Optimal. Leaf size=27 \[ 3+x+\frac {x}{\left (5-x-\frac {3}{2} \left (e^x+x\right )\right ) (-6+\log (2))} \]
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Rubi [F] time = 0.81, 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 {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-9 e^{2 x} (-6+\log (2))-5 \left (-20 x (-6+\log (2))+5 x^2 (-6+\log (2))+4 (-29+\log (32))\right )-6 e^x (59-10 \log (2)+x (-29+\log (32)))}{\left (3 e^x+5 (-2+x)\right )^2 (6-\log (2))} \, dx\\ &=\frac {\int \frac {-9 e^{2 x} (-6+\log (2))-5 \left (-20 x (-6+\log (2))+5 x^2 (-6+\log (2))+4 (-29+\log (32))\right )-6 e^x (59-10 \log (2)+x (-29+\log (32)))}{\left (3 e^x+5 (-2+x)\right )^2} \, dx}{6-\log (2)}\\ &=\frac {\int \left (\frac {10 (-3+x) x}{\left (-10+3 e^x+5 x\right )^2}-\frac {2 (-1+x)}{-10+3 e^x+5 x}+6 \left (1-\frac {\log (2)}{6}\right )\right ) \, dx}{6-\log (2)}\\ &=x-\frac {2 \int \frac {-1+x}{-10+3 e^x+5 x} \, dx}{6-\log (2)}+\frac {10 \int \frac {(-3+x) x}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)}\\ &=x-\frac {2 \int \left (-\frac {1}{-10+3 e^x+5 x}+\frac {x}{-10+3 e^x+5 x}\right ) \, dx}{6-\log (2)}+\frac {10 \int \left (-\frac {3 x}{\left (-10+3 e^x+5 x\right )^2}+\frac {x^2}{\left (-10+3 e^x+5 x\right )^2}\right ) \, dx}{6-\log (2)}\\ &=x+\frac {2 \int \frac {1}{-10+3 e^x+5 x} \, dx}{6-\log (2)}-\frac {2 \int \frac {x}{-10+3 e^x+5 x} \, dx}{6-\log (2)}+\frac {10 \int \frac {x^2}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)}-\frac {30 \int \frac {x}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.33, size = 26, normalized size = 0.96 \begin {gather*} \frac {x \left (-6+\frac {2}{10-3 e^x-5 x}+\log (2)\right )}{-6+\log (2)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.06, size = 56, normalized size = 2.07 \begin {gather*} -\frac {30 \, x^{2} - 3 \, {\left (x \log \relax (2) - 6 \, x\right )} e^{x} - 5 \, {\left (x^{2} - 2 \, x\right )} \log \relax (2) - 58 \, x}{3 \, {\left (\log \relax (2) - 6\right )} e^{x} + 5 \, {\left (x - 2\right )} \log \relax (2) - 30 \, x + 60} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.14, size = 60, normalized size = 2.22 \begin {gather*} \frac {5 \, x^{2} \log \relax (2) + 3 \, x e^{x} \log \relax (2) - 30 \, x^{2} - 18 \, x e^{x} - 10 \, x \log \relax (2) + 58 \, x}{5 \, x \log \relax (2) + 3 \, e^{x} \log \relax (2) - 30 \, x - 18 \, e^{x} - 10 \, \log \relax (2) + 60} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 23, normalized size = 0.85
method | result | size |
risch | \(x -\frac {2 x}{\left (\ln \relax (2)-6\right ) \left (-10+5 x +3 \,{\mathrm e}^{x}\right )}\) | \(23\) |
norman | \(\frac {\frac {6 \left (5 \ln \relax (2)-29\right ) {\mathrm e}^{x}}{5 \left (\ln \relax (2)-6\right )}+5 x^{2}+3 \,{\mathrm e}^{x} x -\frac {4 \left (5 \ln \relax (2)-29\right )}{\ln \relax (2)-6}}{-10+5 x +3 \,{\mathrm e}^{x}}\) | \(54\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.47, size = 52, normalized size = 1.93 \begin {gather*} \frac {5 \, x^{2} {\left (\log \relax (2) - 6\right )} + 3 \, x {\left (\log \relax (2) - 6\right )} e^{x} - 2 \, x {\left (5 \, \log \relax (2) - 29\right )}}{5 \, x {\left (\log \relax (2) - 6\right )} + 3 \, {\left (\log \relax (2) - 6\right )} e^{x} - 10 \, \log \relax (2) + 60} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.55, size = 54, normalized size = 2.00 \begin {gather*} -\frac {x^2\,\left (\ln \left (32\right )-30\right )-x\,\left (10\,\ln \relax (2)-58\right )+x\,{\mathrm {e}}^x\,\left (\ln \relax (8)-18\right )}{30\,x+10\,\ln \relax (2)+18\,{\mathrm {e}}^x-5\,x\,\ln \relax (2)-3\,{\mathrm {e}}^x\,\ln \relax (2)-60} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 31, normalized size = 1.15 \begin {gather*} x - \frac {2 x}{- 30 x + 5 x \log {\relax (2 )} + \left (-18 + 3 \log {\relax (2 )}\right ) e^{x} - 10 \log {\relax (2 )} + 60} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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