Integrand size = 100, antiderivative size = 27 \[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=\frac {2 e^{\frac {\left (e^x+x\right )^2}{x^2}} (3-x)}{(5+3 x)^2} \] Output:
2*exp((exp(x)+x)^2/x^2)/(5+3*x)^2*(3-x)
Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=-\frac {2 e^{\frac {\left (e^x+x\right )^2}{x^2}} (-3+x)}{(5+3 x)^2} \] Input:
Integrate[(E^((E^(2*x) + 2*E^x*x + x^2)/x^2)*(-46*x^3 + 6*x^4 + E^(2*x)*(- 60 + 44*x + 28*x^2 - 12*x^3) + E^x*(-60*x + 44*x^2 + 28*x^3 - 12*x^4)))/(1 25*x^3 + 225*x^4 + 135*x^5 + 27*x^6),x]
Output:
(-2*E^((E^x + x)^2/x^2)*(-3 + x))/(5 + 3*x)^2
Leaf count is larger than twice the leaf count of optimal. \(120\) vs. \(2(27)=54\).
Time = 1.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.44, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2026, 2007, 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.
\(\displaystyle \int \frac {e^{\frac {x^2+2 e^x x+e^{2 x}}{x^2}} \left (6 x^4-46 x^3+e^{2 x} \left (-12 x^3+28 x^2+44 x-60\right )+e^x \left (-12 x^4+28 x^3+44 x^2-60 x\right )\right )}{27 x^6+135 x^5+225 x^4+125 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {x^2+2 e^x x+e^{2 x}}{x^2}} \left (6 x^4-46 x^3+e^{2 x} \left (-12 x^3+28 x^2+44 x-60\right )+e^x \left (-12 x^4+28 x^3+44 x^2-60 x\right )\right )}{x^3 \left (27 x^3+135 x^2+225 x+125\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{\frac {x^2+2 e^x x+e^{2 x}}{x^2}} \left (6 x^4-46 x^3+e^{2 x} \left (-12 x^3+28 x^2+44 x-60\right )+e^x \left (-12 x^4+28 x^3+44 x^2-60 x\right )\right )}{x^3 (3 x+5)^3}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {2 e^{\frac {x^2+2 e^x x+e^{2 x}}{x^2}} \left (e^{2 x} \left (3 x^3-7 x^2-11 x+15\right )+e^x \left (3 x^4-7 x^3-11 x^2+15 x\right )\right )}{x^3 (3 x+5)^3 \left (\frac {e^x x+x+e^x+e^{2 x}}{x^2}-\frac {x^2+2 e^x x+e^{2 x}}{x^3}\right )}\) |
Input:
Int[(E^((E^(2*x) + 2*E^x*x + x^2)/x^2)*(-46*x^3 + 6*x^4 + E^(2*x)*(-60 + 4 4*x + 28*x^2 - 12*x^3) + E^x*(-60*x + 44*x^2 + 28*x^3 - 12*x^4)))/(125*x^3 + 225*x^4 + 135*x^5 + 27*x^6),x]
Output:
(-2*E^((E^(2*x) + 2*E^x*x + x^2)/x^2)*(E^(2*x)*(15 - 11*x - 7*x^2 + 3*x^3) + E^x*(15*x - 11*x^2 - 7*x^3 + 3*x^4)))/(x^3*(5 + 3*x)^3*((E^x + E^(2*x) + x + E^x*x)/x^2 - (E^(2*x) + 2*E^x*x + x^2)/x^3))
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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, 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 = 0.66 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33
method | result | size |
risch | \(-\frac {2 \left (-3+x \right ) {\mathrm e}^{\frac {{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}}{x^{2}}}}{9 x^{2}+30 x +25}\) | \(36\) |
parallelrisch | \(\frac {-60 x \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}}{x^{2}}}+180 \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}}{x^{2}}}}{270 x^{2}+900 x +750}\) | \(57\) |
Input:
int(((-12*x^3+28*x^2+44*x-60)*exp(x)^2+(-12*x^4+28*x^3+44*x^2-60*x)*exp(x) +6*x^4-46*x^3)*exp((exp(x)^2+2*exp(x)*x+x^2)/x^2)/(27*x^6+135*x^5+225*x^4+ 125*x^3),x,method=_RETURNVERBOSE)
Output:
-2*(-3+x)/(9*x^2+30*x+25)*exp((exp(2*x)+2*exp(x)*x+x^2)/x^2)
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=-\frac {2 \, {\left (x - 3\right )} e^{\left (\frac {x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}}{x^{2}}\right )}}{9 \, x^{2} + 30 \, x + 25} \] Input:
integrate(((-12*x^3+28*x^2+44*x-60)*exp(x)^2+(-12*x^4+28*x^3+44*x^2-60*x)* exp(x)+6*x^4-46*x^3)*exp((exp(x)^2+2*exp(x)*x+x^2)/x^2)/(27*x^6+135*x^5+22 5*x^4+125*x^3),x, algorithm="fricas")
Output:
-2*(x - 3)*e^((x^2 + 2*x*e^x + e^(2*x))/x^2)/(9*x^2 + 30*x + 25)
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=\frac {\left (6 - 2 x\right ) e^{\frac {x^{2} + 2 x e^{x} + e^{2 x}}{x^{2}}}}{9 x^{2} + 30 x + 25} \] Input:
integrate(((-12*x**3+28*x**2+44*x-60)*exp(x)**2+(-12*x**4+28*x**3+44*x**2- 60*x)*exp(x)+6*x**4-46*x**3)*exp((exp(x)**2+2*exp(x)*x+x**2)/x**2)/(27*x** 6+135*x**5+225*x**4+125*x**3),x)
Output:
(6 - 2*x)*exp((x**2 + 2*x*exp(x) + exp(2*x))/x**2)/(9*x**2 + 30*x + 25)
Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=-\frac {2 \, {\left (x e - 3 \, e\right )} e^{\left (\frac {2 \, e^{x}}{x} + \frac {e^{\left (2 \, x\right )}}{x^{2}}\right )}}{9 \, x^{2} + 30 \, x + 25} \] Input:
integrate(((-12*x^3+28*x^2+44*x-60)*exp(x)^2+(-12*x^4+28*x^3+44*x^2-60*x)* exp(x)+6*x^4-46*x^3)*exp((exp(x)^2+2*exp(x)*x+x^2)/x^2)/(27*x^6+135*x^5+22 5*x^4+125*x^3),x, algorithm="maxima")
Output:
-2*(x*e - 3*e)*e^(2*e^x/x + e^(2*x)/x^2)/(9*x^2 + 30*x + 25)
\[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=\int { \frac {2 \, {\left (3 \, x^{4} - 23 \, x^{3} - 2 \, {\left (3 \, x^{3} - 7 \, x^{2} - 11 \, x + 15\right )} e^{\left (2 \, x\right )} - 2 \, {\left (3 \, x^{4} - 7 \, x^{3} - 11 \, x^{2} + 15 \, x\right )} e^{x}\right )} e^{\left (\frac {x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}}{x^{2}}\right )}}{27 \, x^{6} + 135 \, x^{5} + 225 \, x^{4} + 125 \, x^{3}} \,d x } \] Input:
integrate(((-12*x^3+28*x^2+44*x-60)*exp(x)^2+(-12*x^4+28*x^3+44*x^2-60*x)* exp(x)+6*x^4-46*x^3)*exp((exp(x)^2+2*exp(x)*x+x^2)/x^2)/(27*x^6+135*x^5+22 5*x^4+125*x^3),x, algorithm="giac")
Output:
integrate(2*(3*x^4 - 23*x^3 - 2*(3*x^3 - 7*x^2 - 11*x + 15)*e^(2*x) - 2*(3 *x^4 - 7*x^3 - 11*x^2 + 15*x)*e^x)*e^((x^2 + 2*x*e^x + e^(2*x))/x^2)/(27*x ^6 + 135*x^5 + 225*x^4 + 125*x^3), x)
Time = 2.98 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=-\frac {2\,\mathrm {e}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{x^2}}\,\left (x-3\right )}{{\left (3\,x+5\right )}^2} \] Input:
int(-(exp((exp(2*x) + 2*x*exp(x) + x^2)/x^2)*(exp(x)*(60*x - 44*x^2 - 28*x ^3 + 12*x^4) - exp(2*x)*(44*x + 28*x^2 - 12*x^3 - 60) + 46*x^3 - 6*x^4))/( 125*x^3 + 225*x^4 + 135*x^5 + 27*x^6),x)
Output:
-(2*exp(1)*exp((2*exp(x))/x)*exp(exp(2*x)/x^2)*(x - 3))/(3*x + 5)^2
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\frac {e^{2 x}+2 e^x x+x^2}{x^2}} \left (-46 x^3+6 x^4+e^{2 x} \left (-60+44 x+28 x^2-12 x^3\right )+e^x \left (-60 x+44 x^2+28 x^3-12 x^4\right )\right )}{125 x^3+225 x^4+135 x^5+27 x^6} \, dx=\frac {2 e^{\frac {e^{2 x}+2 e^{x} x}{x^{2}}} e \left (-x +3\right )}{9 x^{2}+30 x +25} \] Input:
int(((-12*x^3+28*x^2+44*x-60)*exp(x)^2+(-12*x^4+28*x^3+44*x^2-60*x)*exp(x) +6*x^4-46*x^3)*exp((exp(x)^2+2*exp(x)*x+x^2)/x^2)/(27*x^6+135*x^5+225*x^4+ 125*x^3),x)
Output:
(2*e**((e**(2*x) + 2*e**x*x)/x**2)*e*( - x + 3))/(9*x**2 + 30*x + 25)