Integrand size = 66, antiderivative size = 24 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^2+\frac {25 e^8 \left (1+x^2\right )}{4 x^2 (3+x)^2} \] Output:
x^2+1/2*(x^2+1)/(3/5+1/5*x)/(6/5+2/5*x)/exp(ln(x)-4)^2
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^2+\frac {25 e^8 \left (1+x^2\right )}{4 x^2 (3+x)^2} \] Input:
Integrate[(E^8*(-75 - 50*x - 25*x^3 + (x^2*(108*x^2 + 108*x^3 + 36*x^4 + 4 *x^5))/E^8))/(x^2*(54*x + 54*x^2 + 18*x^3 + 2*x^4)),x]
Output:
x^2 + (25*E^8*(1 + x^2))/(4*x^2*(3 + x)^2)
Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {9, 27, 2007, 2123, 2009}
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.
\(\displaystyle \int \frac {e^8 \left (-25 x^3+\frac {\left (4 x^5+36 x^4+108 x^3+108 x^2\right ) x^2}{e^8}-50 x-75\right )}{x^2 \left (2 x^4+18 x^3+54 x^2+54 x\right )} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int -\frac {e^8 \left (25 x^3-\frac {4 \left (x^5+9 x^4+27 x^3+27 x^2\right ) x^2}{e^8}+50 x+75\right )}{2 x^3 \left (x^3+9 x^2+27 x+27\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{2} e^8 \int \frac {25 x^3-\frac {4 \left (x^5+9 x^4+27 x^3+27 x^2\right ) x^2}{e^8}+50 x+75}{x^3 \left (x^3+9 x^2+27 x+27\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -\frac {1}{2} e^8 \int \frac {25 x^3-\frac {4 \left (x^5+9 x^4+27 x^3+27 x^2\right ) x^2}{e^8}+50 x+75}{x^3 (x+3)^3}dx\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle -\frac {1}{2} e^8 \int \left (-\frac {4 x}{e^8}+\frac {25}{27 (x+3)^2}+\frac {250}{9 (x+3)^3}-\frac {25}{27 x^2}+\frac {25}{9 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{2} e^8 \left (-\frac {2 x^2}{e^8}-\frac {25}{18 x^2}-\frac {25}{27 (x+3)}-\frac {125}{9 (x+3)^2}+\frac {25}{27 x}\right )\) |
Input:
Int[(E^8*(-75 - 50*x - 25*x^3 + (x^2*(108*x^2 + 108*x^3 + 36*x^4 + 4*x^5)) /E^8))/(x^2*(54*x + 54*x^2 + 18*x^3 + 2*x^4)),x]
Output:
-1/2*(E^8*(-25/(18*x^2) + 25/(27*x) - (2*x^2)/E^8 - 125/(9*(3 + x)^2) - 25 /(27*(3 + x))))
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Time = 0.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
risch | \(x^{2}+\frac {{\mathrm e}^{8} \left (\frac {25 x^{2}}{4}+\frac {25}{4}\right )}{x^{2} \left (x^{2}+6 x +9\right )}\) | \(28\) |
default | \(x^{2}-\frac {25 \left (-3 x^{4}-3 x^{2}\right ) {\mathrm e}^{8}}{12 x^{4} \left (x^{2}+6 x +9\right )}\) | \(38\) |
parts | \(x^{2}-\frac {25 \left (-3 x^{4}-3 x^{2}\right ) {\mathrm e}^{8}}{12 x^{4} \left (x^{2}+6 x +9\right )}\) | \(38\) |
norman | \(\frac {\left (-54 x^{3} {\mathrm e}^{4}+\frac {{\mathrm e}^{4} \left (25 \,{\mathrm e}^{8}-324\right ) x^{2}}{4}+{\mathrm e}^{4} x^{6}+\frac {25 \,{\mathrm e}^{12}}{4}+6 x^{5} {\mathrm e}^{4}\right ) {\mathrm e}^{-4}}{x^{2} \left (3+x \right )^{2}}\) | \(56\) |
parallelrisch | \(-\frac {\left (-12 \,{\mathrm e}^{-8} x^{8}-72 \,{\mathrm e}^{-8} x^{7}-108 \,{\mathrm e}^{-8} x^{6}-75 x^{2}-75 x^{4}\right ) {\mathrm e}^{8}}{12 x^{4} \left (x^{2}+6 x +9\right )}\) | \(70\) |
Input:
int(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(ln(x)-4)^2-25*x^3-50*x-75)/(2*x^4+ 18*x^3+54*x^2+54*x)/exp(ln(x)-4)^2,x,method=_RETURNVERBOSE)
Output:
x^2+exp(8)*(25/4*x^2+25/4)/x^2/(x^2+6*x+9)
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=\frac {4 \, x^{6} + 24 \, x^{5} + 36 \, x^{4} + 25 \, {\left (x^{2} + 1\right )} e^{8}}{4 \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )}} \] Input:
integrate(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(log(x)-4)^2-25*x^3-50*x-75)/ (2*x^4+18*x^3+54*x^2+54*x)/exp(log(x)-4)^2,x, algorithm="fricas")
Output:
1/4*(4*x^6 + 24*x^5 + 36*x^4 + 25*(x^2 + 1)*e^8)/(x^4 + 6*x^3 + 9*x^2)
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^{2} + \frac {25 x^{2} e^{8} + 25 e^{8}}{4 x^{4} + 24 x^{3} + 36 x^{2}} \] Input:
integrate(((4*x**5+36*x**4+108*x**3+108*x**2)*exp(ln(x)-4)**2-25*x**3-50*x -75)/(2*x**4+18*x**3+54*x**2+54*x)/exp(ln(x)-4)**2,x)
Output:
x**2 + (25*x**2*exp(8) + 25*exp(8))/(4*x**4 + 24*x**3 + 36*x**2)
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=\frac {1}{4} \, {\left (4 \, x^{2} e^{\left (-8\right )} + \frac {25 \, {\left (x^{2} + 1\right )}}{x^{4} + 6 \, x^{3} + 9 \, x^{2}}\right )} e^{8} \] Input:
integrate(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(log(x)-4)^2-25*x^3-50*x-75)/ (2*x^4+18*x^3+54*x^2+54*x)/exp(log(x)-4)^2,x, algorithm="maxima")
Output:
1/4*(4*x^2*e^(-8) + 25*(x^2 + 1)/(x^4 + 6*x^3 + 9*x^2))*e^8
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^{2} + \frac {25 \, {\left (x^{2} e^{8} + e^{8}\right )}}{4 \, {\left (x^{2} + 3 \, x\right )}^{2}} \] Input:
integrate(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(log(x)-4)^2-25*x^3-50*x-75)/ (2*x^4+18*x^3+54*x^2+54*x)/exp(log(x)-4)^2,x, algorithm="giac")
Output:
x^2 + 25/4*(x^2*e^8 + e^8)/(x^2 + 3*x)^2
Time = 3.76 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^2+\frac {\frac {25\,{\mathrm {e}}^8\,x^2}{4}+\frac {25\,{\mathrm {e}}^8}{4}}{x^2\,{\left (x+3\right )}^2} \] Input:
int(-(exp(8 - 2*log(x))*(50*x - exp(2*log(x) - 8)*(108*x^2 + 108*x^3 + 36* x^4 + 4*x^5) + 25*x^3 + 75))/(54*x + 54*x^2 + 18*x^3 + 2*x^4),x)
Output:
x^2 + ((25*exp(8))/4 + (25*x^2*exp(8))/4)/(x^2*(x + 3)^2)
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=\frac {25 e^{8} x^{2}+25 e^{8}+4 x^{6}+24 x^{5}+36 x^{4}}{4 x^{2} \left (x^{2}+6 x +9\right )} \] Input:
int(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(log(x)-4)^2-25*x^3-50*x-75)/(2*x^4 +18*x^3+54*x^2+54*x)/exp(log(x)-4)^2,x)
Output:
(25*e**8*x**2 + 25*e**8 + 4*x**6 + 24*x**5 + 36*x**4)/(4*x**2*(x**2 + 6*x + 9))