Integrand size = 94, antiderivative size = 28 \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=\frac {3}{e-\left (\frac {5}{3}+e^{\frac {-\frac {5}{x}+x^2}{x}}\right ) x} \] Output:
3/(exp(1)-x*(5/3+exp((x^2-5/x)/x)))
Time = 0.66 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=-\frac {9 e^{\frac {5}{x^2}}}{-3 e^{1+\frac {5}{x^2}}+5 e^{\frac {5}{x^2}} x+3 e^x x} \] Input:
Integrate[(45*x^2 + E^((-5 + x^3)/x^2)*(270 + 27*x^2 + 27*x^3))/(9*E^2*x^2 - 30*E*x^3 + 25*x^4 + 9*E^((2*(-5 + x^3))/x^2)*x^4 + E^((-5 + x^3)/x^2)*( -18*E*x^3 + 30*x^4)),x]
Output:
(-9*E^(5/x^2))/(-3*E^(1 + 5/x^2) + 5*E^(5/x^2)*x + 3*E^x*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.
\(\displaystyle \int \frac {45 x^2+e^{\frac {x^3-5}{x^2}} \left (27 x^3+27 x^2+270\right )}{25 x^4-30 e x^3+9 e^2 x^2+9 e^{\frac {2 \left (x^3-5\right )}{x^2}} x^4+e^{\frac {x^3-5}{x^2}} \left (30 x^4-18 e x^3\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {10}{x^2}} \left (45 x^2+e^{\frac {x^3-5}{x^2}} \left (27 x^3+27 x^2+270\right )\right )}{x^2 \left (-5 e^{\frac {5}{x^2}} x+3 e^{\frac {5}{x^2}+1}-3 e^x x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {9 e^{\frac {5}{x^2}} \left (x^3+x^2+10\right )}{x^3 \left (5 e^{\frac {5}{x^2}} x-3 e^{\frac {5}{x^2}+1}+3 e^x x\right )}-\frac {9 e^{\frac {10}{x^2}} \left (5 x^4-3 e x^3-3 e x^2+50 x-30 e\right )}{x^3 \left (5 e^{\frac {5}{x^2}} x-3 e^{\frac {5}{x^2}+1}+3 e^x x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 27 \int \frac {e^{1+\frac {10}{x^2}}}{\left (-5 e^{\frac {5}{x^2}} x-3 e^x x+3 e^{1+\frac {5}{x^2}}\right )^2}dx+27 \int \frac {e^{1+\frac {10}{x^2}}}{x \left (-5 e^{\frac {5}{x^2}} x-3 e^x x+3 e^{1+\frac {5}{x^2}}\right )^2}dx-9 \int \frac {e^{\frac {5}{x^2}}}{-5 e^{\frac {5}{x^2}} x-3 e^x x+3 e^{1+\frac {5}{x^2}}}dx-450 \int \frac {e^{\frac {10}{x^2}}}{x^2 \left (5 e^{\frac {5}{x^2}} x+3 e^x x-3 e^{1+\frac {5}{x^2}}\right )^2}dx-45 \int \frac {e^{\frac {10}{x^2}} x}{\left (5 e^{\frac {5}{x^2}} x+3 e^x x-3 e^{1+\frac {5}{x^2}}\right )^2}dx+9 \int \frac {e^{\frac {5}{x^2}}}{x \left (5 e^{\frac {5}{x^2}} x+3 e^x x-3 e^{1+\frac {5}{x^2}}\right )}dx+270 \int \frac {e^{1+\frac {10}{x^2}}}{x^3 \left (-5 e^{\frac {5}{x^2}} x-3 e^x x+3 e^{1+\frac {5}{x^2}}\right )^2}dx+90 \int \frac {e^{\frac {5}{x^2}}}{x^3 \left (5 e^{\frac {5}{x^2}} x+3 e^x x-3 e^{1+\frac {5}{x^2}}\right )}dx\) |
Input:
Int[(45*x^2 + E^((-5 + x^3)/x^2)*(270 + 27*x^2 + 27*x^3))/(9*E^2*x^2 - 30* E*x^3 + 25*x^4 + 9*E^((2*(-5 + x^3))/x^2)*x^4 + E^((-5 + x^3)/x^2)*(-18*E* x^3 + 30*x^4)),x]
Output:
$Aborted
Time = 0.67 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
norman | \(\frac {9}{-3 \,{\mathrm e}^{\frac {x^{3}-5}{x^{2}}} x +3 \,{\mathrm e}-5 x}\) | \(26\) |
risch | \(\frac {9}{-3 \,{\mathrm e}^{\frac {x^{3}-5}{x^{2}}} x +3 \,{\mathrm e}-5 x}\) | \(26\) |
parallelrisch | \(\frac {9}{-3 \,{\mathrm e}^{\frac {x^{3}-5}{x^{2}}} x +3 \,{\mathrm e}-5 x}\) | \(26\) |
Input:
int(((27*x^3+27*x^2+270)*exp((x^3-5)/x^2)+45*x^2)/(9*x^4*exp((x^3-5)/x^2)^ 2+(-18*x^3*exp(1)+30*x^4)*exp((x^3-5)/x^2)+9*x^2*exp(1)^2-30*x^3*exp(1)+25 *x^4),x,method=_RETURNVERBOSE)
Output:
9/(-3*exp((x^3-5)/x^2)*x+3*exp(1)-5*x)
Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=-\frac {9}{3 \, x e^{\left (\frac {x^{3} - 5}{x^{2}}\right )} + 5 \, x - 3 \, e} \] Input:
integrate(((27*x^3+27*x^2+270)*exp((x^3-5)/x^2)+45*x^2)/(9*x^4*exp((x^3-5) /x^2)^2+(-18*x^3*exp(1)+30*x^4)*exp((x^3-5)/x^2)+9*x^2*exp(1)^2-30*x^3*exp (1)+25*x^4),x, algorithm="fricas")
Output:
-9/(3*x*e^((x^3 - 5)/x^2) + 5*x - 3*e)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=- \frac {9}{3 x e^{\frac {x^{3} - 5}{x^{2}}} + 5 x - 3 e} \] Input:
integrate(((27*x**3+27*x**2+270)*exp((x**3-5)/x**2)+45*x**2)/(9*x**4*exp(( x**3-5)/x**2)**2+(-18*x**3*exp(1)+30*x**4)*exp((x**3-5)/x**2)+9*x**2*exp(1 )**2-30*x**3*exp(1)+25*x**4),x)
Output:
-9/(3*x*exp((x**3 - 5)/x**2) + 5*x - 3*E)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=-\frac {9 \, e^{\left (\frac {5}{x^{2}}\right )}}{3 \, x e^{x} + {\left (5 \, x - 3 \, e\right )} e^{\left (\frac {5}{x^{2}}\right )}} \] Input:
integrate(((27*x^3+27*x^2+270)*exp((x^3-5)/x^2)+45*x^2)/(9*x^4*exp((x^3-5) /x^2)^2+(-18*x^3*exp(1)+30*x^4)*exp((x^3-5)/x^2)+9*x^2*exp(1)^2-30*x^3*exp (1)+25*x^4),x, algorithm="maxima")
Output:
-9*e^(5/x^2)/(3*x*e^x + (5*x - 3*e)*e^(5/x^2))
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=-\frac {9}{3 \, x e^{\left (\frac {x^{3} - 5}{x^{2}}\right )} + 5 \, x - 3 \, e} \] Input:
integrate(((27*x^3+27*x^2+270)*exp((x^3-5)/x^2)+45*x^2)/(9*x^4*exp((x^3-5) /x^2)^2+(-18*x^3*exp(1)+30*x^4)*exp((x^3-5)/x^2)+9*x^2*exp(1)^2-30*x^3*exp (1)+25*x^4),x, algorithm="giac")
Output:
-9/(3*x*e^((x^3 - 5)/x^2) + 5*x - 3*e)
Timed out. \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=\int \frac {45\,x^2+{\mathrm {e}}^{\frac {x^3-5}{x^2}}\,\left (27\,x^3+27\,x^2+270\right )}{9\,x^4\,{\mathrm {e}}^{\frac {2\,\left (x^3-5\right )}{x^2}}-{\mathrm {e}}^{\frac {x^3-5}{x^2}}\,\left (18\,x^3\,\mathrm {e}-30\,x^4\right )+9\,x^2\,{\mathrm {e}}^2-30\,x^3\,\mathrm {e}+25\,x^4} \,d x \] Input:
int((45*x^2 + exp((x^3 - 5)/x^2)*(27*x^2 + 27*x^3 + 270))/(9*x^4*exp((2*(x ^3 - 5))/x^2) - exp((x^3 - 5)/x^2)*(18*x^3*exp(1) - 30*x^4) + 9*x^2*exp(2) - 30*x^3*exp(1) + 25*x^4),x)
Output:
int((45*x^2 + exp((x^3 - 5)/x^2)*(27*x^2 + 27*x^3 + 270))/(9*x^4*exp((2*(x ^3 - 5))/x^2) - exp((x^3 - 5)/x^2)*(18*x^3*exp(1) - 30*x^4) + 9*x^2*exp(2) - 30*x^3*exp(1) + 25*x^4), x)
Time = 0.73 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {45 x^2+e^{\frac {-5+x^3}{x^2}} \left (270+27 x^2+27 x^3\right )}{9 e^2 x^2-30 e x^3+25 x^4+9 e^{\frac {2 \left (-5+x^3\right )}{x^2}} x^4+e^{\frac {-5+x^3}{x^2}} \left (-18 e x^3+30 x^4\right )} \, dx=\frac {9 e^{\frac {5}{x^{2}}}}{3 e^{\frac {5}{x^{2}}} e -5 e^{\frac {5}{x^{2}}} x -3 e^{x} x} \] Input:
int(((27*x^3+27*x^2+270)*exp((x^3-5)/x^2)+45*x^2)/(9*x^4*exp((x^3-5)/x^2)^ 2+(-18*x^3*exp(1)+30*x^4)*exp((x^3-5)/x^2)+9*x^2*exp(1)^2-30*x^3*exp(1)+25 *x^4),x)
Output:
(9*e**(5/x**2))/(3*e**(5/x**2)*e - 5*e**(5/x**2)*x - 3*e**x*x)