Integrand size = 62, antiderivative size = 27 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=\frac {5}{4} \left (e^x-x^3+\log \left (\left (3-x^2 (4+\log (x))\right )^2\right )\right ) \] Output:
5/4*exp(x)-5/4*x^3+5/4*ln((3-(ln(x)+4)*x^2)^2)
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=\frac {5}{4} \left (e^x-x^3+2 \log \left (3-4 x^2-x^2 \log (x)\right )\right ) \] Input:
Integrate[(90*x + 45*x^2 - 60*x^4 + E^x*(-15 + 20*x^2) + (20*x + 5*E^x*x^2 - 15*x^4)*Log[x])/(-12 + 16*x^2 + 4*x^2*Log[x]),x]
Output:
(5*(E^x - x^3 + 2*Log[3 - 4*x^2 - x^2*Log[x]]))/4
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.
\(\displaystyle \int \frac {-60 x^4+45 x^2+e^x \left (20 x^2-15\right )+\left (-15 x^4+5 e^x x^2+20 x\right ) \log (x)+90 x}{16 x^2+4 x^2 \log (x)-12} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {60 x^4-45 x^2-e^x \left (20 x^2-15\right )-\left (-15 x^4+5 e^x x^2+20 x\right ) \log (x)-90 x}{4 \left (-4 x^2+x^2 (-\log (x))+3\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {5 \left (-12 x^4+9 x^2+18 x-e^x \left (3-4 x^2\right )+\left (-3 x^4+e^x x^2+4 x\right ) \log (x)\right )}{-\log (x) x^2-4 x^2+3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5}{4} \int \frac {-12 x^4+9 x^2+18 x-e^x \left (3-4 x^2\right )+\left (-3 x^4+e^x x^2+4 x\right ) \log (x)}{-\log (x) x^2-4 x^2+3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {5}{4} \int \left (\frac {x \left (3 \log (x) x^3+12 x^3-9 x-4 \log (x)-18\right )}{\log (x) x^2+4 x^2-3}-e^x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{4} \left (-12 \int \frac {1}{x \left (\log (x) x^2+4 x^2-3\right )}dx-2 \int \frac {x}{\log (x) x^2+4 x^2-3}dx+x^3-e^x-4 \log (x)\right )\) |
Input:
Int[(90*x + 45*x^2 - 60*x^4 + E^x*(-15 + 20*x^2) + (20*x + 5*E^x*x^2 - 15* x^4)*Log[x])/(-12 + 16*x^2 + 4*x^2*Log[x]),x]
Output:
$Aborted
Time = 1.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
default | \(-\frac {5 x^{3}}{4}+\frac {5 \ln \left (x^{2} \ln \left (x \right )+4 x^{2}-3\right )}{2}+\frac {5 \,{\mathrm e}^{x}}{4}\) | \(27\) |
parallelrisch | \(-\frac {5 x^{3}}{4}+\frac {5 \ln \left (x^{2} \ln \left (x \right )+4 x^{2}-3\right )}{2}+\frac {5 \,{\mathrm e}^{x}}{4}\) | \(27\) |
parts | \(-\frac {5 x^{3}}{4}+\frac {5 \ln \left (x^{2} \ln \left (x \right )+4 x^{2}-3\right )}{2}+\frac {5 \,{\mathrm e}^{x}}{4}\) | \(27\) |
norman | \(-\frac {5 x^{3}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}+\frac {5 \ln \left (4 x^{2} \ln \left (x \right )+16 x^{2}-12\right )}{2}\) | \(28\) |
risch | \(-\frac {5 x^{3}}{4}+5 \ln \left (x \right )+\frac {5 \,{\mathrm e}^{x}}{4}+\frac {5 \ln \left (\ln \left (x \right )+\frac {4 x^{2}-3}{x^{2}}\right )}{2}\) | \(32\) |
Input:
int(((5*exp(x)*x^2-15*x^4+20*x)*ln(x)+(20*x^2-15)*exp(x)-60*x^4+45*x^2+90* x)/(4*x^2*ln(x)+16*x^2-12),x,method=_RETURNVERBOSE)
Output:
-5/4*x^3+5/2*ln(x^2*ln(x)+4*x^2-3)+5/4*exp(x)
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=-\frac {5}{4} \, x^{3} + \frac {5}{4} \, e^{x} + 5 \, \log \left (x\right ) + \frac {5}{2} \, \log \left (\frac {x^{2} \log \left (x\right ) + 4 \, x^{2} - 3}{x^{2}}\right ) \] Input:
integrate(((5*exp(x)*x^2-15*x^4+20*x)*log(x)+(20*x^2-15)*exp(x)-60*x^4+45* x^2+90*x)/(4*x^2*log(x)+16*x^2-12),x, algorithm="fricas")
Output:
-5/4*x^3 + 5/4*e^x + 5*log(x) + 5/2*log((x^2*log(x) + 4*x^2 - 3)/x^2)
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=- \frac {5 x^{3}}{4} + \frac {5 e^{x}}{4} + 5 \log {\left (x \right )} + \frac {5 \log {\left (\log {\left (x \right )} + \frac {4 x^{2} - 3}{x^{2}} \right )}}{2} \] Input:
integrate(((5*exp(x)*x**2-15*x**4+20*x)*ln(x)+(20*x**2-15)*exp(x)-60*x**4+ 45*x**2+90*x)/(4*x**2*ln(x)+16*x**2-12),x)
Output:
-5*x**3/4 + 5*exp(x)/4 + 5*log(x) + 5*log(log(x) + (4*x**2 - 3)/x**2)/2
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=-\frac {5}{4} \, x^{3} + \frac {5}{4} \, e^{x} + 5 \, \log \left (x\right ) + \frac {5}{2} \, \log \left (\frac {x^{2} \log \left (x\right ) + 4 \, x^{2} - 3}{x^{2}}\right ) \] Input:
integrate(((5*exp(x)*x^2-15*x^4+20*x)*log(x)+(20*x^2-15)*exp(x)-60*x^4+45* x^2+90*x)/(4*x^2*log(x)+16*x^2-12),x, algorithm="maxima")
Output:
-5/4*x^3 + 5/4*e^x + 5*log(x) + 5/2*log((x^2*log(x) + 4*x^2 - 3)/x^2)
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=-\frac {5}{4} \, x^{3} + \frac {5}{4} \, e^{x} + \frac {5}{2} \, \log \left (-x^{2} \log \left (x\right ) - 4 \, x^{2} + 3\right ) \] Input:
integrate(((5*exp(x)*x^2-15*x^4+20*x)*log(x)+(20*x^2-15)*exp(x)-60*x^4+45* x^2+90*x)/(4*x^2*log(x)+16*x^2-12),x, algorithm="giac")
Output:
-5/4*x^3 + 5/4*e^x + 5/2*log(-x^2*log(x) - 4*x^2 + 3)
Time = 4.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=\frac {5\,\ln \left (\frac {x^2\,\ln \left (x\right )+4\,x^2-3}{x^2}\right )}{2}+\frac {5\,{\mathrm {e}}^x}{4}+5\,\ln \left (x\right )-\frac {5\,x^3}{4} \] Input:
int((90*x + log(x)*(20*x + 5*x^2*exp(x) - 15*x^4) + exp(x)*(20*x^2 - 15) + 45*x^2 - 60*x^4)/(4*x^2*log(x) + 16*x^2 - 12),x)
Output:
(5*log((x^2*log(x) + 4*x^2 - 3)/x^2))/2 + (5*exp(x))/4 + 5*log(x) - (5*x^3 )/4
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx=\frac {5 e^{x}}{4}+\frac {5 \,\mathrm {log}\left (\mathrm {log}\left (x \right ) x^{2}+4 x^{2}-3\right )}{2}-\frac {5 x^{3}}{4} \] Input:
int(((5*exp(x)*x^2-15*x^4+20*x)*log(x)+(20*x^2-15)*exp(x)-60*x^4+45*x^2+90 *x)/(4*x^2*log(x)+16*x^2-12),x)
Output:
(5*(e**x + 2*log(log(x)*x**2 + 4*x**2 - 3) - x**3))/4