Integrand size = 114, antiderivative size = 28 \[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx=e^{x \left (3-e^{\frac {x}{2 \left (e^4+x\right )}}+\frac {5}{x}+x\right )} \] Output:
exp(x*(3+5/x+x-exp(1/2*x/(x+exp(4)))))
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx=e^{5-\left (-3+e^{\frac {x}{2 \left (e^4+x\right )}}\right ) x+x^2} \] Input:
Integrate[(E^(5 + 3*x - E^(x/(2*E^4 + 2*x))*x + x^2)*(6*x^2 + 4*x^3 + E^8* (6 + 4*x) + E^(x/(2*E^4 + 2*x))*(-2*E^8 - 5*E^4*x - 2*x^2) + E^4*(12*x + 8 *x^2)))/(2*E^8 + 4*E^4*x + 2*x^2),x]
Output:
E^(5 - (-3 + E^(x/(2*(E^4 + x))))*x + x^2)
Time = 4.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {2007, 7257}
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.
\(\displaystyle \int \frac {e^{x^2-e^{\frac {x}{2 x+2 e^4}} x+3 x+5} \left (4 x^3+6 x^2+e^{\frac {x}{2 x+2 e^4}} \left (-2 x^2-5 e^4 x-2 e^8\right )+e^4 \left (8 x^2+12 x\right )+e^8 (4 x+6)\right )}{2 x^2+4 e^4 x+2 e^8} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{x^2-e^{\frac {x}{2 x+2 e^4}} x+3 x+5} \left (4 x^3+6 x^2+e^{\frac {x}{2 x+2 e^4}} \left (-2 x^2-5 e^4 x-2 e^8\right )+e^4 \left (8 x^2+12 x\right )+e^8 (4 x+6)\right )}{\left (\sqrt {2} x+\sqrt {2} e^4\right )^2}dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{x^2-e^{\frac {x}{2 \left (x+e^4\right )}} x+3 x+5}\) |
Input:
Int[(E^(5 + 3*x - E^(x/(2*E^4 + 2*x))*x + x^2)*(6*x^2 + 4*x^3 + E^8*(6 + 4 *x) + E^(x/(2*E^4 + 2*x))*(-2*E^8 - 5*E^4*x - 2*x^2) + E^4*(12*x + 8*x^2)) )/(2*E^8 + 4*E^4*x + 2*x^2),x]
Output:
E^(5 + 3*x - E^(x/(2*(E^4 + x)))*x + x^2)
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 2.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
risch | \({\mathrm e}^{-x \,{\mathrm e}^{\frac {x}{2 \,{\mathrm e}^{4}+2 x}}+x^{2}+3 x +5}\) | \(23\) |
parallelrisch | \({\mathrm e}^{-x \,{\mathrm e}^{\frac {x}{2 \,{\mathrm e}^{4}+2 x}}+x^{2}+3 x +5}\) | \(23\) |
norman | \(\frac {x \,{\mathrm e}^{-x \,{\mathrm e}^{\frac {x}{2 \,{\mathrm e}^{4}+2 x}}+x^{2}+3 x +5}+{\mathrm e}^{4} {\mathrm e}^{-x \,{\mathrm e}^{\frac {x}{2 \,{\mathrm e}^{4}+2 x}}+x^{2}+3 x +5}}{x +{\mathrm e}^{4}}\) | \(64\) |
Input:
int(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*exp(4)^2 +(8*x^2+12*x)*exp(4)+4*x^3+6*x^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3*x+5)/ (2*exp(4)^2+4*x*exp(4)+2*x^2),x,method=_RETURNVERBOSE)
Output:
exp(-x*exp(1/2*x/(x+exp(4)))+x^2+3*x+5)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx=e^{\left (x^{2} - x e^{\left (\frac {x}{2 \, {\left (x + e^{4}\right )}}\right )} + 3 \, x + 5\right )} \] Input:
integrate(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*ex p(4)^2+(8*x^2+12*x)*exp(4)+4*x^3+6*x^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3 *x+5)/(2*exp(4)^2+4*x*exp(4)+2*x^2),x, algorithm="fricas")
Output:
e^(x^2 - x*e^(1/2*x/(x + e^4)) + 3*x + 5)
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx=e^{x^{2} - x e^{\frac {x}{2 x + 2 e^{4}}} + 3 x + 5} \] Input:
integrate(((-2*exp(4)**2-5*x*exp(4)-2*x**2)*exp(x/(2*exp(4)+2*x))+(4*x+6)* exp(4)**2+(8*x**2+12*x)*exp(4)+4*x**3+6*x**2)*exp(-x*exp(x/(2*exp(4)+2*x)) +x**2+3*x+5)/(2*exp(4)**2+4*x*exp(4)+2*x**2),x)
Output:
exp(x**2 - x*exp(x/(2*x + 2*exp(4))) + 3*x + 5)
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx=e^{\left (x^{2} - x e^{\left (-\frac {e^{4}}{2 \, {\left (x + e^{4}\right )}} + \frac {1}{2}\right )} + 3 \, x + 5\right )} \] Input:
integrate(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*ex p(4)^2+(8*x^2+12*x)*exp(4)+4*x^3+6*x^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3 *x+5)/(2*exp(4)^2+4*x*exp(4)+2*x^2),x, algorithm="maxima")
Output:
e^(x^2 - x*e^(-1/2*e^4/(x + e^4) + 1/2) + 3*x + 5)
\[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx=\int { \frac {{\left (4 \, x^{3} + 6 \, x^{2} + 2 \, {\left (2 \, x + 3\right )} e^{8} + 4 \, {\left (2 \, x^{2} + 3 \, x\right )} e^{4} - {\left (2 \, x^{2} + 5 \, x e^{4} + 2 \, e^{8}\right )} e^{\left (\frac {x}{2 \, {\left (x + e^{4}\right )}}\right )}\right )} e^{\left (x^{2} - x e^{\left (\frac {x}{2 \, {\left (x + e^{4}\right )}}\right )} + 3 \, x + 5\right )}}{2 \, {\left (x^{2} + 2 \, x e^{4} + e^{8}\right )}} \,d x } \] Input:
integrate(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*ex p(4)^2+(8*x^2+12*x)*exp(4)+4*x^3+6*x^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3 *x+5)/(2*exp(4)^2+4*x*exp(4)+2*x^2),x, algorithm="giac")
Output:
integrate(1/2*(4*x^3 + 6*x^2 + 2*(2*x + 3)*e^8 + 4*(2*x^2 + 3*x)*e^4 - (2* x^2 + 5*x*e^4 + 2*e^8)*e^(1/2*x/(x + e^4)))*e^(x^2 - x*e^(1/2*x/(x + e^4)) + 3*x + 5)/(x^2 + 2*x*e^4 + e^8), x)
Time = 2.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx={\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{\frac {x}{2\,x+2\,{\mathrm {e}}^4}}} \] Input:
int((exp(3*x - x*exp(x/(2*x + 2*exp(4))) + x^2 + 5)*(exp(4)*(12*x + 8*x^2) - exp(x/(2*x + 2*exp(4)))*(2*exp(8) + 5*x*exp(4) + 2*x^2) + 6*x^2 + 4*x^3 + exp(8)*(4*x + 6)))/(2*exp(8) + 4*x*exp(4) + 2*x^2),x)
Output:
exp(3*x)*exp(x^2)*exp(5)*exp(-x*exp(x/(2*x + 2*exp(4))))
Time = 2.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 e^8+4 e^4 x+2 x^2} \, dx=\frac {e^{x^{2}+3 x} e^{5}}{e^{e^{\frac {x}{2 e^{4}+2 x}} x}} \] Input:
int(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*exp(4)^2 +(8*x^2+12*x)*exp(4)+4*x^3+6*x^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3*x+5)/ (2*exp(4)^2+4*x*exp(4)+2*x^2),x)
Output:
(e**(x**2 + 3*x)*e**5)/e**(e**(x/(2*e**4 + 2*x))*x)