Integrand size = 108, antiderivative size = 28 \[ \int \frac {-54 x+3 x^2+12 e^x x^2+\left (6 x^2+e^x \left (-12 x^2+12 x^3\right )\right ) \log (x)}{81-54 x+9 x^2+\left (-36 x+12 x^2+e^x \left (72 x-24 x^2\right )\right ) \log (x)+\left (4 x^2-16 e^x x^2+16 e^{2 x} x^2\right ) \log ^2(x)} \, dx=5-\frac {x^2}{3-x-\frac {1}{3} \left (2-4 e^x\right ) x \log (x)} \]
Time = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-54 x+3 x^2+12 e^x x^2+\left (6 x^2+e^x \left (-12 x^2+12 x^3\right )\right ) \log (x)}{81-54 x+9 x^2+\left (-36 x+12 x^2+e^x \left (72 x-24 x^2\right )\right ) \log (x)+\left (4 x^2-16 e^x x^2+16 e^{2 x} x^2\right ) \log ^2(x)} \, dx=-\frac {3 x^2}{9-3 x-2 x \log (x)+4 e^x x \log (x)} \]
Integrate[(-54*x + 3*x^2 + 12*E^x*x^2 + (6*x^2 + E^x*(-12*x^2 + 12*x^3))*L og[x])/(81 - 54*x + 9*x^2 + (-36*x + 12*x^2 + E^x*(72*x - 24*x^2))*Log[x] + (4*x^2 - 16*E^x*x^2 + 16*E^(2*x)*x^2)*Log[x]^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.
\(\displaystyle \int \frac {12 e^x x^2+3 x^2+\left (6 x^2+e^x \left (12 x^3-12 x^2\right )\right ) \log (x)-54 x}{9 x^2+\left (-16 e^x x^2+16 e^{2 x} x^2+4 x^2\right ) \log ^2(x)+\left (12 x^2+e^x \left (72 x-24 x^2\right )-36 x\right ) \log (x)-54 x+81} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3 x \left (4 e^x x+x+2 \left (2 e^x (x-1)+1\right ) x \log (x)-18\right )}{\left (-3 x+2 \left (2 e^x-1\right ) x \log (x)+9\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int -\frac {x \left (-4 e^x x-2 \left (1-2 e^x (1-x)\right ) \log (x) x-x+18\right )}{\left (-2 \left (1-2 e^x\right ) \log (x) x-3 x+9\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -3 \int \frac {x \left (-4 e^x x-2 \left (1-2 e^x (1-x)\right ) \log (x) x-x+18\right )}{\left (-2 \left (1-2 e^x\right ) \log (x) x-3 x+9\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -3 \int \left (-\frac {x (x \log (x)-\log (x)+1)}{\log (x) \left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )}-\frac {x \left (2 \log ^2(x) x^2+3 \log (x) x^2-9 \log (x) x+3 x-9 \log (x)-9\right )}{\log (x) \left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (-3 \int \frac {x^3}{\left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )^2}dx-2 \int \frac {x^3 \log (x)}{\left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )^2}dx+9 \int \frac {x^2}{\left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )^2}dx-3 \int \frac {x^2}{\log (x) \left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )^2}dx-\int \frac {x^2}{4 e^x \log (x) x-2 \log (x) x-3 x+9}dx+9 \int \frac {x}{\left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )^2}dx+9 \int \frac {x}{\log (x) \left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )^2}dx+\int \frac {x}{4 e^x \log (x) x-2 \log (x) x-3 x+9}dx-\int \frac {x}{\log (x) \left (4 e^x \log (x) x-2 \log (x) x-3 x+9\right )}dx\right )\) |
Int[(-54*x + 3*x^2 + 12*E^x*x^2 + (6*x^2 + E^x*(-12*x^2 + 12*x^3))*Log[x]) /(81 - 54*x + 9*x^2 + (-36*x + 12*x^2 + E^x*(72*x - 24*x^2))*Log[x] + (4*x ^2 - 16*E^x*x^2 + 16*E^(2*x)*x^2)*Log[x]^2),x]
3.14.32.3.1 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[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
risch | \(-\frac {3 x^{2}}{4 x \,{\mathrm e}^{x} \ln \left (x \right )-2 x \ln \left (x \right )-3 x +9}\) | \(25\) |
parallelrisch | \(-\frac {3 x^{2}}{4 x \,{\mathrm e}^{x} \ln \left (x \right )-2 x \ln \left (x \right )-3 x +9}\) | \(25\) |
int((((12*x^3-12*x^2)*exp(x)+6*x^2)*ln(x)+12*exp(x)*x^2+3*x^2-54*x)/((16*e xp(x)^2*x^2-16*exp(x)*x^2+4*x^2)*ln(x)^2+((-24*x^2+72*x)*exp(x)+12*x^2-36* x)*ln(x)+9*x^2-54*x+81),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-54 x+3 x^2+12 e^x x^2+\left (6 x^2+e^x \left (-12 x^2+12 x^3\right )\right ) \log (x)}{81-54 x+9 x^2+\left (-36 x+12 x^2+e^x \left (72 x-24 x^2\right )\right ) \log (x)+\left (4 x^2-16 e^x x^2+16 e^{2 x} x^2\right ) \log ^2(x)} \, dx=-\frac {3 \, x^{2}}{2 \, {\left (2 \, x e^{x} - x\right )} \log \left (x\right ) - 3 \, x + 9} \]
integrate((((12*x^3-12*x^2)*exp(x)+6*x^2)*log(x)+12*exp(x)*x^2+3*x^2-54*x) /((16*exp(x)^2*x^2-16*exp(x)*x^2+4*x^2)*log(x)^2+((-24*x^2+72*x)*exp(x)+12 *x^2-36*x)*log(x)+9*x^2-54*x+81),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-54 x+3 x^2+12 e^x x^2+\left (6 x^2+e^x \left (-12 x^2+12 x^3\right )\right ) \log (x)}{81-54 x+9 x^2+\left (-36 x+12 x^2+e^x \left (72 x-24 x^2\right )\right ) \log (x)+\left (4 x^2-16 e^x x^2+16 e^{2 x} x^2\right ) \log ^2(x)} \, dx=- \frac {3 x^{2}}{4 x e^{x} \log {\left (x \right )} - 2 x \log {\left (x \right )} - 3 x + 9} \]
integrate((((12*x**3-12*x**2)*exp(x)+6*x**2)*ln(x)+12*exp(x)*x**2+3*x**2-5 4*x)/((16*exp(x)**2*x**2-16*exp(x)*x**2+4*x**2)*ln(x)**2+((-24*x**2+72*x)* exp(x)+12*x**2-36*x)*ln(x)+9*x**2-54*x+81),x)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-54 x+3 x^2+12 e^x x^2+\left (6 x^2+e^x \left (-12 x^2+12 x^3\right )\right ) \log (x)}{81-54 x+9 x^2+\left (-36 x+12 x^2+e^x \left (72 x-24 x^2\right )\right ) \log (x)+\left (4 x^2-16 e^x x^2+16 e^{2 x} x^2\right ) \log ^2(x)} \, dx=-\frac {3 \, x^{2}}{4 \, x e^{x} \log \left (x\right ) - 2 \, x \log \left (x\right ) - 3 \, x + 9} \]
integrate((((12*x^3-12*x^2)*exp(x)+6*x^2)*log(x)+12*exp(x)*x^2+3*x^2-54*x) /((16*exp(x)^2*x^2-16*exp(x)*x^2+4*x^2)*log(x)^2+((-24*x^2+72*x)*exp(x)+12 *x^2-36*x)*log(x)+9*x^2-54*x+81),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-54 x+3 x^2+12 e^x x^2+\left (6 x^2+e^x \left (-12 x^2+12 x^3\right )\right ) \log (x)}{81-54 x+9 x^2+\left (-36 x+12 x^2+e^x \left (72 x-24 x^2\right )\right ) \log (x)+\left (4 x^2-16 e^x x^2+16 e^{2 x} x^2\right ) \log ^2(x)} \, dx=-\frac {3 \, x^{2}}{4 \, x e^{x} \log \left (x\right ) - 2 \, x \log \left (x\right ) - 3 \, x + 9} \]
integrate((((12*x^3-12*x^2)*exp(x)+6*x^2)*log(x)+12*exp(x)*x^2+3*x^2-54*x) /((16*exp(x)^2*x^2-16*exp(x)*x^2+4*x^2)*log(x)^2+((-24*x^2+72*x)*exp(x)+12 *x^2-36*x)*log(x)+9*x^2-54*x+81),x, algorithm=\
Timed out. \[ \int \frac {-54 x+3 x^2+12 e^x x^2+\left (6 x^2+e^x \left (-12 x^2+12 x^3\right )\right ) \log (x)}{81-54 x+9 x^2+\left (-36 x+12 x^2+e^x \left (72 x-24 x^2\right )\right ) \log (x)+\left (4 x^2-16 e^x x^2+16 e^{2 x} x^2\right ) \log ^2(x)} \, dx=\int -\frac {54\,x-12\,x^2\,{\mathrm {e}}^x-3\,x^2+\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (12\,x^2-12\,x^3\right )-6\,x^2\right )}{\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (72\,x-24\,x^2\right )-36\,x+12\,x^2\right )-54\,x+{\ln \left (x\right )}^2\,\left (16\,x^2\,{\mathrm {e}}^{2\,x}-16\,x^2\,{\mathrm {e}}^x+4\,x^2\right )+9\,x^2+81} \,d x \]
int(-(54*x - 12*x^2*exp(x) - 3*x^2 + log(x)*(exp(x)*(12*x^2 - 12*x^3) - 6* x^2))/(log(x)*(exp(x)*(72*x - 24*x^2) - 36*x + 12*x^2) - 54*x + log(x)^2*( 16*x^2*exp(2*x) - 16*x^2*exp(x) + 4*x^2) + 9*x^2 + 81),x)