\(\int \frac {-4 x+3 x^2+e^3 (2 x-x^2)+e^5 (4-12 x+13 x^2-6 x^3+x^4+e^6 (1-2 x+x^2)+e^3 (-4+10 x-8 x^2+2 x^3))}{4-12 x+13 x^2-6 x^3+x^4+e^6 (1-2 x+x^2)+e^3 (-4+10 x-8 x^2+2 x^3)} \, dx\) [2012]

Optimal result

Integrand size = 127, antiderivative size = 31 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=x \left (e^5+\frac {x^2}{\left (2-e^3-x\right ) \left (-x+x^2\right )}\right ) \] Output:

x*(1/(2-x-exp(3))*x^2/(x^2-x)+exp(5))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=\frac {1}{\left (1-e^3\right ) (-1+x)}+e^5 (-1+x)+\frac {\left (-2+e^3\right )^2}{\left (-1+e^3\right ) \left (-2+e^3+x\right )} \] Input:

Integrate[(-4*x + 3*x^2 + E^3*(2*x - x^2) + E^5*(4 - 12*x + 13*x^2 - 6*x^3 
 + x^4 + E^6*(1 - 2*x + x^2) + E^3*(-4 + 10*x - 8*x^2 + 2*x^3)))/(4 - 12*x 
 + 13*x^2 - 6*x^3 + x^4 + E^6*(1 - 2*x + x^2) + E^3*(-4 + 10*x - 8*x^2 + 2 
*x^3)),x]
 

Output:

1/((1 - E^3)*(-1 + x)) + E^5*(-1 + x) + (-2 + E^3)^2/((-1 + E^3)*(-2 + E^3 
 + x))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(31)=62\).

Time = 0.43 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.55, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2459, 1380, 2345, 27, 24}

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.

\(\displaystyle \int \frac {3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (x^4-6 x^3+13 x^2+e^6 \left (x^2-2 x+1\right )+e^3 \left (2 x^3-8 x^2+10 x-4\right )-12 x+4\right )-4 x}{x^4-6 x^3+13 x^2+e^6 \left (x^2-2 x+1\right )+e^3 \left (2 x^3-8 x^2+10 x-4\right )-12 x+4} \, dx\)

\(\Big \downarrow \) 2459

\(\displaystyle \int \frac {e^5 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^4+\frac {1}{2} \left (6-2 e^3-e^5+2 e^8-e^{11}\right ) \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^2+\left (5-4 e^3+e^6\right ) \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )+\frac {1}{16} \left (1-e^3\right )^2 \left (12-4 e^3+e^5-2 e^8+e^{11}\right )}{\left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^4-\frac {1}{2} \left (1-e^3\right )^2 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^2+\frac {1}{16} \left (1-e^3\right )^4}d\left (x+\frac {1}{4} \left (2 e^3-6\right )\right )\)

\(\Big \downarrow \) 1380

\(\displaystyle \int \frac {16 e^5 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^4+8 \left (6-2 e^3-e^5+2 e^8-e^{11}\right ) \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^2+16 \left (5-4 e^3+e^6\right ) \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )+\left (1-e^3\right )^2 \left (12-4 e^3+e^5-2 e^8+e^{11}\right )}{\left (\left (1-e^3\right )^2-4 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^2\right )^2}d\left (x+\frac {1}{4} \left (2 e^3-6\right )\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {2 \left (2 \left (3-e^3\right ) \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )+e^6-4 e^3+5\right )}{\left (e^3-1\right )^2-4 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^2}-\frac {\int -\frac {2 e^5 \left (1-e^3\right )^2 \left (\left (1-e^3\right )^2-4 \left (x+\frac {1}{4} \left (-6+2 e^3\right )\right )^2\right )}{\left (-1+e^3\right )^2-4 \left (x+\frac {1}{4} \left (-6+2 e^3\right )\right )^2}d\left (x+\frac {1}{4} \left (-6+2 e^3\right )\right )}{2 \left (1-e^3\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle e^5 \int 1d\left (x+\frac {1}{4} \left (-6+2 e^3\right )\right )+\frac {2 \left (2 \left (3-e^3\right ) \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )+e^6-4 e^3+5\right )}{\left (e^3-1\right )^2-4 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^2}\)

\(\Big \downarrow \) 24

\(\displaystyle e^5 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )+\frac {2 \left (2 \left (3-e^3\right ) \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )+e^6-4 e^3+5\right )}{\left (e^3-1\right )^2-4 \left (x+\frac {1}{4} \left (2 e^3-6\right )\right )^2}\)

Input:

Int[(-4*x + 3*x^2 + E^3*(2*x - x^2) + E^5*(4 - 12*x + 13*x^2 - 6*x^3 + x^4 
 + E^6*(1 - 2*x + x^2) + E^3*(-4 + 10*x - 8*x^2 + 2*x^3)))/(4 - 12*x + 13* 
x^2 - 6*x^3 + x^4 + E^6*(1 - 2*x + x^2) + E^3*(-4 + 10*x - 8*x^2 + 2*x^3)) 
,x]
 

Output:

E^5*((-6 + 2*E^3)/4 + x) + (2*(5 - 4*E^3 + E^6 + 2*(3 - E^3)*((-6 + 2*E^3) 
/4 + x)))/((-1 + E^3)^2 - 4*((-6 + 2*E^3)/4 + x)^2)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

rule 2459

Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 
]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x 
 -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial 
Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - 
> x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ 
[Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] &&  !(MonomialQ[Qx, x] 
&& IGtQ[p, 0])
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19

method	result	size
risch	\(x \,{\mathrm e}^{5}+\frac {\left ({\mathrm e}^{3}-3\right ) x -{\mathrm e}^{3}+2}{x \,{\mathrm e}^{3}+x^{2}-{\mathrm e}^{3}-3 x +2}\)	\(37\)
norman	\(\frac {x^{3} {\mathrm e}^{5}+2+\left (-{\mathrm e}^{6} {\mathrm e}^{5}+5 \,{\mathrm e}^{3} {\mathrm e}^{5}+{\mathrm e}^{3}-7 \,{\mathrm e}^{5}-3\right ) x +{\mathrm e}^{6} {\mathrm e}^{5}-5 \,{\mathrm e}^{3} {\mathrm e}^{5}-{\mathrm e}^{3}+6 \,{\mathrm e}^{5}}{\left (-1+x \right ) \left (x +{\mathrm e}^{3}-2\right )}\)	\(67\)
gosper	\(-\frac {{\mathrm e}^{6} {\mathrm e}^{5} x -x^{3} {\mathrm e}^{5}-{\mathrm e}^{6} {\mathrm e}^{5}-5 \,{\mathrm e}^{3} {\mathrm e}^{5} x +5 \,{\mathrm e}^{3} {\mathrm e}^{5}-x \,{\mathrm e}^{3}+7 x \,{\mathrm e}^{5}+{\mathrm e}^{3}-6 \,{\mathrm e}^{5}+3 x -2}{x \,{\mathrm e}^{3}+x^{2}-{\mathrm e}^{3}-3 x +2}\)	\(78\)
parallelrisch	\(-\frac {{\mathrm e}^{6} {\mathrm e}^{5} x -x^{3} {\mathrm e}^{5}-{\mathrm e}^{6} {\mathrm e}^{5}-5 \,{\mathrm e}^{3} {\mathrm e}^{5} x +5 \,{\mathrm e}^{3} {\mathrm e}^{5}-x \,{\mathrm e}^{3}+7 x \,{\mathrm e}^{5}+{\mathrm e}^{3}-6 \,{\mathrm e}^{5}+3 x -2}{x \,{\mathrm e}^{3}+x^{2}-{\mathrm e}^{3}-3 x +2}\)	\(78\)

Input:

int((((x^2-2*x+1)*exp(3)^2+(2*x^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13*x^2-12 
*x+4)*exp(5)+(-x^2+2*x)*exp(3)+3*x^2-4*x)/((x^2-2*x+1)*exp(3)^2+(2*x^3-8*x 
^2+10*x-4)*exp(3)+x^4-6*x^3+13*x^2-12*x+4),x,method=_RETURNVERBOSE)
 

Output:

x*exp(5)+((exp(3)-3)*x-exp(3)+2)/(x*exp(3)+x^2-exp(3)-3*x+2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=\frac {{\left (x^{2} - x\right )} e^{8} + {\left (x^{3} - 3 \, x^{2} + 2 \, x\right )} e^{5} + {\left (x - 1\right )} e^{3} - 3 \, x + 2}{x^{2} + {\left (x - 1\right )} e^{3} - 3 \, x + 2} \] Input:

integrate((((x^2-2*x+1)*exp(3)^2+(2*x^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13* 
x^2-12*x+4)*exp(5)+(-x^2+2*x)*exp(3)+3*x^2-4*x)/((x^2-2*x+1)*exp(3)^2+(2*x 
^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13*x^2-12*x+4),x, algorithm="fricas")
 

Output:

((x^2 - x)*e^8 + (x^3 - 3*x^2 + 2*x)*e^5 + (x - 1)*e^3 - 3*x + 2)/(x^2 + ( 
x - 1)*e^3 - 3*x + 2)
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=x e^{5} + \frac {x \left (-3 + e^{3}\right ) - e^{3} + 2}{x^{2} + x \left (-3 + e^{3}\right ) - e^{3} + 2} \] Input:

integrate((((x**2-2*x+1)*exp(3)**2+(2*x**3-8*x**2+10*x-4)*exp(3)+x**4-6*x* 
*3+13*x**2-12*x+4)*exp(5)+(-x**2+2*x)*exp(3)+3*x**2-4*x)/((x**2-2*x+1)*exp 
(3)**2+(2*x**3-8*x**2+10*x-4)*exp(3)+x**4-6*x**3+13*x**2-12*x+4),x)
 

Output:

x*exp(5) + (x*(-3 + exp(3)) - exp(3) + 2)/(x**2 + x*(-3 + exp(3)) - exp(3) 
 + 2)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=x e^{5} + \frac {x {\left (e^{3} - 3\right )} - e^{3} + 2}{x^{2} + x {\left (e^{3} - 3\right )} - e^{3} + 2} \] Input:

integrate((((x^2-2*x+1)*exp(3)^2+(2*x^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13* 
x^2-12*x+4)*exp(5)+(-x^2+2*x)*exp(3)+3*x^2-4*x)/((x^2-2*x+1)*exp(3)^2+(2*x 
^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13*x^2-12*x+4),x, algorithm="maxima")
 

Output:

x*e^5 + (x*(e^3 - 3) - e^3 + 2)/(x^2 + x*(e^3 - 3) - e^3 + 2)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=x e^{5} + \frac {x e^{3} - 3 \, x - e^{3} + 2}{x^{2} + x e^{3} - 3 \, x - e^{3} + 2} \] Input:

integrate((((x^2-2*x+1)*exp(3)^2+(2*x^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13* 
x^2-12*x+4)*exp(5)+(-x^2+2*x)*exp(3)+3*x^2-4*x)/((x^2-2*x+1)*exp(3)^2+(2*x 
^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13*x^2-12*x+4),x, algorithm="giac")
 

Output:

x*e^5 + (x*e^3 - 3*x - e^3 + 2)/(x^2 + x*e^3 - 3*x - e^3 + 2)
 

Mupad [B] (verification not implemented)

Time = 3.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=\frac {x\,\left ({\mathrm {e}}^3-3\right )-{\mathrm {e}}^3+2}{x^2+\left ({\mathrm {e}}^3-3\right )\,x-{\mathrm {e}}^3+2}+x\,{\mathrm {e}}^5 \] Input:

int((exp(3)*(2*x - x^2) - 4*x + exp(5)*(exp(3)*(10*x - 8*x^2 + 2*x^3 - 4) 
- 12*x + exp(6)*(x^2 - 2*x + 1) + 13*x^2 - 6*x^3 + x^4 + 4) + 3*x^2)/(exp( 
3)*(10*x - 8*x^2 + 2*x^3 - 4) - 12*x + exp(6)*(x^2 - 2*x + 1) + 13*x^2 - 6 
*x^3 + x^4 + 4),x)
 

Output:

(x*(exp(3) - 3) - exp(3) + 2)/(x*(exp(3) - 3) - exp(3) + x^2 + 2) + x*exp( 
5)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.48 \[ \int \frac {-4 x+3 x^2+e^3 \left (2 x-x^2\right )+e^5 \left (4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )\right )}{4-12 x+13 x^2-6 x^3+x^4+e^6 \left (1-2 x+x^2\right )+e^3 \left (-4+10 x-8 x^2+2 x^3\right )} \, dx=\frac {e^{11} x^{2}-e^{11}+e^{8} x^{3}-5 e^{8} x^{2}+4 e^{8}-3 e^{5} x^{3}+7 e^{5} x^{2}-4 e^{5}-e^{3} x^{2}+3 x^{2}}{e^{6} x -e^{6}+e^{3} x^{2}-6 e^{3} x +5 e^{3}-3 x^{2}+9 x -6} \] Input:

int((((x^2-2*x+1)*exp(3)^2+(2*x^3-8*x^2+10*x-4)*exp(3)+x^4-6*x^3+13*x^2-12 
*x+4)*exp(5)+(-x^2+2*x)*exp(3)+3*x^2-4*x)/((x^2-2*x+1)*exp(3)^2+(2*x^3-8*x 
^2+10*x-4)*exp(3)+x^4-6*x^3+13*x^2-12*x+4),x)
 

Output:

(e**11*x**2 - e**11 + e**8*x**3 - 5*e**8*x**2 + 4*e**8 - 3*e**5*x**3 + 7*e 
**5*x**2 - 4*e**5 - e**3*x**2 + 3*x**2)/(e**6*x - e**6 + e**3*x**2 - 6*e** 
3*x + 5*e**3 - 3*x**2 + 9*x - 6)