Integrand size = 96, antiderivative size = 28 \[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=e^{-4+\frac {5}{3} x^2 \left (3-16 \left (e^{x/3}+x\right )^2\right )} x \] Output:
x*exp(5/3*(3-4*(x+exp(1/3*x))*(4*x+4*exp(1/3*x)))*x^2-4)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=e^{\frac {1}{3} \left (-12+\left (15-80 e^{2 x/3}\right ) x^2-160 e^{x/3} x^3-80 x^4\right )} x \] Input:
Integrate[(E^((-12 + 15*x^2 - 80*E^((2*x)/3)*x^2 - 160*E^(x/3)*x^3 - 80*x^ 4)/3)*(9 + 90*x^2 - 960*x^4 + E^((2*x)/3)*(-480*x^2 - 160*x^3) + E^(x/3)*( -1440*x^3 - 160*x^4)))/9,x]
Output:
E^((-12 + (15 - 80*E^((2*x)/3))*x^2 - 160*E^(x/3)*x^3 - 80*x^4)/3)*x
Leaf count is larger than twice the leaf count of optimal. \(147\) vs. \(2(28)=56\).
Time = 0.82 (sec) , antiderivative size = 147, normalized size of antiderivative = 5.25, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {27, 2726}
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.
\(\displaystyle \int \frac {1}{9} \left (-960 x^4+90 x^2+e^{x/3} \left (-160 x^4-1440 x^3\right )+e^{2 x/3} \left (-160 x^3-480 x^2\right )+9\right ) \exp \left (\frac {1}{3} \left (-80 x^4-160 e^{x/3} x^3-80 e^{2 x/3} x^2+15 x^2-12\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \exp \left (\frac {1}{3} \left (-80 x^4-160 e^{x/3} x^3-80 e^{2 x/3} x^2+15 x^2-12\right )\right ) \left (-960 x^4+90 x^2-160 e^{2 x/3} \left (x^3+3 x^2\right )-160 e^{x/3} \left (x^4+9 x^3\right )+9\right )dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {\left (-96 x^4+9 x^2-16 e^{x/3} \left (x^4+9 x^3\right )-16 e^{2 x/3} \left (x^3+3 x^2\right )\right ) \exp \left (\frac {1}{3} \left (-80 x^4-160 e^{x/3} x^3-80 e^{2 x/3} x^2+15 x^2-12\right )\right )}{-16 e^{x/3} x^3-96 x^3-144 e^{x/3} x^2-16 e^{2 x/3} x^2-48 e^{2 x/3} x+9 x}\) |
Input:
Int[(E^((-12 + 15*x^2 - 80*E^((2*x)/3)*x^2 - 160*E^(x/3)*x^3 - 80*x^4)/3)* (9 + 90*x^2 - 960*x^4 + E^((2*x)/3)*(-480*x^2 - 160*x^3) + E^(x/3)*(-1440* x^3 - 160*x^4)))/9,x]
Output:
(E^((-12 + 15*x^2 - 80*E^((2*x)/3)*x^2 - 160*E^(x/3)*x^3 - 80*x^4)/3)*(9*x ^2 - 96*x^4 - 16*E^((2*x)/3)*(3*x^2 + x^3) - 16*E^(x/3)*(9*x^3 + x^4)))/(9 *x - 48*E^((2*x)/3)*x - 144*E^(x/3)*x^2 - 16*E^((2*x)/3)*x^2 - 96*x^3 - 16 *E^(x/3)*x^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 0.43 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
risch | \(x \,{\mathrm e}^{-\frac {80 x^{2} {\mathrm e}^{\frac {2 x}{3}}}{3}-\frac {160 x^{3} {\mathrm e}^{\frac {x}{3}}}{3}-\frac {80 x^{4}}{3}+5 x^{2}-4}\) | \(34\) |
norman | \(x \,{\mathrm e}^{-\frac {80 x^{2} {\mathrm e}^{\frac {2 x}{3}}}{3}-\frac {160 x^{3} {\mathrm e}^{\frac {x}{3}}}{3}-\frac {80 x^{4}}{3}+5 x^{2}-4}\) | \(36\) |
parallelrisch | \(x \,{\mathrm e}^{-\frac {80 x^{2} {\mathrm e}^{\frac {2 x}{3}}}{3}-\frac {160 x^{3} {\mathrm e}^{\frac {x}{3}}}{3}-\frac {80 x^{4}}{3}+5 x^{2}-4}\) | \(36\) |
Input:
int(1/9*((-160*x^3-480*x^2)*exp(1/3*x)^2+(-160*x^4-1440*x^3)*exp(1/3*x)-96 0*x^4+90*x^2+9)*exp(-80/3*x^2*exp(1/3*x)^2-160/3*x^3*exp(1/3*x)-80/3*x^4+5 *x^2-4),x,method=_RETURNVERBOSE)
Output:
x*exp(-80/3*x^2*exp(2/3*x)-160/3*x^3*exp(1/3*x)-80/3*x^4+5*x^2-4)
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=x e^{\left (-\frac {80}{3} \, x^{4} - \frac {160}{3} \, x^{3} e^{\left (\frac {1}{3} \, x\right )} - \frac {80}{3} \, x^{2} e^{\left (\frac {2}{3} \, x\right )} + 5 \, x^{2} - 4\right )} \] Input:
integrate(1/9*((-160*x^3-480*x^2)*exp(1/3*x)^2+(-160*x^4-1440*x^3)*exp(1/3 *x)-960*x^4+90*x^2+9)*exp(-80/3*x^2*exp(1/3*x)^2-160/3*x^3*exp(1/3*x)-80/3 *x^4+5*x^2-4),x, algorithm="fricas")
Output:
x*e^(-80/3*x^4 - 160/3*x^3*e^(1/3*x) - 80/3*x^2*e^(2/3*x) + 5*x^2 - 4)
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=x e^{- \frac {80 x^{4}}{3} - \frac {160 x^{3} e^{\frac {x}{3}}}{3} - \frac {80 x^{2} e^{\frac {2 x}{3}}}{3} + 5 x^{2} - 4} \] Input:
integrate(1/9*((-160*x**3-480*x**2)*exp(1/3*x)**2+(-160*x**4-1440*x**3)*ex p(1/3*x)-960*x**4+90*x**2+9)*exp(-80/3*x**2*exp(1/3*x)**2-160/3*x**3*exp(1 /3*x)-80/3*x**4+5*x**2-4),x)
Output:
x*exp(-80*x**4/3 - 160*x**3*exp(x/3)/3 - 80*x**2*exp(2*x/3)/3 + 5*x**2 - 4 )
Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=x e^{\left (-\frac {80}{3} \, x^{4} - \frac {160}{3} \, x^{3} e^{\left (\frac {1}{3} \, x\right )} - \frac {80}{3} \, x^{2} e^{\left (\frac {2}{3} \, x\right )} + 5 \, x^{2} - 4\right )} \] Input:
integrate(1/9*((-160*x^3-480*x^2)*exp(1/3*x)^2+(-160*x^4-1440*x^3)*exp(1/3 *x)-960*x^4+90*x^2+9)*exp(-80/3*x^2*exp(1/3*x)^2-160/3*x^3*exp(1/3*x)-80/3 *x^4+5*x^2-4),x, algorithm="maxima")
Output:
x*e^(-80/3*x^4 - 160/3*x^3*e^(1/3*x) - 80/3*x^2*e^(2/3*x) + 5*x^2 - 4)
\[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=\int { -\frac {1}{9} \, {\left (960 \, x^{4} - 90 \, x^{2} + 160 \, {\left (x^{3} + 3 \, x^{2}\right )} e^{\left (\frac {2}{3} \, x\right )} + 160 \, {\left (x^{4} + 9 \, x^{3}\right )} e^{\left (\frac {1}{3} \, x\right )} - 9\right )} e^{\left (-\frac {80}{3} \, x^{4} - \frac {160}{3} \, x^{3} e^{\left (\frac {1}{3} \, x\right )} - \frac {80}{3} \, x^{2} e^{\left (\frac {2}{3} \, x\right )} + 5 \, x^{2} - 4\right )} \,d x } \] Input:
integrate(1/9*((-160*x^3-480*x^2)*exp(1/3*x)^2+(-160*x^4-1440*x^3)*exp(1/3 *x)-960*x^4+90*x^2+9)*exp(-80/3*x^2*exp(1/3*x)^2-160/3*x^3*exp(1/3*x)-80/3 *x^4+5*x^2-4),x, algorithm="giac")
Output:
integrate(-1/9*(960*x^4 - 90*x^2 + 160*(x^3 + 3*x^2)*e^(2/3*x) + 160*(x^4 + 9*x^3)*e^(1/3*x) - 9)*e^(-80/3*x^4 - 160/3*x^3*e^(1/3*x) - 80/3*x^2*e^(2 /3*x) + 5*x^2 - 4), x)
Time = 1.81 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=x\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^{-\frac {80\,x^4}{3}}\,{\mathrm {e}}^{-\frac {80\,x^2\,{\mathrm {e}}^{\frac {2\,x}{3}}}{3}}\,{\mathrm {e}}^{-\frac {160\,x^3\,{\mathrm {e}}^{x/3}}{3}} \] Input:
int(-(exp(5*x^2 - (160*x^3*exp(x/3))/3 - (80*x^2*exp((2*x)/3))/3 - (80*x^4 )/3 - 4)*(exp((2*x)/3)*(480*x^2 + 160*x^3) + exp(x/3)*(1440*x^3 + 160*x^4) - 90*x^2 + 960*x^4 - 9))/9,x)
Output:
x*exp(-4)*exp(5*x^2)*exp(-(80*x^4)/3)*exp(-(80*x^2*exp((2*x)/3))/3)*exp(-( 160*x^3*exp(x/3))/3)
Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {1}{9} e^{\frac {1}{3} \left (-12+15 x^2-80 e^{2 x/3} x^2-160 e^{x/3} x^3-80 x^4\right )} \left (9+90 x^2-960 x^4+e^{2 x/3} \left (-480 x^2-160 x^3\right )+e^{x/3} \left (-1440 x^3-160 x^4\right )\right ) \, dx=\frac {e^{5 x^{2}} x}{e^{\frac {80 e^{\frac {2 x}{3}} x^{2}}{3}+\frac {160 e^{\frac {x}{3}} x^{3}}{3}+\frac {80 x^{4}}{3}} e^{4}} \] Input:
int(1/9*((-160*x^3-480*x^2)*exp(1/3*x)^2+(-160*x^4-1440*x^3)*exp(1/3*x)-96 0*x^4+90*x^2+9)*exp(-80/3*x^2*exp(1/3*x)^2-160/3*x^3*exp(1/3*x)-80/3*x^4+5 *x^2-4),x)
Output:
(e**(5*x**2)*x)/(e**((80*e**((2*x)/3)*x**2 + 160*e**(x/3)*x**3 + 80*x**4)/ 3)*e**4)