Optimal. Leaf size=22 \[ 1+x-\frac {1}{5} \log \left (-2-e^5+x+\log \left (\frac {3 x}{2}\right )\right ) \]
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Rubi [A]
time = 0.44, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps
used = 7, number of rules used = 5, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 6873, 12,
6874, 6816} \begin {gather*} x-\frac {1}{5} \log \left (-x-\log \left (\frac {3 x}{2}\right )+e^5+2\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 6816
Rule 6873
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+\left (-11-5 e^5\right ) x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )}{-10 x-5 e^5 x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )} \, dx\\ &=\int \frac {-1+\left (-11-5 e^5\right ) x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )}{\left (-10-5 e^5\right ) x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )} \, dx\\ &=\int \frac {1-\left (-11-5 e^5\right ) x-5 x^2-5 x \log \left (\frac {3 x}{2}\right )}{5 x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \frac {1-\left (-11-5 e^5\right ) x-5 x^2-5 x \log \left (\frac {3 x}{2}\right )}{x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \left (5+\frac {1+x}{x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )}\right ) \, dx\\ &=x+\frac {1}{5} \int \frac {1+x}{x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )} \, dx\\ &=x-\frac {1}{5} \log \left (2+e^5-x-\log \left (\frac {3 x}{2}\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.05, size = 27, normalized size = 1.23 \begin {gather*} \frac {1}{5} \left (5 x-\log \left (2+e^5-x-\log \left (\frac {3 x}{2}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.97, size = 21, normalized size = 0.95
method | result | size |
risch | \(x -\frac {\ln \left (x -{\mathrm e}^{5}+\ln \left (\frac {3 x}{2}\right )-2\right )}{5}\) | \(17\) |
norman | \(x -\frac {\ln \left ({\mathrm e}^{5}-\ln \left (\frac {3 x}{2}\right )-x +2\right )}{5}\) | \(19\) |
derivativedivides | \(x -\frac {\ln \left (-3 \ln \left (\frac {3 x}{2}\right )+3 \,{\mathrm e}^{5}-3 x +6\right )}{5}\) | \(21\) |
default | \(x -\frac {\ln \left (-3 \ln \left (\frac {3 x}{2}\right )+3 \,{\mathrm e}^{5}-3 x +6\right )}{5}\) | \(21\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.58, size = 20, normalized size = 0.91 \begin {gather*} x - \frac {1}{5} \, \log \left (x - e^{5} + \log \left (3\right ) - \log \left (2\right ) + \log \left (x\right ) - 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 16, normalized size = 0.73 \begin {gather*} x - \frac {1}{5} \, \log \left (x - e^{5} + \log \left (\frac {3}{2} \, x\right ) - 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.06, size = 17, normalized size = 0.77 \begin {gather*} x - \frac {\log {\left (x + \log {\left (\frac {3 x}{2} \right )} - e^{5} - 2 \right )}}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 18, normalized size = 0.82 \begin {gather*} x - \frac {1}{5} \, \log \left (-x + e^{5} - \log \left (\frac {3}{2} \, x\right ) + 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.33, size = 16, normalized size = 0.73 \begin {gather*} x-\frac {\ln \left (x+\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^5-2\right )}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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