Optimal. Leaf size=27 \[ x-\frac {-x+e^{2 x} (10+\log (-2 x))}{16+4 x} \]
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Rubi [A]
time = 0.92, antiderivative size = 41, normalized size of antiderivative = 1.52, number of steps
used = 20, number of rules used = 9, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1608, 27, 12,
6874, 697, 2209, 2208, 2228, 2634} \begin {gather*} x-\frac {5 e^{2 x}}{2 (x+4)}-\frac {1}{x+4}-\frac {e^{2 x} \log (-2 x)}{4 (x+4)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 697
Rule 1608
Rule 2208
Rule 2209
Rule 2228
Rule 2634
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{x \left (64+32 x+4 x^2\right )} \, dx\\ &=\int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{4 x (4+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{x (4+x)^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {4 \left (17+8 x+x^2\right )}{(4+x)^2}-\frac {e^{2 x} \left (4+71 x+20 x^2+7 x \log (-2 x)+2 x^2 \log (-2 x)\right )}{x (4+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {e^{2 x} \left (4+71 x+20 x^2+7 x \log (-2 x)+2 x^2 \log (-2 x)\right )}{x (4+x)^2} \, dx\right )+\int \frac {17+8 x+x^2}{(4+x)^2} \, dx\\ &=-\left (\frac {1}{4} \int \left (\frac {e^{2 x} \left (4+71 x+20 x^2\right )}{x (4+x)^2}+\frac {e^{2 x} (7+2 x) \log (-2 x)}{(4+x)^2}\right ) \, dx\right )+\int \left (1+\frac {1}{(4+x)^2}\right ) \, dx\\ &=x-\frac {1}{4+x}-\frac {1}{4} \int \frac {e^{2 x} \left (4+71 x+20 x^2\right )}{x (4+x)^2} \, dx-\frac {1}{4} \int \frac {e^{2 x} (7+2 x) \log (-2 x)}{(4+x)^2} \, dx\\ &=x-\frac {1}{4+x}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}+\frac {1}{4} \int \frac {e^{2 x}}{x (4+x)} \, dx-\frac {1}{4} \int \left (\frac {e^{2 x}}{4 x}-\frac {10 e^{2 x}}{(4+x)^2}+\frac {79 e^{2 x}}{4 (4+x)}\right ) \, dx\\ &=x-\frac {1}{4+x}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}-\frac {1}{16} \int \frac {e^{2 x}}{x} \, dx+\frac {1}{4} \int \left (\frac {e^{2 x}}{4 x}-\frac {e^{2 x}}{4 (4+x)}\right ) \, dx+\frac {5}{2} \int \frac {e^{2 x}}{(4+x)^2} \, dx-\frac {79}{16} \int \frac {e^{2 x}}{4+x} \, dx\\ &=x-\frac {1}{4+x}-\frac {5 e^{2 x}}{2 (4+x)}-\frac {\text {Ei}(2 x)}{16}-\frac {79 \text {Ei}(2 (4+x))}{16 e^8}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}+\frac {1}{16} \int \frac {e^{2 x}}{x} \, dx-\frac {1}{16} \int \frac {e^{2 x}}{4+x} \, dx+5 \int \frac {e^{2 x}}{4+x} \, dx\\ &=x-\frac {1}{4+x}-\frac {5 e^{2 x}}{2 (4+x)}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.12, size = 36, normalized size = 1.33 \begin {gather*} -\frac {4+10 e^{2 x}-16 x-4 x^2+e^{2 x} \log (-2 x)}{4 (4+x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.04, size = 34, normalized size = 1.26
method | result | size |
norman | \(\frac {x^{2}-17-\frac {\ln \left (-2 x \right ) {\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{2 x}}{2}}{4+x}\) | \(28\) |
default | \(\frac {-\ln \left (-2 x \right ) {\mathrm e}^{2 x}-10 \,{\mathrm e}^{2 x}}{4 x +16}+x -\frac {1}{4+x}\) | \(34\) |
risch | \(-\frac {{\mathrm e}^{2 x} \ln \left (-2 x \right )}{4 \left (4+x \right )}+\frac {2 x^{2}+8 x -5 \,{\mathrm e}^{2 x}-2}{2 x +8}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 35, normalized size = 1.30 \begin {gather*} x - \frac {{\left (\log \left (2\right ) + 10\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (-x\right )}{4 \, {\left (x + 4\right )}} - \frac {1}{x + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, x^{2} - e^{\left (2 \, x\right )} \log \left (-2 \, x\right ) + 16 \, x - 10 \, e^{\left (2 \, x\right )} - 4}{4 \, {\left (x + 4\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.12, size = 26, normalized size = 0.96 \begin {gather*} x + \frac {\left (- \log {\left (- 2 x \right )} - 10\right ) e^{2 x}}{4 x + 16} - \frac {1}{x + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, x^{2} - e^{\left (2 \, x\right )} \log \left (-2 \, x\right ) + 16 \, x - 10 \, e^{\left (2 \, x\right )} - 4}{4 \, {\left (x + 4\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.28, size = 33, normalized size = 1.22 \begin {gather*} \frac {17\,x-10\,{\mathrm {e}}^{2\,x}+4\,x^2-\ln \left (-2\,x\right )\,{\mathrm {e}}^{2\,x}}{4\,\left (x+4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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