Optimal. Leaf size=34 \[ x+\frac {3 x \left (i \pi +\log \left (-5+5 e^3\right )\right )}{20 \left (\frac {3}{5}-e^{2 x}\right )} \]
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Rubi [B] time = 0.79, antiderivative size = 205, normalized size of antiderivative = 6.03, number of steps used = 18, number of rules used = 13, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {6741, 12, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 31, 29} \begin {gather*} \frac {1}{4} x^2 \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )+x-\frac {1}{16} (1-2 x)^2 \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )-\frac {1}{8} (1-2 x) \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{8} \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (3-5 e^{2 x}\right )-\frac {1}{4} x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {3 x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )}{4 \left (3-5 e^{2 x}\right )}-\frac {1}{4} x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 29
Rule 31
Rule 36
Rule 2184
Rule 2185
Rule 2190
Rule 2191
Rule 2279
Rule 2282
Rule 2391
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{4 \left (3-5 e^{2 x}\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{\left (3-5 e^{2 x}\right )^2} \, dx\\ &=\frac {1}{4} \int \left (4+\frac {3 (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{3-5 e^{2 x}}+\frac {18 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{\left (3-5 e^{2 x}\right )^2}\right ) \, dx\\ &=x+\frac {1}{4} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {1-2 x}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (9 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {x}{\left (3-5 e^{2 x}\right )^2} \, dx\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {1}{4} \left (5 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {e^{2 x} (1-2 x)}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {x}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (15 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {e^{2 x} x}{\left (3-5 e^{2 x}\right )^2} \, dx\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{4} \left (-i \pi -\log \left (-5 \left (1-e^3\right )\right )\right ) \int \log \left (1-\frac {5 e^{2 x}}{3}\right ) \, dx-\frac {1}{4} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {1}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (5 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {e^{2 x} x}{3-5 e^{2 x}} \, dx\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{8} \left (-i \pi -\log \left (-5 \left (1-e^3\right )\right )\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {5 x}{3}\right )}{x} \, dx,x,e^{2 x}\right )+\frac {1}{4} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \int \log \left (1-\frac {5 e^{2 x}}{3}\right ) \, dx-\frac {1}{8} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(3-5 x) x} \, dx,x,e^{2 x}\right )\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \text {Li}_2\left (\frac {5 e^{2 x}}{3}\right )-\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )+\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {5 x}{3}\right )}{x} \, dx,x,e^{2 x}\right )-\frac {1}{8} \left (5 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{3-5 x} \, dx,x,e^{2 x}\right )\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (3-5 e^{2 x}\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.27, size = 35, normalized size = 1.03 \begin {gather*} \frac {1}{4} x \left (4+\frac {3 i \pi +\log (125)+3 \log \left (-1+e^3\right )}{3-5 e^{2 x}}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.98, size = 33, normalized size = 0.97 \begin {gather*} \frac {20 \, x e^{\left (2 \, x\right )} - 3 \, x \log \left (-5 \, e^{3} + 5\right ) - 12 \, x}{4 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 33, normalized size = 0.97 \begin {gather*} \frac {20 \, x e^{\left (2 \, x\right )} - 3 \, x \log \left (-5 \, e^{3} + 5\right ) - 12 \, x}{4 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 26, normalized size = 0.76
method | result | size |
risch | \(x -\frac {3 x \left (\ln \relax (5)+\ln \left (-{\mathrm e}^{3}+1\right )\right )}{4 \left (5 \,{\mathrm e}^{2 x}-3\right )}\) | \(26\) |
norman | \(\frac {\left (-3-\frac {3 \ln \relax (5)}{4}-\frac {3 \ln \left (-{\mathrm e}^{3}+1\right )}{4}\right ) x +5 x \,{\mathrm e}^{2 x}}{5 \,{\mathrm e}^{2 x}-3}\) | \(37\) |
default | \(\ln \left ({\mathrm e}^{x}\right )+\frac {\ln \relax (5) \ln \left ({\mathrm e}^{x}\right )}{4}+\frac {\ln \left (-{\mathrm e}^{3}+1\right ) \ln \left ({\mathrm e}^{x}\right )}{4}-\frac {5 \ln \left (-{\mathrm e}^{3}+1\right ) x \,{\mathrm e}^{2 x}}{4 \left (5 \,{\mathrm e}^{2 x}-3\right )}-\frac {5 \ln \relax (5) x \,{\mathrm e}^{2 x}}{4 \left (5 \,{\mathrm e}^{2 x}-3\right )}\) | \(67\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.40, size = 93, normalized size = 2.74 \begin {gather*} \frac {1}{8} \, {\left (2 \, x - \frac {3}{5 \, e^{\left (2 \, x\right )} - 3} - \log \left (5 \, e^{\left (2 \, x\right )} - 3\right )\right )} \log \left (-5 \, e^{3} + 5\right ) - \frac {1}{8} \, {\left (\frac {10 \, x e^{\left (2 \, x\right )}}{5 \, e^{\left (2 \, x\right )} - 3} - \log \left (e^{\left (2 \, x\right )} - \frac {3}{5}\right )\right )} \log \left (-5 \, e^{3} + 5\right ) + x + \frac {3 \, \log \left (-5 \, e^{3} + 5\right )}{8 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.21, size = 22, normalized size = 0.65 \begin {gather*} x-\frac {3\,x\,\ln \left (5-5\,{\mathrm {e}}^3\right )}{20\,{\mathrm {e}}^{2\,x}-12} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.22, size = 32, normalized size = 0.94 \begin {gather*} x + \frac {3 x \log {\relax (5 )} + 3 x \log {\left (-1 + e^{3} \right )} + 3 i \pi x}{12 - 20 e^{2 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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