Optimal. Leaf size=20 \[ \frac {4+\frac {e^5 \log (\log (-3-x))}{x}}{x} \]
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Rubi [F] time = 0.63, 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 {e^5 x+\left (-12 x-4 x^2\right ) \log (-3-x)+e^5 (-6-2 x) \log (-3-x) \log (\log (-3-x))}{\left (3 x^3+x^4\right ) \log (-3-x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 x+\left (-12 x-4 x^2\right ) \log (-3-x)+e^5 (-6-2 x) \log (-3-x) \log (\log (-3-x))}{x^3 (3+x) \log (-3-x)} \, dx\\ &=\int \frac {-4 x+\frac {e^5 x}{(3+x) \log (-3-x)}-2 e^5 \log (\log (-3-x))}{x^3} \, dx\\ &=\int \left (\frac {e^5-12 \log (-3-x)-4 x \log (-3-x)}{x^2 (3+x) \log (-3-x)}-\frac {2 e^5 \log (\log (-3-x))}{x^3}\right ) \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\right )+\int \frac {e^5-12 \log (-3-x)-4 x \log (-3-x)}{x^2 (3+x) \log (-3-x)} \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\right )+\int \frac {-4+\frac {e^5}{(3+x) \log (-3-x)}}{x^2} \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\right )+\int \left (-\frac {4}{x^2}+\frac {e^5}{x^2 (3+x) \log (-3-x)}\right ) \, dx\\ &=\frac {4}{x}+e^5 \int \frac {1}{x^2 (3+x) \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}+e^5 \int \left (\frac {1}{3 x^2 \log (-3-x)}-\frac {1}{9 x \log (-3-x)}+\frac {1}{9 (3+x) \log (-3-x)}\right ) \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{9} e^5 \int \frac {1}{(3+x) \log (-3-x)} \, dx+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{9} e^5 \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-3-x\right )+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{9} e^5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-3-x)\right )+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}+\frac {1}{9} e^5 \log (\log (-3-x))-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 20, normalized size = 1.00 \begin {gather*} \frac {4}{x}+\frac {e^5 \log (\log (-3-x))}{x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.61, size = 18, normalized size = 0.90 \begin {gather*} \frac {e^{5} \log \left (\log \left (-x - 3\right )\right ) + 4 \, x}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 18, normalized size = 0.90 \begin {gather*} \frac {e^{5} \log \left (\log \left (-x - 3\right )\right ) + 4 \, x}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 20, normalized size = 1.00
method | result | size |
risch | \(\frac {{\mathrm e}^{5} \ln \left (\ln \left (-3-x \right )\right )}{x^{2}}+\frac {4}{x}\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 18, normalized size = 0.90 \begin {gather*} \frac {e^{5} \log \left (\log \left (-x - 3\right )\right ) + 4 \, x}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.84, size = 18, normalized size = 0.90 \begin {gather*} \frac {4\,x+\ln \left (\ln \left (-x-3\right )\right )\,{\mathrm {e}}^5}{x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.37, size = 17, normalized size = 0.85 \begin {gather*} \frac {4}{x} + \frac {e^{5} \log {\left (\log {\left (- x - 3 \right )} \right )}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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