\left (5 \, x - 4\right )} \log \left (3\right ) - 50 \, x - 5 \, \log \left (5\right ) + 40}{5 \, {\left (x^{3} e^{\left (2 \, x - 2\right )} - \log \left (3\right ) - 5\right )}} + 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.65, size = 42, normalized size = 1.56 \begin {gather*} e^{\frac {2 x^{3} e^{2 x - 2} - 10 - 5 \log {\left (5 \right )} - 2 \log {\left (3 \right )}}{5 x^{3} e^{2 x - 2} - 25 - 5 \log {\left (3 \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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}}^{2\,x-2}\,{\mathrm {e}}^{\frac {2\,\ln \left (3\right )+5\,\ln \left (5\right )-2\,x^3\,{\mathrm {e}}^{2\,x-2}+10}{5\,\ln \left (3\right )-5\,x^3\,{\mathrm {e}}^{2\,x-2}+25}}\,\ln \left (5\right )\,\left (2\,x^3+3\,x^2\right )}{10\,\ln \left (3\right )-{\mathrm {e}}^{2\,x-2}\,\left (2\,x^3\,\ln \left (3\right )+10\,x^3\right )+{\ln \left (3\right )}^2+x^6\,{\mathrm {e}}^{4\,x-4}+25} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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