Optimal. Leaf size=23 \[ \frac {15 \log (x) \log \left (\frac {5}{-e^3+2 x}\right )}{4 x} \]
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Rubi [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in
optimal.
time = 1.20, antiderivative size = 115, normalized size of antiderivative = 5.00, number of steps
used = 24, number of rules used = 14, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.203, Rules used = {1607, 6820,
12, 14, 2442, 36, 31, 29, 6874, 2379, 2438, 2423, 2439, 2338} \begin {gather*} \frac {15 \text {PolyLog}\left (2,\frac {e^3}{2 x}\right )}{2 e^3}+\frac {15 \text {PolyLog}\left (2,\frac {2 x}{e^3}\right )}{2 e^3}-\frac {15 \log ^2(x)}{4 e^3}-\frac {15 \log \left (1-\frac {e^3}{2 x}\right ) \log (x)}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}-\frac {45 \log (x)}{2 e^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 29
Rule 31
Rule 36
Rule 1607
Rule 2338
Rule 2379
Rule 2423
Rule 2438
Rule 2439
Rule 2442
Rule 6820
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{\left (4 e^3-8 x\right ) x^2} \, dx\\ &=\int \frac {15 \left (-\log \left (-\frac {5}{e^3-2 x}\right ) (-1+\log (x))+\frac {2 x \log (x)}{e^3-2 x}\right )}{4 x^2} \, dx\\ &=\frac {15}{4} \int \frac {-\log \left (-\frac {5}{e^3-2 x}\right ) (-1+\log (x))+\frac {2 x \log (x)}{e^3-2 x}}{x^2} \, dx\\ &=\frac {15}{4} \int \left (\frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2}-\frac {\left (2 x-e^3 \log \left (-\frac {5}{e^3-2 x}\right )+2 x \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{x^2 \left (-e^3+2 x\right )}\right ) \, dx\\ &=\frac {15}{4} \int \frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2} \, dx-\frac {15}{4} \int \frac {\left (2 x-e^3 \log \left (-\frac {5}{e^3-2 x}\right )+2 x \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{x^2 \left (-e^3+2 x\right )} \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15}{4} \int \frac {\left (-2 x+\left (e^3-2 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{\left (e^3-2 x\right ) x^2} \, dx+\frac {15}{2} \int \frac {1}{\left (e^3-2 x\right ) x} \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15}{4} \int \frac {\left (-\frac {2 x}{e^3-2 x}+\log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{x^2} \, dx+\frac {15 \int \frac {1}{x} \, dx}{2 e^3}+\frac {15 \int \frac {1}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}-\frac {15}{4} \int \left (-\frac {2 \log (x)}{\left (e^3-2 x\right ) x}+\frac {\log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{x^2}\right ) \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}-\frac {15}{4} \int \frac {\log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{x^2} \, dx+\frac {15}{2} \int \frac {\log (x)}{\left (e^3-2 x\right ) x} \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}-\frac {15 \log ^2(x)}{2 e^3}+\frac {15}{4} \int \left (-\frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2}-\frac {2 \log \left (e^3-2 x\right )}{e^3 x}+\frac {2 \log (x)}{e^3 x}\right ) \, dx+\frac {15 \int \frac {\log (x)}{x} \, dx}{2 e^3}+\frac {15 \int \frac {\log (x)}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}-\frac {15 \log ^2(x)}{4 e^3}-\frac {15}{4} \int \frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2} \, dx-\frac {15 \int \frac {\log \left (e^3-2 x\right )}{x} \, dx}{2 e^3}+\frac {15 \int \frac {\log (x)}{x} \, dx}{2 e^3}+\frac {15 \int \frac {\log \left (\frac {2 x}{e^3}\right )}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 \log (x)}{e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}+\frac {15 \text {Li}_2\left (1-\frac {2 x}{e^3}\right )}{2 e^3}-\frac {15}{2} \int \frac {1}{\left (e^3-2 x\right ) x} \, dx-\frac {15 \int \frac {\log \left (1-\frac {2 x}{e^3}\right )}{x} \, dx}{2 e^3}\\ &=-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 \log (x)}{e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}+\frac {15 \text {Li}_2\left (\frac {2 x}{e^3}\right )}{2 e^3}+\frac {15 \text {Li}_2\left (1-\frac {2 x}{e^3}\right )}{2 e^3}-\frac {15 \int \frac {1}{x} \, dx}{2 e^3}-\frac {15 \int \frac {1}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}-\frac {45 \log (x)}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}+\frac {15 \text {Li}_2\left (\frac {2 x}{e^3}\right )}{2 e^3}+\frac {15 \text {Li}_2\left (1-\frac {2 x}{e^3}\right )}{2 e^3}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.05, size = 21, normalized size = 0.91 \begin {gather*} \frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 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. \(147\) vs.
\(2(20)=40\).
time = 2.34, size = 148, normalized size = 6.43
method | result | size |
risch | \(-\frac {15 \ln \left (x \right ) \ln \left ({\mathrm e}^{3}-2 x \right )}{4 x}+\frac {15 \left (-2 i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-2 x}\right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-2 x}\right )^{3}+2 i \pi +2 \ln \left (5\right )\right ) \ln \left (x \right )}{8 x}\) | \(68\) |
default | \(-\frac {15 \left (\ln \left (x \right ) \ln \left ({\mathrm e}^{3}-2 x \right )-\ln \left (-\frac {1}{{\mathrm e}^{3}-2 x}\right )-\left (\ln \left (-\frac {1}{{\mathrm e}^{3}-2 x}\right )+\ln \left ({\mathrm e}^{3}-2 x \right )\right ) \ln \left (x \right )+2 \,{\mathrm e}^{-3} \ln \left (x \right ) x -2 \,{\mathrm e}^{-3} x \ln \left ({\mathrm e}^{3}-2 x \right )\right )}{4 x}+\frac {15 \ln \left (5\right ) \ln \left (x \right )}{4 x}+\frac {15 \ln \left (\frac {{\mathrm e}^{3}}{-{\mathrm e}^{3}+2 x}+1\right ) {\mathrm e}^{-3}}{2}-\frac {15 \ln \left (\frac {1}{-{\mathrm e}^{3}+2 x}\right )}{2 \left (-{\mathrm e}^{3}+2 x \right ) \left (\frac {{\mathrm e}^{3}}{-{\mathrm e}^{3}+2 x}+1\right )}\) | \(148\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.57, size = 24, normalized size = 1.04 \begin {gather*} \frac {15 \, {\left (\log \left (5\right ) \log \left (x\right ) - \log \left (2 \, x - e^{3}\right ) \log \left (x\right )\right )}}{4 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 20, normalized size = 0.87 \begin {gather*} \frac {15 \, \log \left (x\right ) \log \left (\frac {5}{2 \, x - e^{3}}\right )}{4 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.32, size = 19, normalized size = 0.83 \begin {gather*} \frac {15 \log {\left (x \right )} \log {\left (- \frac {5}{- 2 x + e^{3}} \right )}}{4 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 69 vs.
\(2 (20) = 40\).
time = 0.44, size = 69, normalized size = 3.00 \begin {gather*} \frac {15 \, {\left (\pi ^{2} \mathrm {sgn}\left (2 \, x - e^{3}\right ) \mathrm {sgn}\left (x\right ) - \pi ^{2} \mathrm {sgn}\left (2 \, x - e^{3}\right ) + 3 \, \pi ^{2} \mathrm {sgn}\left (x\right ) - 3 \, \pi ^{2} + 4 \, \log \left (5\right ) \log \left ({\left | x \right |}\right ) - 4 \, \log \left ({\left | 2 \, x - e^{3} \right |}\right ) \log \left ({\left | x \right |}\right )\right )}}{16 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.50, size = 21, normalized size = 0.91 \begin {gather*} \frac {15\,\ln \left (x\right )\,\left (\ln \left (5\right )+\ln \left (\frac {1}{2\,x-{\mathrm {e}}^3}\right )\right )}{4\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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