Integrand size = 70, antiderivative size = 27 \[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {e^{4-x+x^2} x^3}{9 (-5+x) \log (\log (x))} \] Output:
1/9*x^3/(-5+x)*exp(-ln(ln(ln(x)))+x^2-x+4)
Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {e^{4-x+x^2} x^3}{9 (-5+x) \log (\log (x))} \] Input:
Integrate[(E^(4 - x + x^2)*(5*x^2 - x^3 + (-15*x^2 + 7*x^3 - 11*x^4 + 2*x^ 5)*Log[x]*Log[Log[x]]))/((225 - 90*x + 9*x^2)*Log[x]*Log[Log[x]]^2),x]
Output:
(E^(4 - x + x^2)*x^3)/(9*(-5 + x)*Log[Log[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.
\(\displaystyle \int \frac {e^{x^2-x+4} \left (-x^3+5 x^2+\left (2 x^5-11 x^4+7 x^3-15 x^2\right ) \log (x) \log (\log (x))\right )}{\left (9 x^2-90 x+225\right ) \log (x) \log ^2(\log (x))} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 36 \int \frac {e^{x^2-x+4} \left (-x^3+5 x^2-\left (-2 x^5+11 x^4-7 x^3+15 x^2\right ) \log (x) \log (\log (x))\right )}{324 (5-x)^2 \log (x) \log ^2(\log (x))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {e^{x^2-x+4} \left (-x^3+5 x^2-\left (-2 x^5+11 x^4-7 x^3+15 x^2\right ) \log (x) \log (\log (x))\right )}{(5-x)^2 \log (x) \log ^2(\log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{9} \int \left (\frac {e^{x^2-x+4} x^2 \left (2 x^3-11 x^2+7 x-15\right )}{(x-5)^2 \log (\log (x))}-\frac {e^{x^2-x+4} x^2}{(x-5) \log (x) \log ^2(\log (x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (-5 \int \frac {e^{x^2-x+4}}{\log (x) \log ^2(\log (x))}dx-25 \int \frac {e^{x^2-x+4}}{(x-5) \log (x) \log ^2(\log (x))}dx-\int \frac {e^{x^2-x+4} x}{\log (x) \log ^2(\log (x))}dx+230 \int \frac {e^{x^2-x+4}}{\log (\log (x))}dx-125 \int \frac {e^{x^2-x+4}}{(x-5)^2 \log (\log (x))}dx+1125 \int \frac {e^{x^2-x+4}}{(x-5) \log (\log (x))}dx+47 \int \frac {e^{x^2-x+4} x}{\log (\log (x))}dx+9 \int \frac {e^{x^2-x+4} x^2}{\log (\log (x))}dx+2 \int \frac {e^{x^2-x+4} x^3}{\log (\log (x))}dx\right )\) |
Input:
Int[(E^(4 - x + x^2)*(5*x^2 - x^3 + (-15*x^2 + 7*x^3 - 11*x^4 + 2*x^5)*Log [x]*Log[Log[x]]))/((225 - 90*x + 9*x^2)*Log[x]*Log[Log[x]]^2),x]
Output:
$Aborted
Time = 117.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {x^{3} {\mathrm e}^{x^{2}-x +4}}{9 \left (-5+x \right ) \ln \left (\ln \left (x \right )\right )}\) | \(25\) |
parallelrisch | \(\frac {{\mathrm e}^{-\ln \left (\ln \left (\ln \left (x \right )\right )\right )+x^{2}-x +4} x^{3}}{-45+9 x}\) | \(26\) |
Input:
int(((2*x^5-11*x^4+7*x^3-15*x^2)*ln(x)*ln(ln(x))-x^3+5*x^2)*exp(-ln(ln(ln( x)))+x^2-x+4)/(9*x^2-90*x+225)/ln(x)/ln(ln(x)),x,method=_RETURNVERBOSE)
Output:
1/9*x^3/(-5+x)/ln(ln(x))*exp(x^2-x+4)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {x^{3} e^{\left (x^{2} - x - \log \left (\log \left (\log \left (x\right )\right )\right ) + 4\right )}}{9 \, {\left (x - 5\right )}} \] Input:
integrate(((2*x^5-11*x^4+7*x^3-15*x^2)*log(x)*log(log(x))-x^3+5*x^2)*exp(- log(log(log(x)))+x^2-x+4)/(9*x^2-90*x+225)/log(x)/log(log(x)),x, algorithm ="fricas")
Output:
1/9*x^3*e^(x^2 - x - log(log(log(x))) + 4)/(x - 5)
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {x^{3} e^{x^{2} - x + 4}}{9 x \log {\left (\log {\left (x \right )} \right )} - 45 \log {\left (\log {\left (x \right )} \right )}} \] Input:
integrate(((2*x**5-11*x**4+7*x**3-15*x**2)*ln(x)*ln(ln(x))-x**3+5*x**2)*ex p(-ln(ln(ln(x)))+x**2-x+4)/(9*x**2-90*x+225)/ln(x)/ln(ln(x)),x)
Output:
x**3*exp(x**2 - x + 4)/(9*x*log(log(x)) - 45*log(log(x)))
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {x^{3} e^{\left (x^{2} - x + 4\right )}}{9 \, {\left (x - 5\right )} \log \left (\log \left (x\right )\right )} \] Input:
integrate(((2*x^5-11*x^4+7*x^3-15*x^2)*log(x)*log(log(x))-x^3+5*x^2)*exp(- log(log(log(x)))+x^2-x+4)/(9*x^2-90*x+225)/log(x)/log(log(x)),x, algorithm ="maxima")
Output:
1/9*x^3*e^(x^2 - x + 4)/((x - 5)*log(log(x)))
\[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=\int { -\frac {{\left (x^{3} - {\left (2 \, x^{5} - 11 \, x^{4} + 7 \, x^{3} - 15 \, x^{2}\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right ) - 5 \, x^{2}\right )} e^{\left (x^{2} - x - \log \left (\log \left (\log \left (x\right )\right )\right ) + 4\right )}}{9 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right )} \,d x } \] Input:
integrate(((2*x^5-11*x^4+7*x^3-15*x^2)*log(x)*log(log(x))-x^3+5*x^2)*exp(- log(log(log(x)))+x^2-x+4)/(9*x^2-90*x+225)/log(x)/log(log(x)),x, algorithm ="giac")
Output:
undef
Time = 4.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=-\frac {x^3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^4}{9\,\left (5\,\ln \left (\ln \left (x\right )\right )-x\,\ln \left (\ln \left (x\right )\right )\right )} \] Input:
int(-(exp(x^2 - log(log(log(x))) - x + 4)*(x^3 - 5*x^2 + log(log(x))*log(x )*(15*x^2 - 7*x^3 + 11*x^4 - 2*x^5)))/(log(log(x))*log(x)*(9*x^2 - 90*x + 225)),x)
Output:
-(x^3*exp(-x)*exp(x^2)*exp(4))/(9*(5*log(log(x)) - x*log(log(x))))
Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {e^{x^{2}} e^{4} x^{3}}{9 e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \left (-5+x \right )} \] Input:
int(((2*x^5-11*x^4+7*x^3-15*x^2)*log(x)*log(log(x))-x^3+5*x^2)*exp(-log(lo g(log(x)))+x^2-x+4)/(9*x^2-90*x+225)/log(x)/log(log(x)),x)
Output:
(e**(x**2)*e**4*x**3)/(9*e**x*log(log(x))*(x - 5))