Integrand size = 98, antiderivative size = 31 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=\frac {e^3-x^2 (1+\log (x))}{-e^{60 x^2}+e^4 x} \] Output:
(exp(3)-(1+ln(x))*x^2)/(x*exp(4)-exp(60*x^2))
Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=-\frac {e^3-x^2-x^2 \log (x)}{e^{60 x^2}-e^4 x} \] Input:
Integrate[(E^4*(-E^3 - 2*x^2) + E^(60*x^2)*(3*x + 120*E^3*x - 120*x^3) + ( -(E^4*x^2) + E^(60*x^2)*(2*x - 120*x^3))*Log[x])/(E^(120*x^2) - 2*E^(4 + 6 0*x^2)*x + E^8*x^2),x]
Output:
-((E^3 - x^2 - x^2*Log[x])/(E^(60*x^2) - E^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.
\(\displaystyle \int \frac {e^4 \left (-2 x^2-e^3\right )+e^{60 x^2} \left (-120 x^3+120 e^3 x+3 x\right )+\left (e^{60 x^2} \left (2 x-120 x^3\right )-e^4 x^2\right ) \log (x)}{e^8 x^2-2 e^{60 x^2+4} x+e^{120 x^2}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^4 \left (-2 x^2-e^3\right )+e^{60 x^2} \left (-120 x^3+120 e^3 x+3 x\right )+\left (e^{60 x^2} \left (2 x-120 x^3\right )-e^4 x^2\right ) \log (x)}{\left (e^{60 x^2}-e^4 x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \left (-120 x^2-120 x^2 \log (x)+2 \log (x)+3 \left (1+40 e^3\right )\right )}{e^{60 x^2}-e^4 x}+\frac {e^4 \left (120 x^2-1\right ) \left (-x^2+x^2 (-\log (x))+e^3\right )}{\left (e^4 x-e^{60 x^2}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -e^7 \int \frac {1}{\left (e^4 x-e^{60 x^2}\right )^2}dx+120 e^7 \int \frac {x^2}{\left (e^4 x-e^{60 x^2}\right )^2}dx+e^4 \int \frac {x^2}{\left (e^4 x-e^{60 x^2}\right )^2}dx-3 \left (1+40 e^3\right ) \int \frac {x}{e^4 x-e^{60 x^2}}dx-e^4 \int \frac {\int \frac {x^2}{\left (e^{60 x^2}-e^4 x\right )^2}dx}{x}dx+2 \int \frac {\int \frac {x}{e^4 x-e^{60 x^2}}dx}{x}dx+e^4 \log (x) \int \frac {x^2}{\left (e^4 x-e^{60 x^2}\right )^2}dx-2 \log (x) \int \frac {x}{e^4 x-e^{60 x^2}}dx-120 e^4 \int \frac {x^4}{\left (e^4 x-e^{60 x^2}\right )^2}dx+120 e^4 \int \frac {\int \frac {x^4}{\left (e^{60 x^2}-e^4 x\right )^2}dx}{x}dx-120 e^4 \log (x) \int \frac {x^4}{\left (e^4 x-e^{60 x^2}\right )^2}dx+120 \int \frac {x^3}{e^4 x-e^{60 x^2}}dx-120 \int \frac {\int \frac {x^3}{e^4 x-e^{60 x^2}}dx}{x}dx+120 \log (x) \int \frac {x^3}{e^4 x-e^{60 x^2}}dx\) |
Input:
Int[(E^4*(-E^3 - 2*x^2) + E^(60*x^2)*(3*x + 120*E^3*x - 120*x^3) + (-(E^4* x^2) + E^(60*x^2)*(2*x - 120*x^3))*Log[x])/(E^(120*x^2) - 2*E^(4 + 60*x^2) *x + E^8*x^2),x]
Output:
$Aborted
Time = 1.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(\frac {-x^{2} \ln \left (x \right )-x^{2}+{\mathrm e}^{3}}{x \,{\mathrm e}^{4}-{\mathrm e}^{60 x^{2}}}\) | \(32\) |
risch | \(-\frac {x^{2} \ln \left (x \right )}{x \,{\mathrm e}^{4}-{\mathrm e}^{60 x^{2}}}+\frac {{\mathrm e}^{3}-x^{2}}{x \,{\mathrm e}^{4}-{\mathrm e}^{60 x^{2}}}\) | \(48\) |
Input:
int((((-120*x^3+2*x)*exp(60*x^2)-x^2*exp(4))*ln(x)+(120*x*exp(3)-120*x^3+3 *x)*exp(60*x^2)+(-exp(3)-2*x^2)*exp(4))/(exp(60*x^2)^2-2*x*exp(4)*exp(60*x ^2)+x^2*exp(4)^2),x,method=_RETURNVERBOSE)
Output:
(-x^2*ln(x)-x^2+exp(3))/(x*exp(4)-exp(60*x^2))
Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=-\frac {x^{2} e^{4} \log \left (x\right ) + x^{2} e^{4} - e^{7}}{x e^{8} - e^{\left (60 \, x^{2} + 4\right )}} \] Input:
integrate((((-120*x^3+2*x)*exp(60*x^2)-x^2*exp(4))*log(x)+(120*x*exp(3)-12 0*x^3+3*x)*exp(60*x^2)+(-exp(3)-2*x^2)*exp(4))/(exp(60*x^2)^2-2*x*exp(4)*e xp(60*x^2)+x^2*exp(4)^2),x, algorithm="fricas")
Output:
-(x^2*e^4*log(x) + x^2*e^4 - e^7)/(x*e^8 - e^(60*x^2 + 4))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=\frac {x^{2} \log {\left (x \right )} + x^{2} - e^{3}}{- x e^{4} + e^{60 x^{2}}} \] Input:
integrate((((-120*x**3+2*x)*exp(60*x**2)-x**2*exp(4))*ln(x)+(120*x*exp(3)- 120*x**3+3*x)*exp(60*x**2)+(-exp(3)-2*x**2)*exp(4))/(exp(60*x**2)**2-2*x*e xp(4)*exp(60*x**2)+x**2*exp(4)**2),x)
Output:
(x**2*log(x) + x**2 - exp(3))/(-x*exp(4) + exp(60*x**2))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=-\frac {x^{2} \log \left (x\right ) + x^{2} - e^{3}}{x e^{4} - e^{\left (60 \, x^{2}\right )}} \] Input:
integrate((((-120*x^3+2*x)*exp(60*x^2)-x^2*exp(4))*log(x)+(120*x*exp(3)-12 0*x^3+3*x)*exp(60*x^2)+(-exp(3)-2*x^2)*exp(4))/(exp(60*x^2)^2-2*x*exp(4)*e xp(60*x^2)+x^2*exp(4)^2),x, algorithm="maxima")
Output:
-(x^2*log(x) + x^2 - e^3)/(x*e^4 - e^(60*x^2))
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=-\frac {x^{2} \log \left (x\right ) + x^{2} - e^{3}}{x e^{4} - e^{\left (60 \, x^{2}\right )}} \] Input:
integrate((((-120*x^3+2*x)*exp(60*x^2)-x^2*exp(4))*log(x)+(120*x*exp(3)-12 0*x^3+3*x)*exp(60*x^2)+(-exp(3)-2*x^2)*exp(4))/(exp(60*x^2)^2-2*x*exp(4)*e xp(60*x^2)+x^2*exp(4)^2),x, algorithm="giac")
Output:
-(x^2*log(x) + x^2 - e^3)/(x*e^4 - e^(60*x^2))
Time = 1.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=\frac {x^2\,\ln \left (x\right )-{\mathrm {e}}^3+x^2}{{\mathrm {e}}^{60\,x^2}-x\,{\mathrm {e}}^4} \] Input:
int((exp(60*x^2)*(3*x + 120*x*exp(3) - 120*x^3) + log(x)*(exp(60*x^2)*(2*x - 120*x^3) - x^2*exp(4)) - exp(4)*(exp(3) + 2*x^2))/(exp(120*x^2) + x^2*e xp(8) - 2*x*exp(4)*exp(60*x^2)),x)
Output:
(x^2*log(x) - exp(3) + x^2)/(exp(60*x^2) - x*exp(4))
Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^4 \left (-e^3-2 x^2\right )+e^{60 x^2} \left (3 x+120 e^3 x-120 x^3\right )+\left (-e^4 x^2+e^{60 x^2} \left (2 x-120 x^3\right )\right ) \log (x)}{e^{120 x^2}-2 e^{4+60 x^2} x+e^8 x^2} \, dx=\frac {\mathrm {log}\left (x \right ) x^{2}-e^{3}+x^{2}}{e^{60 x^{2}}-e^{4} x} \] Input:
int((((-120*x^3+2*x)*exp(60*x^2)-x^2*exp(4))*log(x)+(120*x*exp(3)-120*x^3+ 3*x)*exp(60*x^2)+(-exp(3)-2*x^2)*exp(4))/(exp(60*x^2)^2-2*x*exp(4)*exp(60* x^2)+x^2*exp(4)^2),x)
Output:
(log(x)*x**2 - e**3 + x**2)/(e**(60*x**2) - e**4*x)