Integrand size = 66, antiderivative size = 22 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} e^{-\frac {1}{2}-(-5+x) x^2 \log (\log (\log (x)))} \] Output:
1/20/exp(1/2+ln(ln(ln(x)))*x^2*(-5+x))
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {\log ^{-\left ((-5+x) x^2\right )}(\log (x))}{20 \sqrt {e}} \] Input:
Integrate[(E^((-1 - (-10*x^2 + 2*x^3)*Log[Log[Log[x]]])/2)*(5*x - x^2 + (1 0*x - 3*x^2)*Log[x]*Log[Log[x]]*Log[Log[Log[x]]]))/(20*Log[x]*Log[Log[x]]) ,x]
Output:
1/(20*Sqrt[E]*Log[Log[x]]^((-5 + x)*x^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.
\(\displaystyle \int \frac {\left (-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))+5 x\right ) \exp \left (\frac {1}{2} \left (-\left (\left (2 x^3-10 x^2\right ) \log (\log (\log (x)))\right )-1\right )\right )}{20 \log (x) \log (\log (x))} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{20} \int \frac {\log ^{\frac {1}{2} \left (10 x^2-2 x^3\right )-1}(\log (x)) \left (-x^2+5 x+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{\sqrt {e} \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\log ^{-x^3+5 x^2-1}(\log (x)) \left (-x^2+5 x+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{\log (x)}dx}{20 \sqrt {e}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (-\frac {(x-5) x \log ^{-x^3+5 x^2-1}(\log (x))}{\log (x)}-x (3 x-10) \log (\log (\log (x))) \log ^{5 x^2-x^3}(\log (x))\right )dx}{20 \sqrt {e}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {10 \int x \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x)))dx-3 \int x^2 \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x)))dx+5 \int \frac {x \log ^{-x^3+5 x^2-1}(\log (x))}{\log (x)}dx-\int \frac {x^2 \log ^{-x^3+5 x^2-1}(\log (x))}{\log (x)}dx}{20 \sqrt {e}}\) |
Input:
Int[(E^((-1 - (-10*x^2 + 2*x^3)*Log[Log[Log[x]]])/2)*(5*x - x^2 + (10*x - 3*x^2)*Log[x]*Log[Log[x]]*Log[Log[Log[x]]]))/(20*Log[x]*Log[Log[x]]),x]
Output:
$Aborted
Time = 1.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82
method | result | size |
risch | \(\frac {\ln \left (\ln \left (x \right )\right )^{-x^{2} \left (-5+x \right )} {\mathrm e}^{-\frac {1}{2}}}{20}\) | \(18\) |
parallelrisch | \(\frac {{\mathrm e}^{\left (-x^{3}+5 x^{2}\right ) \ln \left (\ln \left (\ln \left (x \right )\right )\right )-\frac {1}{2}}}{20}\) | \(25\) |
Input:
int(1/20*((-3*x^2+10*x)*ln(x)*ln(ln(x))*ln(ln(ln(x)))-x^2+5*x)/ln(x)/ln(ln (x))/exp(1/2*(2*x^3-10*x^2)*ln(ln(ln(x)))+1/2),x,method=_RETURNVERBOSE)
Output:
1/20/(ln(ln(x))^(x^2*(-5+x)))*exp(-1/2)
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} \, e^{\left (-{\left (x^{3} - 5 \, x^{2}\right )} \log \left (\log \left (\log \left (x\right )\right )\right ) - \frac {1}{2}\right )} \] Input:
integrate(1/20*((-3*x^2+10*x)*log(x)*log(log(x))*log(log(log(x)))-x^2+5*x) /log(x)/log(log(x))/exp(1/2*(2*x^3-10*x^2)*log(log(log(x)))+1/2),x, algori thm="fricas")
Output:
1/20*e^(-(x^3 - 5*x^2)*log(log(log(x))) - 1/2)
Time = 0.75 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {e^{- \left (x^{3} - 5 x^{2}\right ) \log {\left (\log {\left (\log {\left (x \right )} \right )} \right )} - \frac {1}{2}}}{20} \] Input:
integrate(1/20*((-3*x**2+10*x)*ln(x)*ln(ln(x))*ln(ln(ln(x)))-x**2+5*x)/ln( x)/ln(ln(x))/exp(1/2*(2*x**3-10*x**2)*ln(ln(ln(x)))+1/2),x)
Output:
exp(-(x**3 - 5*x**2)*log(log(log(x))) - 1/2)/20
Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} \, e^{\left (-x^{3} \log \left (\log \left (\log \left (x\right )\right )\right ) + 5 \, x^{2} \log \left (\log \left (\log \left (x\right )\right )\right ) - \frac {1}{2}\right )} \] Input:
integrate(1/20*((-3*x^2+10*x)*log(x)*log(log(x))*log(log(log(x)))-x^2+5*x) /log(x)/log(log(x))/exp(1/2*(2*x^3-10*x^2)*log(log(log(x)))+1/2),x, algori thm="maxima")
Output:
1/20*e^(-x^3*log(log(log(x))) + 5*x^2*log(log(log(x))) - 1/2)
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} \, e^{\left (-x^{3} \log \left (\log \left (\log \left (x\right )\right )\right ) + 5 \, x^{2} \log \left (\log \left (\log \left (x\right )\right )\right ) - \frac {1}{2}\right )} \] Input:
integrate(1/20*((-3*x^2+10*x)*log(x)*log(log(x))*log(log(log(x)))-x^2+5*x) /log(x)/log(log(x))/exp(1/2*(2*x^3-10*x^2)*log(log(log(x)))+1/2),x, algori thm="giac")
Output:
1/20*e^(-x^3*log(log(log(x))) + 5*x^2*log(log(log(x))) - 1/2)
Time = 2.65 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {{\ln \left (\ln \left (x\right )\right )}^{5\,x^2-x^3}\,{\mathrm {e}}^{-\frac {1}{2}}}{20} \] Input:
int((exp((log(log(log(x)))*(10*x^2 - 2*x^3))/2 - 1/2)*(x/4 - x^2/20 + (log (log(x))*log(log(log(x)))*log(x)*(10*x - 3*x^2))/20))/(log(log(x))*log(x)) ,x)
Output:
(log(log(x))^(5*x^2 - x^3)*exp(-1/2))/20
\[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {\sqrt {e}\, \left (-\left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{5 x^{2}} x^{2}}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{x^{3}} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (x \right )}d x \right )-3 \left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{5 x^{2}} \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right )\right ) x^{2}}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{x^{3}}}d x \right )+10 \left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{5 x^{2}} \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right )\right ) x}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{x^{3}}}d x \right )+5 \left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{5 x^{2}} x}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{x^{3}} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (x \right )}d x \right )\right )}{20 e} \] Input:
int(1/20*((-3*x^2+10*x)*log(x)*log(log(x))*log(log(log(x)))-x^2+5*x)/log(x )/log(log(x))/exp(1/2*(2*x^3-10*x^2)*log(log(log(x)))+1/2),x)
Output:
(sqrt(e)*( - int((log(log(x))**(5*x**2)*x**2)/(log(log(x))**(x**3)*log(log (x))*log(x)),x) - 3*int((log(log(x))**(5*x**2)*log(log(log(x)))*x**2)/log( log(x))**(x**3),x) + 10*int((log(log(x))**(5*x**2)*log(log(log(x)))*x)/log (log(x))**(x**3),x) + 5*int((log(log(x))**(5*x**2)*x)/(log(log(x))**(x**3) *log(log(x))*log(x)),x)))/(20*e)