Integrand size = 46, antiderivative size = 20 \[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\frac {\left (e^x+\frac {2+\log (\log (2))}{x}\right ) \log (\log (x))}{x^2} \] Output:
(exp(x)+(ln(ln(2))+2)/x)/x^2*ln(ln(x))
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\frac {\left (2+e^x x+\log (\log (2))\right ) \log (\log (x))}{x^3} \] Input:
Integrate[(2 + E^x*x + Log[Log[2]] + ((-6 + E^x*(-2*x + x^2))*Log[x] - 3*L og[x]*Log[Log[2]])*Log[Log[x]])/(x^4*Log[x]),x]
Output:
((2 + E^x*x + Log[Log[2]])*Log[Log[x]])/x^3
Time = 0.97 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {\left (\left (e^x \left (x^2-2 x\right )-6\right ) \log (x)-3 \log (\log (2)) \log (x)\right ) \log (\log (x))+e^x x+2+\log (\log (2))}{x^4 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (\left (e^x \left (x^2-2 x\right )-6\right ) \log (x)-3 \log (\log (2)) \log (x)\right ) \log (\log (x))+e^x x+2 \left (1+\frac {1}{2} \log (\log (2))\right )}{x^4 \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^x (x \log (x) \log (\log (x))-2 \log (x) \log (\log (x))+1)}{x^3 \log (x)}-\frac {(2+\log (\log (2))) (3 \log (x) \log (\log (x))-1)}{x^4 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(2+\log (\log (2))) \log (\log (x))}{x^3}+\frac {e^x \log (\log (x))}{x^2}\) |
Input:
Int[(2 + E^x*x + Log[Log[2]] + ((-6 + E^x*(-2*x + x^2))*Log[x] - 3*Log[x]* Log[Log[2]])*Log[Log[x]])/(x^4*Log[x]),x]
Output:
(E^x*Log[Log[x]])/x^2 + ((2 + Log[Log[2]])*Log[Log[x]])/x^3
Time = 2.60 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85
method | result | size |
risch | \(\frac {\left ({\mathrm e}^{x} x +\ln \left (\ln \left (2\right )\right )+2\right ) \ln \left (\ln \left (x \right )\right )}{x^{3}}\) | \(17\) |
parallelrisch | \(\frac {\ln \left (\ln \left (x \right )\right ) {\mathrm e}^{x} x +\ln \left (\ln \left (x \right )\right ) \ln \left (\ln \left (2\right )\right )+2 \ln \left (\ln \left (x \right )\right )}{x^{3}}\) | \(25\) |
Input:
int(((-3*ln(x)*ln(ln(2))+((x^2-2*x)*exp(x)-6)*ln(x))*ln(ln(x))+ln(ln(2))+e xp(x)*x+2)/x^4/ln(x),x,method=_RETURNVERBOSE)
Output:
(exp(x)*x+ln(ln(2))+2)/x^3*ln(ln(x))
Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\frac {{\left (x e^{x} + \log \left (\log \left (2\right )\right ) + 2\right )} \log \left (\log \left (x\right )\right )}{x^{3}} \] Input:
integrate(((-3*log(x)*log(log(2))+((x^2-2*x)*exp(x)-6)*log(x))*log(log(x)) +log(log(2))+exp(x)*x+2)/x^4/log(x),x, algorithm="fricas")
Output:
(x*e^x + log(log(2)) + 2)*log(log(x))/x^3
Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\frac {e^{x} \log {\left (\log {\left (x \right )} \right )}}{x^{2}} + \frac {\left (\log {\left (\log {\left (2 \right )} \right )} + 2\right ) \log {\left (\log {\left (x \right )} \right )}}{x^{3}} \] Input:
integrate(((-3*ln(x)*ln(ln(2))+((x**2-2*x)*exp(x)-6)*ln(x))*ln(ln(x))+ln(l n(2))+exp(x)*x+2)/x**4/ln(x),x)
Output:
exp(x)*log(log(x))/x**2 + (log(log(2)) + 2)*log(log(x))/x**3
\[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\int { \frac {x e^{x} + {\left ({\left ({\left (x^{2} - 2 \, x\right )} e^{x} - 6\right )} \log \left (x\right ) - 3 \, \log \left (x\right ) \log \left (\log \left (2\right )\right )\right )} \log \left (\log \left (x\right )\right ) + \log \left (\log \left (2\right )\right ) + 2}{x^{4} \log \left (x\right )} \,d x } \] Input:
integrate(((-3*log(x)*log(log(2))+((x^2-2*x)*exp(x)-6)*log(x))*log(log(x)) +log(log(2))+exp(x)*x+2)/x^4/log(x),x, algorithm="maxima")
Output:
-(log(log(2)) + 2)*integrate(1/(x^4*log(x)), x) + Ei(-3*log(x))*log(log(2) ) + (x*e^x + log(log(2)) + 2)*log(log(x))/x^3 + 2*Ei(-3*log(x))
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\frac {x e^{x} \log \left (\log \left (x\right )\right ) + \log \left (\log \left (2\right )\right ) \log \left (\log \left (x\right )\right ) + 2 \, \log \left (\log \left (x\right )\right )}{x^{3}} \] Input:
integrate(((-3*log(x)*log(log(2))+((x^2-2*x)*exp(x)-6)*log(x))*log(log(x)) +log(log(2))+exp(x)*x+2)/x^4/log(x),x, algorithm="giac")
Output:
(x*e^x*log(log(x)) + log(log(2))*log(log(x)) + 2*log(log(x)))/x^3
Timed out. \[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\int \frac {\ln \left (\ln \left (2\right )\right )-\ln \left (\ln \left (x\right )\right )\,\left (3\,\ln \left (\ln \left (2\right )\right )\,\ln \left (x\right )+\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (2\,x-x^2\right )+6\right )\right )+x\,{\mathrm {e}}^x+2}{x^4\,\ln \left (x\right )} \,d x \] Input:
int((log(log(2)) - log(log(x))*(3*log(log(2))*log(x) + log(x)*(exp(x)*(2*x - x^2) + 6)) + x*exp(x) + 2)/(x^4*log(x)),x)
Output:
int((log(log(2)) - log(log(x))*(3*log(log(2))*log(x) + log(x)*(exp(x)*(2*x - x^2) + 6)) + x*exp(x) + 2)/(x^4*log(x)), x)
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2+e^x x+\log (\log (2))+\left (\left (-6+e^x \left (-2 x+x^2\right )\right ) \log (x)-3 \log (x) \log (\log (2))\right ) \log (\log (x))}{x^4 \log (x)} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \left (e^{x} x +\mathrm {log}\left (\mathrm {log}\left (2\right )\right )+2\right )}{x^{3}} \] Input:
int(((-3*log(x)*log(log(2))+((x^2-2*x)*exp(x)-6)*log(x))*log(log(x))+log(l og(2))+exp(x)*x+2)/x^4/log(x),x)
Output:
(log(log(x))*(e**x*x + log(log(2)) + 2))/x**3