Integrand size = 61, antiderivative size = 19 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=\frac {e^{12}}{-1-\frac {1+\frac {1}{e^5}}{x}+x} \] Output:
exp(12)/(x-(exp(-5)+1)/x-1)
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=-\frac {e^{17} x}{1+e^5 \left (1+x-x^2\right )} \] Input:
Integrate[(E^12*(-E^5 + E^10*(-1 - x^2)))/(1 + E^5*(2 + 2*x - 2*x^2) + E^1 0*(1 + 2*x - x^2 - 2*x^3 + x^4)),x]
Output:
-((E^17*x)/(1 + E^5*(1 + x - x^2)))
Time = 0.53 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {27, 25, 2459, 1380, 27, 2345, 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.
\(\displaystyle \int \frac {e^{12} \left (e^{10} \left (-x^2-1\right )-e^5\right )}{e^5 \left (-2 x^2+2 x+2\right )+e^{10} \left (x^4-2 x^3-x^2+2 x+1\right )+1} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{12} \int -\frac {e^{10} \left (x^2+1\right )+e^5}{2 e^5 \left (-x^2+x+1\right )+e^{10} \left (x^4-2 x^3-x^2+2 x+1\right )+1}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e^{12} \int \frac {e^{10} \left (x^2+1\right )+e^5}{2 e^5 \left (-x^2+x+1\right )+e^{10} \left (x^4-2 x^3-x^2+2 x+1\right )+1}dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle -e^{12} \int \frac {e^{10} \left (x-\frac {1}{2}\right )^2+e^{10} \left (x-\frac {1}{2}\right )+\frac {1}{4} e^5 \left (4+5 e^5\right )}{e^{10} \left (x-\frac {1}{2}\right )^4-\frac {1}{2} e^5 \left (4+5 e^5\right ) \left (x-\frac {1}{2}\right )^2+\frac {1}{16} \left (4+5 e^5\right )^2}d\left (x-\frac {1}{2}\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle -e^{22} \int \frac {4 \left (4 e^{10} \left (x-\frac {1}{2}\right )^2+4 e^{10} \left (x-\frac {1}{2}\right )+e^5 \left (4+5 e^5\right )\right )}{e^{10} \left (-4 e^5 \left (x-\frac {1}{2}\right )^2+5 e^5+4\right )^2}d\left (x-\frac {1}{2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 e^{12} \int \frac {4 e^{10} \left (x-\frac {1}{2}\right )^2+4 e^{10} \left (x-\frac {1}{2}\right )+e^5 \left (4+5 e^5\right )}{\left (-4 e^5 \left (x-\frac {1}{2}\right )^2+5 e^5+4\right )^2}d\left (x-\frac {1}{2}\right )\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -4 e^{12} \left (\frac {e^5 \left (2 \left (x-\frac {1}{2}\right )+1\right )}{2 \left (-4 e^5 \left (x-\frac {1}{2}\right )^2+5 e^5+4\right )}-\frac {\int 0d\left (x-\frac {1}{2}\right )}{2 \left (4+5 e^5\right )}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {2 e^{17} \left (2 \left (x-\frac {1}{2}\right )+1\right )}{-4 e^5 \left (x-\frac {1}{2}\right )^2+5 e^5+4}\) |
Input:
Int[(E^12*(-E^5 + E^10*(-1 - x^2)))/(1 + E^5*(2 + 2*x - 2*x^2) + E^10*(1 + 2*x - x^2 - 2*x^3 + x^4)),x]
Output:
(-2*E^17*(1 + 2*(-1/2 + x)))/(4 + 5*E^5 - 4*E^5*(-1/2 + x)^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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])
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26
method | result | size |
gosper | \(\frac {x \,{\mathrm e}^{17}}{x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-{\mathrm e}^{5}-1}\) | \(24\) |
risch | \(\frac {x \,{\mathrm e}^{17}}{x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-{\mathrm e}^{5}-1}\) | \(24\) |
norman | \(\frac {{\mathrm e}^{5} x \,{\mathrm e}^{12}}{x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-{\mathrm e}^{5}-1}\) | \(26\) |
parallelrisch | \(\frac {{\mathrm e}^{5} x \,{\mathrm e}^{12}}{x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-{\mathrm e}^{5}-1}\) | \(26\) |
Input:
int(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5)^2+(-2 *x^2+2*x+2)*exp(5)+1),x,method=_RETURNVERBOSE)
Output:
x*exp(17)/(x^2*exp(5)-x*exp(5)-exp(5)-1)
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=\frac {x e^{17}}{{\left (x^{2} - x - 1\right )} e^{5} - 1} \] Input:
integrate(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5) ^2+(-2*x^2+2*x+2)*exp(5)+1),x, algorithm="fricas")
Output:
x*e^17/((x^2 - x - 1)*e^5 - 1)
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=\frac {x e^{17}}{x^{2} e^{5} - x e^{5} - e^{5} - 1} \] Input:
integrate(((-x**2-1)*exp(5)**2-exp(5))*exp(12)/((x**4-2*x**3-x**2+2*x+1)*e xp(5)**2+(-2*x**2+2*x+2)*exp(5)+1),x)
Output:
x*exp(17)/(x**2*exp(5) - x*exp(5) - exp(5) - 1)
Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=\frac {x e^{17}}{x^{2} e^{5} - x e^{5} - e^{5} - 1} \] Input:
integrate(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5) ^2+(-2*x^2+2*x+2)*exp(5)+1),x, algorithm="maxima")
Output:
x*e^17/(x^2*e^5 - x*e^5 - e^5 - 1)
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=\frac {x e^{17}}{x^{2} e^{5} - x e^{5} - e^{5} - 1} \] Input:
integrate(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5) ^2+(-2*x^2+2*x+2)*exp(5)+1),x, algorithm="giac")
Output:
x*e^17/(x^2*e^5 - x*e^5 - e^5 - 1)
Time = 1.81 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=-\frac {x\,{\mathrm {e}}^{17}}{-{\mathrm {e}}^5\,x^2+{\mathrm {e}}^5\,x+{\mathrm {e}}^5+1} \] Input:
int(-(exp(12)*(exp(5) + exp(10)*(x^2 + 1)))/(exp(5)*(2*x - 2*x^2 + 2) + ex p(10)*(2*x - x^2 - 2*x^3 + x^4 + 1) + 1),x)
Output:
-(x*exp(17))/(exp(5) + x*exp(5) - x^2*exp(5) + 1)
Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int \frac {e^{12} \left (-e^5+e^{10} \left (-1-x^2\right )\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx=\frac {e^{12} \left (e^{5} x^{2}-e^{5}-1\right )}{e^{5} x^{2}-e^{5} x -e^{5}-1} \] Input:
int(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5)^2+(-2 *x^2+2*x+2)*exp(5)+1),x)
Output:
(e**12*(e**5*x**2 - e**5 - 1))/(e**5*x**2 - e**5*x - e**5 - 1)