Integrand size = 41, antiderivative size = 19 \[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx=3+e^{\frac {e^{-3-3 x} x}{\log (4 x)}} \] Output:
exp(exp(-ln(ln(4*x))+ln(x)-3*x-3))+3
Time = 1.83 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx=e^{\frac {e^{-3-3 x} x}{\log (4 x)}} \] Input:
Integrate[(E^(-3 - 3*x + (E^(-3 - 3*x)*x)/Log[4*x])*(-1 + (1 - 3*x)*Log[4* x]))/Log[4*x]^2,x]
Output:
E^((E^(-3 - 3*x)*x)/Log[4*x])
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.
\(\displaystyle \int \frac {e^{-3 x+\frac {e^{-3 x-3} x}{\log (4 x)}-3} ((1-3 x) \log (4 x)-1)}{\log ^2(4 x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(1-3 x) e^{-3 x+\frac {e^{-3 x-3} x}{\log (4 x)}-3}}{\log (4 x)}-\frac {e^{-3 x+\frac {e^{-3 x-3} x}{\log (4 x)}-3}}{\log ^2(4 x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{\frac {e^{-3 x-3} x}{\log (4 x)}-3 x-3}}{\log ^2(4 x)}dx+\int \frac {e^{\frac {e^{-3 x-3} x}{\log (4 x)}-3 x-3}}{\log (4 x)}dx-3 \int \frac {e^{\frac {e^{-3 x-3} x}{\log (4 x)}-3 x-3} x}{\log (4 x)}dx\) |
Input:
Int[(E^(-3 - 3*x + (E^(-3 - 3*x)*x)/Log[4*x])*(-1 + (1 - 3*x)*Log[4*x]))/L og[4*x]^2,x]
Output:
$Aborted
Time = 1.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{-\ln \left (\ln \left (4 x \right )\right )+\ln \left (x \right )-3 x -3}}\) | \(17\) |
risch | \({\mathrm e}^{\frac {x \,{\mathrm e}^{-3 x -3}}{\ln \left (x \right )+2 \ln \left (2\right )}}\) | \(19\) |
Input:
int(((1-3*x)*ln(4*x)-1)*exp(-ln(ln(4*x))+ln(x)-3*x-3)*exp(exp(-ln(ln(4*x)) +ln(x)-3*x-3))/x/ln(4*x),x,method=_RETURNVERBOSE)
Output:
exp(exp(-ln(ln(4*x))+ln(x)-3*x-3))
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx=e^{\left (e^{\left (-3 \, x + \log \left (x\right ) - \log \left (2 \, \log \left (2\right ) + \log \left (x\right )\right ) - 3\right )}\right )} \] Input:
integrate(((1-3*x)*log(4*x)-1)*exp(-log(log(4*x))+log(x)-3*x-3)*exp(exp(-l og(log(4*x))+log(x)-3*x-3))/x/log(4*x),x, algorithm="fricas")
Output:
e^(e^(-3*x + log(x) - log(2*log(2) + log(x)) - 3))
Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx=e^{\frac {x e^{- 3 x - 3}}{\log {\left (4 x \right )}}} \] Input:
integrate(((1-3*x)*ln(4*x)-1)*exp(-ln(ln(4*x))+ln(x)-3*x-3)*exp(exp(-ln(ln (4*x))+ln(x)-3*x-3))/x/ln(4*x),x)
Output:
exp(x*exp(-3*x - 3)/log(4*x))
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx=e^{\left (\frac {x}{2 \, e^{\left (3 \, x + 3\right )} \log \left (2\right ) + e^{\left (3 \, x + 3\right )} \log \left (x\right )}\right )} \] Input:
integrate(((1-3*x)*log(4*x)-1)*exp(-log(log(4*x))+log(x)-3*x-3)*exp(exp(-l og(log(4*x))+log(x)-3*x-3))/x/log(4*x),x, algorithm="maxima")
Output:
e^(x/(2*e^(3*x + 3)*log(2) + e^(3*x + 3)*log(x)))
\[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx=\int { -\frac {{\left ({\left (3 \, x - 1\right )} \log \left (4 \, x\right ) + 1\right )} e^{\left (-3 \, x + e^{\left (-3 \, x + \log \left (x\right ) - \log \left (\log \left (4 \, x\right )\right ) - 3\right )} + \log \left (x\right ) - \log \left (\log \left (4 \, x\right )\right ) - 3\right )}}{x \log \left (4 \, x\right )} \,d x } \] Input:
integrate(((1-3*x)*log(4*x)-1)*exp(-log(log(4*x))+log(x)-3*x-3)*exp(exp(-l og(log(4*x))+log(x)-3*x-3))/x/log(4*x),x, algorithm="giac")
Output:
integrate(-((3*x - 1)*log(4*x) + 1)*e^(-3*x + e^(-3*x + log(x) - log(log(4 *x)) - 3) + log(x) - log(log(4*x)) - 3)/(x*log(4*x)), x)
Time = 2.83 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx={\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-3\,x}\,{\mathrm {e}}^{-3}}{\ln \left (4\,x\right )}} \] Input:
int(-(exp(log(x) - log(log(4*x)) - 3*x - 3)*exp(exp(log(x) - log(log(4*x)) - 3*x - 3))*(log(4*x)*(3*x - 1) + 1))/(x*log(4*x)),x)
Output:
exp((x*exp(-3*x)*exp(-3))/log(4*x))
\[ \int \frac {e^{-3-3 x+\frac {e^{-3-3 x} x}{\log (4 x)}} (-1+(1-3 x) \log (4 x))}{\log ^2(4 x)} \, dx=\frac {-\left (\int \frac {e^{\frac {x}{e^{3 x} \mathrm {log}\left (4 x \right ) e^{3}}}}{e^{3 x} \mathrm {log}\left (4 x \right )^{2}}d x \right )+\int \frac {e^{\frac {x}{e^{3 x} \mathrm {log}\left (4 x \right ) e^{3}}}}{e^{3 x} \mathrm {log}\left (4 x \right )}d x -3 \left (\int \frac {e^{\frac {x}{e^{3 x} \mathrm {log}\left (4 x \right ) e^{3}}} x}{e^{3 x} \mathrm {log}\left (4 x \right )}d x \right )}{e^{3}} \] Input:
int(((1-3*x)*log(4*x)-1)*exp(-log(log(4*x))+log(x)-3*x-3)*exp(exp(-log(log (4*x))+log(x)-3*x-3))/x/log(4*x),x)
Output:
( - int(e**(x/(e**(3*x)*log(4*x)*e**3))/(e**(3*x)*log(4*x)**2),x) + int(e* *(x/(e**(3*x)*log(4*x)*e**3))/(e**(3*x)*log(4*x)),x) - 3*int((e**(x/(e**(3 *x)*log(4*x)*e**3))*x)/(e**(3*x)*log(4*x)),x))/e**3