Optimal. Leaf size=21 \[ \frac {5 \log (x) \left (e^{5-x}+\log (5 x)\right )}{2 x} \]
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Rubi [A]
time = 0.08, antiderivative size = 30, normalized size of antiderivative = 1.43, number of steps
used = 9, number of rules used = 6, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 14, 2326,
2341, 2340, 2413} \begin {gather*} \frac {5 \log (5 x) \log (x)}{2 x}+\frac {5 e^{5-x} \log (x)}{2 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2326
Rule 2340
Rule 2341
Rule 2413
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{x^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {5 e^{5-x} (-1+\log (x)+x \log (x))}{x^2}-\frac {5 (-\log (x)-\log (5 x)+\log (x) \log (5 x))}{x^2}\right ) \, dx\\ &=-\left (\frac {5}{2} \int \frac {e^{5-x} (-1+\log (x)+x \log (x))}{x^2} \, dx\right )-\frac {5}{2} \int \frac {-\log (x)-\log (5 x)+\log (x) \log (5 x)}{x^2} \, dx\\ &=\frac {5 e^{5-x} \log (x)}{2 x}-\frac {5}{2} \int \left (-\frac {\log (x)}{x^2}+\frac {(-1+\log (x)) \log (5 x)}{x^2}\right ) \, dx\\ &=\frac {5 e^{5-x} \log (x)}{2 x}+\frac {5}{2} \int \frac {\log (x)}{x^2} \, dx-\frac {5}{2} \int \frac {(-1+\log (x)) \log (5 x)}{x^2} \, dx\\ &=-\frac {5}{2 x}-\frac {5 \log (x)}{2 x}+\frac {5 e^{5-x} \log (x)}{2 x}+\frac {5 \log (x) \log (5 x)}{2 x}-\frac {5}{2} \int \frac {\log (x)}{x^2} \, dx\\ &=\frac {5 e^{5-x} \log (x)}{2 x}+\frac {5 \log (x) \log (5 x)}{2 x}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.60, size = 26, normalized size = 1.24 \begin {gather*} \frac {5 e^{-x} \log (x) \left (e^5+e^x \log (5 x)\right )}{2 x} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs.
\(2(18)=36\).
time = 0.55, size = 56, normalized size = 2.67
method | result | size |
risch | \(\frac {5 \ln \left (x \right )^{2}}{2 x}+\frac {5 \left (2 \ln \left (5\right )+2 \,{\mathrm e}^{5-x}\right ) \ln \left (x \right )}{4 x}\) | \(31\) |
default | \(\frac {5 \ln \left (x \right ) \ln \left (5\right )}{2 x}+\frac {5 \ln \left (5\right )}{2 x}+\frac {5 \ln \left (x \right )^{2}}{2 x}+\frac {5 \ln \left (x \right )}{2 x}+\frac {5 \ln \left (x \right ) {\mathrm e}^{5-x}}{2 x}-\frac {5 \ln \left (5 x \right )}{2 x}\) | \(56\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 22, normalized size = 1.05 \begin {gather*} \frac {5 \, {\left ({\left (e^{\left (-x + 5\right )} + \log \left (5\right )\right )} \log \left (x\right ) + \log \left (x\right )^{2}\right )}}{2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.12, size = 34, normalized size = 1.62 \begin {gather*} \frac {5 e^{5 - x} \log {\left (x \right )}}{2 x} + \frac {5 \log {\left (x \right )}^{2}}{2 x} + \frac {5 \log {\left (5 \right )} \log {\left (x \right )}}{2 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 24, normalized size = 1.14 \begin {gather*} \frac {5 \, {\left (e^{\left (-x + 5\right )} \log \left (x\right ) + \log \left (5\right ) \log \left (x\right ) + \log \left (x\right )^{2}\right )}}{2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.15, size = 24, normalized size = 1.14 \begin {gather*} \frac {5\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left ({\mathrm {e}}^5+{\mathrm {e}}^x\,\ln \left (x\right )+{\mathrm {e}}^x\,\ln \left (5\right )\right )}{2\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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