Integrand size = 48, antiderivative size = 26 \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\frac {32 e^{-2 x} \log ^2(x)}{3 x^2 \log ^2\left (\frac {x^2}{4}\right )} \] Output:
32/3*exp(ln(ln(x)/x/ln(1/4*x^2))-x)^2
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\frac {32 e^{-2 x} \log ^2(x)}{3 x^2 \log ^2\left (\frac {x^2}{4}\right )} \] Input:
Integrate[(Log[x]*(-128*Log[x] + (64 + (-64 - 64*x)*Log[x])*Log[x^2/4]))/( 3*E^(2*x)*x^3*Log[x^2/4]^3),x]
Output:
(32*Log[x]^2)/(3*E^(2*x)*x^2*Log[x^2/4]^2)
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-2 x} \log (x) \left (((-64 x-64) \log (x)+64) \log \left (\frac {x^2}{4}\right )-128 \log (x)\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {64 e^{-2 x} \log (x) \left (2 \log (x)-(1-(x+1) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{x^3 \log ^3\left (\frac {x^2}{4}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {64}{3} \int \frac {e^{-2 x} \log (x) \left (2 \log (x)-(1-(x+1) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{x^3 \log ^3\left (\frac {x^2}{4}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {64}{3} \int \left (\frac {2 e^{-2 x} \log ^2(x)}{x^3 \log ^3\left (\frac {x^2}{4}\right )}+\frac {e^{-2 x} (x \log (x)+\log (x)-1) \log (x)}{x^3 \log ^2\left (\frac {x^2}{4}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {64}{3} \left (\int \frac {e^{-2 x} \log ^2(x)}{x^2 \log ^2\left (\frac {x^2}{4}\right )}dx-\int \frac {e^{-2 x} \log (x)}{x^3 \log ^2\left (\frac {x^2}{4}\right )}dx+\int \frac {e^{-2 x} \log ^2(x)}{x^3 \log ^2\left (\frac {x^2}{4}\right )}dx+2 \int \frac {e^{-2 x} \log ^2(x)}{x^3 \log ^3\left (\frac {x^2}{4}\right )}dx\right )\) |
Input:
Int[(Log[x]*(-128*Log[x] + (64 + (-64 - 64*x)*Log[x])*Log[x^2/4]))/(3*E^(2 *x)*x^3*Log[x^2/4]^3),x]
Output:
$Aborted
Time = 3.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\frac {32 \ln \left (x \right )^{2} {\mathrm e}^{-2 x}}{3 \ln \left (\frac {x^{2}}{4}\right )^{2} x^{2}}\) | \(25\) |
Input:
int(1/3*(((-64*x-64)*ln(x)+64)*ln(1/4*x^2)-128*ln(x))*exp(ln(ln(x)/x/ln(1/ 4*x^2))-x)^2/x/ln(x)/ln(1/4*x^2),x,method=_RETURNVERBOSE)
Output:
32/3*exp(ln(ln(x)/x/ln(1/4*x^2))-x)^2
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\frac {32}{3} \, e^{\left (-2 \, x + 2 \, \log \left (-\frac {\log \left (x\right )}{2 \, {\left (x \log \left (2\right ) - x \log \left (x\right )\right )}}\right )\right )} \] Input:
integrate(1/3*(((-64*x-64)*log(x)+64)*log(1/4*x^2)-128*log(x))*exp(log(log (x)/x/log(1/4*x^2))-x)^2/x/log(x)/log(1/4*x^2),x, algorithm="fricas")
Output:
32/3*e^(-2*x + 2*log(-1/2*log(x)/(x*log(2) - x*log(x))))
Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\frac {8 e^{- 2 x} \log {\left (x \right )}^{2}}{3 x^{2} \log {\left (x \right )}^{2} - 6 x^{2} \log {\left (2 \right )} \log {\left (x \right )} + 3 x^{2} \log {\left (2 \right )}^{2}} \] Input:
integrate(1/3*(((-64*x-64)*ln(x)+64)*ln(1/4*x**2)-128*ln(x))*exp(ln(ln(x)/ x/ln(1/4*x**2))-x)**2/x/ln(x)/ln(1/4*x**2),x)
Output:
8*exp(-2*x)*log(x)**2/(3*x**2*log(x)**2 - 6*x**2*log(2)*log(x) + 3*x**2*lo g(2)**2)
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\frac {8 \, e^{\left (-2 \, x\right )} \log \left (x\right )^{2}}{3 \, {\left (x^{2} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (2\right ) \log \left (x\right ) + x^{2} \log \left (x\right )^{2}\right )}} \] Input:
integrate(1/3*(((-64*x-64)*log(x)+64)*log(1/4*x^2)-128*log(x))*exp(log(log (x)/x/log(1/4*x^2))-x)^2/x/log(x)/log(1/4*x^2),x, algorithm="maxima")
Output:
8/3*e^(-2*x)*log(x)^2/(x^2*log(2)^2 - 2*x^2*log(2)*log(x) + x^2*log(x)^2)
Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\frac {32}{3} \, e^{\left (-2 \, x + 2 \, \log \left (\frac {\log \left (x\right )}{x \log \left (\frac {1}{4} \, x^{2}\right )}\right )\right )} \] Input:
integrate(1/3*(((-64*x-64)*log(x)+64)*log(1/4*x^2)-128*log(x))*exp(log(log (x)/x/log(1/4*x^2))-x)^2/x/log(x)/log(1/4*x^2),x, algorithm="giac")
Output:
32/3*e^(-2*x + 2*log(log(x)/(x*log(1/4*x^2))))
Timed out. \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\int -\frac {{\mathrm {e}}^{2\,\ln \left (\frac {\ln \left (x\right )}{x\,\ln \left (\frac {x^2}{4}\right )}\right )-2\,x}\,\left (128\,\ln \left (x\right )+\ln \left (\frac {x^2}{4}\right )\,\left (\ln \left (x\right )\,\left (64\,x+64\right )-64\right )\right )}{3\,x\,\ln \left (\frac {x^2}{4}\right )\,\ln \left (x\right )} \,d x \] Input:
int(-(exp(2*log(log(x)/(x*log(x^2/4))) - 2*x)*(128*log(x) + log(x^2/4)*(lo g(x)*(64*x + 64) - 64)))/(3*x*log(x^2/4)*log(x)),x)
Output:
int(-(exp(2*log(log(x)/(x*log(x^2/4))) - 2*x)*(128*log(x) + log(x^2/4)*(lo g(x)*(64*x + 64) - 64)))/(3*x*log(x^2/4)*log(x)), x)
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx=\frac {32 \mathrm {log}\left (x \right )^{2}}{3 e^{2 x} \mathrm {log}\left (\frac {x^{2}}{4}\right )^{2} x^{2}} \] Input:
int(1/3*(((-64*x-64)*log(x)+64)*log(1/4*x^2)-128*log(x))*exp(log(log(x)/x/ log(1/4*x^2))-x)^2/x/log(x)/log(1/4*x^2),x)
Output:
(32*log(x)**2)/(3*e**(2*x)*log(x**2/4)**2*x**2)