\(\int \frac {-18+12 x^4-2 x^8+e^x (6 x^4-2 x^8)+e^x (24 x^4+6 x^5-2 x^9) \log (x)}{9 x-6 x^5+x^9} \, dx\) [4935]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 64, antiderivative size = 26 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=\log \left (\frac {1}{9} x^{-2+\frac {2 e^x x}{\frac {3}{x^3}-x}}\right ) \]

[Out]

ln(1/9*exp(exp(x)/(3/x^3-x)*x*ln(x))^2/x^2)

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1608, 28, 6874, 2326} \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=\frac {2 e^x x^3 \left (3 x \log (x)-x^5 \log (x)\right )}{\left (3-x^4\right )^2}-2 \log (x) \]

[In]

Int[(-18 + 12*x^4 - 2*x^8 + E^x*(6*x^4 - 2*x^8) + E^x*(24*x^4 + 6*x^5 - 2*x^9)*Log[x])/(9*x - 6*x^5 + x^9),x]

[Out]

-2*Log[x] + (2*E^x*x^3*(3*x*Log[x] - x^5*Log[x]))/(3 - x^4)^2

Rule 28

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

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2326

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{x \left (9-6 x^4+x^8\right )} \, dx \\ & = \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{x \left (-3+x^4\right )^2} \, dx \\ & = \int \left (-\frac {2}{x}-\frac {2 e^x x^3 \left (-3+x^4-12 \log (x)-3 x \log (x)+x^5 \log (x)\right )}{\left (-3+x^4\right )^2}\right ) \, dx \\ & = -2 \log (x)-2 \int \frac {e^x x^3 \left (-3+x^4-12 \log (x)-3 x \log (x)+x^5 \log (x)\right )}{\left (-3+x^4\right )^2} \, dx \\ & = -2 \log (x)+\frac {2 e^x x^3 \left (3 x \log (x)-x^5 \log (x)\right )}{\left (3-x^4\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \left (-3+\left (1+e^x\right ) x^4\right ) \log (x)}{-3+x^4} \]

[In]

Integrate[(-18 + 12*x^4 - 2*x^8 + E^x*(6*x^4 - 2*x^8) + E^x*(24*x^4 + 6*x^5 - 2*x^9)*Log[x])/(9*x - 6*x^5 + x^
9),x]

[Out]

(-2*(-3 + (1 + E^x)*x^4)*Log[x])/(-3 + x^4)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85

method result size
risch \(-\frac {2 x^{4} {\mathrm e}^{x} \ln \left (x \right )}{x^{4}-3}-2 \ln \left (x \right )\) \(22\)
parallelrisch \(\frac {-6 x^{4} {\mathrm e}^{x} \ln \left (x \right )-6 x^{4} \ln \left (x \right )+18 \ln \left (x \right )}{3 x^{4}-9}\) \(31\)

[In]

int(((-2*x^9+6*x^5+24*x^4)*exp(x)*ln(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x,method=_RETUR
NVERBOSE)

[Out]

-2*x^4/(x^4-3)*exp(x)*ln(x)-2*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \, {\left (x^{4} e^{x} + x^{4} - 3\right )} \log \left (x\right )}{x^{4} - 3} \]

[In]

integrate(((-2*x^9+6*x^5+24*x^4)*exp(x)*log(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x, algor
ithm="fricas")

[Out]

-2*(x^4*e^x + x^4 - 3)*log(x)/(x^4 - 3)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=- \frac {2 x^{4} e^{x} \log {\left (x \right )}}{x^{4} - 3} - 2 \log {\left (x \right )} \]

[In]

integrate(((-2*x**9+6*x**5+24*x**4)*exp(x)*ln(x)+(-2*x**8+6*x**4)*exp(x)-2*x**8+12*x**4-18)/(x**9-6*x**5+9*x),
x)

[Out]

-2*x**4*exp(x)*log(x)/(x**4 - 3) - 2*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \, x^{4} e^{x} \log \left (x\right )}{x^{4} - 3} - 2 \, \log \left (x\right ) \]

[In]

integrate(((-2*x^9+6*x^5+24*x^4)*exp(x)*log(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x, algor
ithm="maxima")

[Out]

-2*x^4*e^x*log(x)/(x^4 - 3) - 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2 \, {\left (x^{4} e^{x} \log \left (x\right ) + x^{4} \log \left (x\right ) - 3 \, \log \left (x\right )\right )}}{x^{4} - 3} \]

[In]

integrate(((-2*x^9+6*x^5+24*x^4)*exp(x)*log(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x, algor
ithm="giac")

[Out]

-2*(x^4*e^x*log(x) + x^4*log(x) - 3*log(x))/(x^4 - 3)

Mupad [B] (verification not implemented)

Time = 11.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{9 x-6 x^5+x^9} \, dx=-\frac {2\,\ln \left (x\right )\,\left (x^4\,{\mathrm {e}}^x+x^4-3\right )}{x^4-3} \]

[In]

int((exp(x)*(6*x^4 - 2*x^8) + 12*x^4 - 2*x^8 + exp(x)*log(x)*(24*x^4 + 6*x^5 - 2*x^9) - 18)/(9*x - 6*x^5 + x^9
),x)

[Out]

-(2*log(x)*(x^4*exp(x) + x^4 - 3))/(x^4 - 3)