Optimal. Leaf size=20 \[ 1-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x} \]
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Rubi [A] time = 0.20, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1593, 6742, 72, 2551} \begin {gather*} -\frac {3 e^4 \log \left (e^{2 x} (x+2)\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 72
Rule 1593
Rule 2551
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 \left (-15 x-6 x^2\right )+e^4 (6+3 x) \log \left (e^{2 x} (2+x)\right )}{x^2 (2+x)} \, dx\\ &=\int \left (-\frac {3 e^4 (5+2 x)}{x (2+x)}+\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x^2}\right ) \, dx\\ &=-\left (\left (3 e^4\right ) \int \frac {5+2 x}{x (2+x)} \, dx\right )+\left (3 e^4\right ) \int \frac {\log \left (e^{2 x} (2+x)\right )}{x^2} \, dx\\ &=-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x}+\left (3 e^4\right ) \int \frac {5+2 x}{x (2+x)} \, dx-\left (3 e^4\right ) \int \left (\frac {5}{2 x}-\frac {1}{2 (2+x)}\right ) \, dx\\ &=-\frac {15}{2} e^4 \log (x)+\frac {3}{2} e^4 \log (2+x)-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x}+\left (3 e^4\right ) \int \left (\frac {5}{2 x}-\frac {1}{2 (2+x)}\right ) \, dx\\ &=-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 18, normalized size = 0.90 \begin {gather*} -\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 16, normalized size = 0.80 \begin {gather*} -\frac {3 \, e^{4} \log \left ({\left (x + 2\right )} e^{\left (2 \, x\right )}\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 11, normalized size = 0.55 \begin {gather*} -\frac {3 \, e^{4} \log \left (x + 2\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.43, size = 19, normalized size = 0.95
method | result | size |
norman | \(-\frac {3 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{2 x} \left (2+x \right )\right )}{x}\) | \(19\) |
default | \(-\frac {3 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{2 x} \left (2+x \right )\right )}{x}+6 \,{\mathrm e}^{4} \ln \left (2 x +4\right )+\frac {15 \,{\mathrm e}^{4} \ln \left (-2 x \right )}{2}-\frac {15 \,{\mathrm e}^{4} \ln \left (-2 x -4\right )}{2}-\frac {15 \,{\mathrm e}^{4} \ln \relax (x )}{2}+\frac {3 \,{\mathrm e}^{4} \ln \left (2+x \right )}{2}\) | \(62\) |
risch | \(\frac {3 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{-2 x}\right )}{x}-\frac {3 \,{\mathrm e}^{4} \left (-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )+i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )^{2}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )^{3}+2 \ln \left (2+x \right )\right )}{2 x}\) | \(124\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.40, size = 46, normalized size = 2.30 \begin {gather*} \frac {15}{2} \, {\left (\log \left (x + 2\right ) - \log \relax (x)\right )} e^{4} - 6 \, e^{4} \log \left (x + 2\right ) + \frac {15}{2} \, e^{4} \log \relax (x) - \frac {3 \, {\left (x e^{4} + 2 \, e^{4}\right )} \log \left (x + 2\right )}{2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.34, size = 11, normalized size = 0.55 \begin {gather*} -\frac {3\,\ln \left (x+2\right )\,{\mathrm {e}}^4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.21, size = 17, normalized size = 0.85 \begin {gather*} - \frac {3 e^{4} \log {\left (\left (x + 2\right ) e^{2 x} \right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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