Integrand size = 57, antiderivative size = 22 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=\frac {3 (-2+x)}{8-e^3+x+\frac {9 x^2}{25}} \] Output:
3*(-2+x)/(x+8+9/25*x^2-exp(3))
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=-\frac {75 (2-x)}{200-25 e^3+25 x+9 x^2} \] Input:
Integrate[(18750 - 1875*E^3 + 2700*x - 675*x^2)/(40000 + 625*E^6 + 10000*x + 4225*x^2 + 450*x^3 + 81*x^4 + E^3*(-10000 - 1250*x - 450*x^2)),x]
Output:
(-75*(2 - x))/(200 - 25*E^3 + 25*x + 9*x^2)
Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {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.
\(\displaystyle \int \frac {-675 x^2+2700 x-1875 e^3+18750}{81 x^4+450 x^3+4225 x^2+e^3 \left (-450 x^2-1250 x-10000\right )+10000 x+625 e^6+40000} \, dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \int \frac {-675 \left (x+\frac {25}{18}\right )^2+4575 \left (x+\frac {25}{18}\right )+\frac {625}{12} \left (263-36 e^3\right )}{81 \left (x+\frac {25}{18}\right )^4+\frac {25}{2} \left (263-36 e^3\right ) \left (x+\frac {25}{18}\right )^2+\frac {625 \left (263-36 e^3\right )^2}{1296}}d\left (x+\frac {25}{18}\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 81 \int \frac {100 \left (-324 \left (x+\frac {25}{18}\right )^2+2196 \left (x+\frac {25}{18}\right )+25 \left (263-36 e^3\right )\right )}{3 \left (324 \left (x+\frac {25}{18}\right )^2+25 \left (263-36 e^3\right )\right )^2}d\left (x+\frac {25}{18}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2700 \int \frac {-324 \left (x+\frac {25}{18}\right )^2+2196 \left (x+\frac {25}{18}\right )+25 \left (263-36 e^3\right )}{\left (324 \left (x+\frac {25}{18}\right )^2+25 \left (263-36 e^3\right )\right )^2}d\left (x+\frac {25}{18}\right )\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle 2700 \left (-\frac {\int 0d\left (x+\frac {25}{18}\right )}{50 \left (263-36 e^3\right )}-\frac {61-18 \left (x+\frac {25}{18}\right )}{18 \left (324 \left (x+\frac {25}{18}\right )^2+25 \left (263-36 e^3\right )\right )}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {150 \left (61-18 \left (x+\frac {25}{18}\right )\right )}{324 \left (x+\frac {25}{18}\right )^2+25 \left (263-36 e^3\right )}\) |
Input:
Int[(18750 - 1875*E^3 + 2700*x - 675*x^2)/(40000 + 625*E^6 + 10000*x + 422 5*x^2 + 450*x^3 + 81*x^4 + E^3*(-10000 - 1250*x - 450*x^2)),x]
Output:
(-150*(61 - 18*(25/18 + x)))/(25*(263 - 36*E^3) + 324*(25/18 + 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 = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {-3 x +6}{-\frac {9 x^{2}}{25}+{\mathrm e}^{3}-x -8}\) | \(21\) |
gosper | \(-\frac {75 \left (-2+x \right )}{-9 x^{2}+25 \,{\mathrm e}^{3}-25 x -200}\) | \(22\) |
norman | \(\frac {-75 x +150}{-9 x^{2}+25 \,{\mathrm e}^{3}-25 x -200}\) | \(23\) |
parallelrisch | \(-\frac {-1350+675 x}{9 \left (-9 x^{2}+25 \,{\mathrm e}^{3}-25 x -200\right )}\) | \(24\) |
Input:
int((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250*x-100 00)*exp(3)+81*x^4+450*x^3+4225*x^2+10000*x+40000),x,method=_RETURNVERBOSE)
Output:
(-3*x+6)/(-9/25*x^2+exp(3)-x-8)
Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=\frac {75 \, {\left (x - 2\right )}}{9 \, x^{2} + 25 \, x - 25 \, e^{3} + 200} \] Input:
integrate((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250 *x-10000)*exp(3)+81*x^4+450*x^3+4225*x^2+10000*x+40000),x, algorithm="fric as")
Output:
75*(x - 2)/(9*x^2 + 25*x - 25*e^3 + 200)
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=- \frac {150 - 75 x}{9 x^{2} + 25 x - 25 e^{3} + 200} \] Input:
integrate((-1875*exp(3)-675*x**2+2700*x+18750)/(625*exp(3)**2+(-450*x**2-1 250*x-10000)*exp(3)+81*x**4+450*x**3+4225*x**2+10000*x+40000),x)
Output:
-(150 - 75*x)/(9*x**2 + 25*x - 25*exp(3) + 200)
Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=\frac {75 \, {\left (x - 2\right )}}{9 \, x^{2} + 25 \, x - 25 \, e^{3} + 200} \] Input:
integrate((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250 *x-10000)*exp(3)+81*x^4+450*x^3+4225*x^2+10000*x+40000),x, algorithm="maxi ma")
Output:
75*(x - 2)/(9*x^2 + 25*x - 25*e^3 + 200)
Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=\frac {75 \, {\left (x - 2\right )}}{9 \, x^{2} + 25 \, x - 25 \, e^{3} + 200} \] Input:
integrate((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250 *x-10000)*exp(3)+81*x^4+450*x^3+4225*x^2+10000*x+40000),x, algorithm="giac ")
Output:
75*(x - 2)/(9*x^2 + 25*x - 25*e^3 + 200)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=\frac {75\,x-150}{9\,x^2+25\,x-25\,{\mathrm {e}}^3+200} \] Input:
int((2700*x - 1875*exp(3) - 675*x^2 + 18750)/(10000*x + 625*exp(6) - exp(3 )*(1250*x + 450*x^2 + 10000) + 4225*x^2 + 450*x^3 + 81*x^4 + 40000),x)
Output:
(75*x - 150)/(25*x - 25*exp(3) + 9*x^2 + 200)
Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 \left (-10000-1250 x-450 x^2\right )} \, dx=\frac {-75 e^{3}+27 x^{2}+750}{25 e^{3}-9 x^{2}-25 x -200} \] Input:
int((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250*x-100 00)*exp(3)+81*x^4+450*x^3+4225*x^2+10000*x+40000),x)
Output:
(3*( - 25*e**3 + 9*x**2 + 250))/(25*e**3 - 9*x**2 - 25*x - 200)