Integrand size = 74, antiderivative size = 22 \[ \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx=e^{\frac {4 \left (\frac {2}{3}+x+\log (x)\right )}{x (-x+\log (3))}} \]
Time = 5.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx=e^{-\frac {8+12 x}{3 x^2-x \log (27)}} x^{-\frac {4}{x^2-x \log (3)}} \]
Integrate[(E^((8 + 12*x + 12*Log[x])/(-3*x^2 + 3*x*Log[3]))*(4*x + 12*x^2 + 4*Log[3] + (24*x - 12*Log[3])*Log[x]))/(3*x^4 - 6*x^3*Log[3] + 3*x^2*Log [3]^2),x]
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.
\(\displaystyle \int \frac {e^{\frac {12 x+12 \log (x)+8}{3 x \log (3)-3 x^2}} \left (12 x^2+4 x+(24 x-12 \log (3)) \log (x)+4 \log (3)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {12 x+12 \log (x)+8}{3 x \log (3)-3 x^2}} \left (12 x^2+4 x+(24 x-12 \log (3)) \log (x)+4 \log (3)\right )}{x^2 \left (3 x^2-6 x \log (3)+3 \log ^2(3)\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 12 \int \frac {e^{-\frac {4 (3 x+2)}{3 \left (x^2-x \log (3)\right )}} x^{-2-\frac {12}{3 x^2-3 x \log (3)}} \left (12 x^2+4 x+12 (2 x-\log (3)) \log (x)+\log (81)\right )}{36 (x-\log (3))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {e^{-\frac {4 (3 x+2)}{3 \left (x^2-x \log (3)\right )}} x^{-2-\frac {4}{x^2-x \log (3)}} \left (12 x^2+4 x+12 (2 x-\log (3)) \log (x)+\log (81)\right )}{(x-\log (3))^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{3} \int \frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}} \left (12 x^2+4 x+12 (2 x-\log (3)) \log (x)+\log (81)\right )}{(x-\log (3))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (\frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} \left (12 x^2+4 x+\log (81)\right ) x^{-2-\frac {4}{x^2-x \log (3)}}}{(x-\log (3))^2}+\frac {12 e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} (2 x-\log (3)) \log (x) x^{-2-\frac {4}{x^2-x \log (3)}}}{(x-\log (3))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (2 \left (6 \log ^2(3)+\log (81)\right ) \int \frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}}}{(x-\log (3))^2}dx+12 \int e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}}dx+12 \log (3) \log (x) \int \frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}}}{(x-\log (3))^2}dx+24 \log (x) \int \frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}}}{x-\log (3)}dx+4 (1+\log (729)) \int \frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}}}{x-\log (3)}dx-12 \log (3) \int \frac {\int \frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}}}{(x-\log (3))^2}dx}{x}dx-24 \int \frac {\int \frac {e^{-\frac {4 (3 x+2)}{3 x (x-\log (3))}} x^{-2-\frac {4}{x^2-x \log (3)}}}{x-\log (3)}dx}{x}dx\right )\) |
Int[(E^((8 + 12*x + 12*Log[x])/(-3*x^2 + 3*x*Log[3]))*(4*x + 12*x^2 + 4*Lo g[3] + (24*x - 12*Log[3])*Log[x]))/(3*x^4 - 6*x^3*Log[3] + 3*x^2*Log[3]^2) ,x]
3.13.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*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] && !AlgebraicFu nctionQ[u, x]
Time = 1.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09
method | result | size |
risch | \({\mathrm e}^{\frac {4 \ln \left (x \right )+4 x +\frac {8}{3}}{x \left (\ln \left (3\right )-x \right )}}\) | \(24\) |
parallelrisch | \({\mathrm e}^{\frac {4 \ln \left (x \right )+4 x +\frac {8}{3}}{x \left (\ln \left (3\right )-x \right )}}\) | \(24\) |
norman | \(\frac {x \ln \left (3\right ) {\mathrm e}^{\frac {12 \ln \left (x \right )+12 x +8}{3 x \ln \left (3\right )-3 x^{2}}}-x^{2} {\mathrm e}^{\frac {12 \ln \left (x \right )+12 x +8}{3 x \ln \left (3\right )-3 x^{2}}}}{x \left (\ln \left (3\right )-x \right )}\) | \(71\) |
int(((-12*ln(3)+24*x)*ln(x)+4*ln(3)+12*x^2+4*x)*exp((12*ln(x)+12*x+8)/(3*x *ln(3)-3*x^2))/(3*x^2*ln(3)^2-6*x^3*ln(3)+3*x^4),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx=e^{\left (-\frac {4 \, {\left (3 \, x + 3 \, \log \left (x\right ) + 2\right )}}{3 \, {\left (x^{2} - x \log \left (3\right )\right )}}\right )} \]
integrate(((-12*log(3)+24*x)*log(x)+4*log(3)+12*x^2+4*x)*exp((12*log(x)+12 *x+8)/(3*x*log(3)-3*x^2))/(3*x^2*log(3)^2-6*x^3*log(3)+3*x^4),x, algorithm =\
Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx=e^{\frac {12 x + 12 \log {\left (x \right )} + 8}{- 3 x^{2} + 3 x \log {\left (3 \right )}}} \]
integrate(((-12*ln(3)+24*x)*ln(x)+4*ln(3)+12*x**2+4*x)*exp((12*ln(x)+12*x+ 8)/(3*x*ln(3)-3*x**2))/(3*x**2*ln(3)**2-6*x**3*ln(3)+3*x**4),x)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (23) = 46\).
Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.91 \[ \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx=e^{\left (-\frac {4 \, \log \left (x\right )}{x \log \left (3\right ) - \log \left (3\right )^{2}} - \frac {8}{3 \, {\left (x \log \left (3\right ) - \log \left (3\right )^{2}\right )}} - \frac {4}{x - \log \left (3\right )} + \frac {4 \, \log \left (x\right )}{x \log \left (3\right )} + \frac {8}{3 \, x \log \left (3\right )}\right )} \]
integrate(((-12*log(3)+24*x)*log(x)+4*log(3)+12*x^2+4*x)*exp((12*log(x)+12 *x+8)/(3*x*log(3)-3*x^2))/(3*x^2*log(3)^2-6*x^3*log(3)+3*x^4),x, algorithm =\
e^(-4*log(x)/(x*log(3) - log(3)^2) - 8/3/(x*log(3) - log(3)^2) - 4/(x - lo g(3)) + 4*log(x)/(x*log(3)) + 8/3/(x*log(3)))
Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx=e^{\left (-\frac {4 \, x}{x^{2} - x \log \left (3\right )} - \frac {4 \, \log \left (x\right )}{x^{2} - x \log \left (3\right )} - \frac {8}{3 \, {\left (x^{2} - x \log \left (3\right )\right )}}\right )} \]
integrate(((-12*log(3)+24*x)*log(x)+4*log(3)+12*x^2+4*x)*exp((12*log(x)+12 *x+8)/(3*x*log(3)-3*x^2))/(3*x^2*log(3)^2-6*x^3*log(3)+3*x^4),x, algorithm =\
Time = 12.91 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx=x^{\frac {4}{x\,\ln \left (3\right )-x^2}}\,{\mathrm {e}}^{-\frac {4}{x-\ln \left (3\right )}}\,{\mathrm {e}}^{\frac {8}{3\,x\,\ln \left (3\right )-3\,x^2}} \]