Optimal. Leaf size=29 \[ \frac {1}{5} \log (x) \log \left (e^e \left (3 e^{4 x}-\frac {2}{5-x}\right ) x\right ) \]
________________________________________________________________________________________
Rubi [F] time = 0.91, 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 {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {\left (-10+3 e^{4 x} (-5+x)^2 (1+4 x)\right ) \log (x)}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)}+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{5 x} \, dx\\ &=\frac {1}{5} \int \frac {\frac {\left (-10+3 e^{4 x} (-5+x)^2 (1+4 x)\right ) \log (x)}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)}+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (-\frac {2 (-19+4 x) \log (x)}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )}+\frac {\log (x)+4 x \log (x)+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {\log (x)+4 x \log (x)+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx-\frac {2}{5} \int \frac {(-19+4 x) \log (x)}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx\\ &=\frac {1}{5} \int \left (\frac {(1+4 x) \log (x)}{x}+\frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x}\right ) \, dx+\frac {2}{5} \int \frac {4 \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx+\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ &=\frac {1}{5} \int \frac {(1+4 x) \log (x)}{x} \, dx+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \left (\frac {4 \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x}+\frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x}\right ) \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ &=\frac {1}{5} \int \frac {\log (x)}{x} \, dx+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx+\frac {4}{5} \int \log (x) \, dx+\frac {8}{5} \int \frac {\int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ &=-\frac {4 x}{5}+\frac {4}{5} x \log (x)+\frac {\log ^2(x)}{10}+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx+\frac {8}{5} \int \frac {\int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.29, size = 37, normalized size = 1.28 \begin {gather*} \frac {1}{5} \left (e \log (x)+\log (x) \log \left (\frac {x \left (2-15 e^{4 x}+3 e^{4 x} x\right )}{-5+x}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.47, size = 31, normalized size = 1.07 \begin {gather*} \frac {1}{5} \, \log \relax (x) \log \left (\frac {{\left (3 \, {\left (x^{2} - 5 \, x\right )} e^{\left (4 \, x\right )} + 2 \, x\right )} e^{e}}{x - 5}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.30, size = 41, normalized size = 1.41 \begin {gather*} \frac {1}{5} \, e \log \relax (x) + \frac {1}{5} \, \log \left (3 \, x e^{\left (4 \, x\right )} - 15 \, e^{\left (4 \, x\right )} + 2\right ) \log \relax (x) - \frac {1}{5} \, \log \left (x - 5\right ) \log \relax (x) + \frac {1}{5} \, \log \relax (x)^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [C] time = 0.24, size = 405, normalized size = 13.97
method | result | size |
risch | \(\frac {\ln \relax (x ) \ln \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{5}-\frac {\ln \relax (x ) \ln \left (x -5\right )}{5}+\frac {\ln \relax (x )^{2}}{5}+\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2}}{10}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )}{10}+\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2}}{10}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{3}}{10}+\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2}}{10}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{3}}{10}+\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )^{2}}{10}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{x -5}\right )}{10}+\frac {\ln \relax (3) \ln \relax (x )}{5}+\frac {{\mathrm e} \ln \relax (x )}{5}\) | \(405\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.41, size = 37, normalized size = 1.28 \begin {gather*} \frac {1}{5} \, e \log \relax (x) + \frac {1}{5} \, \log \left (3 \, {\left (x - 5\right )} e^{\left (4 \, x\right )} + 2\right ) \log \relax (x) - \frac {1}{5} \, \log \left (x - 5\right ) \log \relax (x) + \frac {1}{5} \, \log \relax (x)^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 3.54, size = 33, normalized size = 1.14 \begin {gather*} \frac {\ln \relax (x)\,\left (\ln \left (\frac {2\,x-{\mathrm {e}}^{4\,x}\,\left (15\,x-3\,x^2\right )}{x-5}\right )+\mathrm {e}\right )}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.96, size = 31, normalized size = 1.07 \begin {gather*} \frac {\log {\relax (x )} \log {\left (\frac {\left (2 x + \left (3 x^{2} - 15 x\right ) e^{4 x}\right ) e^{e}}{x - 5} \right )}}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________