Integrand size = 94, antiderivative size = 24 \[ \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx=\log \left (4 \left (-1-\frac {3}{-e+e^{\frac {4}{\log (x)}}+2 x}\right )\right ) \]
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx=-\log \left (-e+e^{\frac {4}{\log (x)}}+2 x\right )+\log \left (3-e+e^{\frac {4}{\log (x)}}+2 x\right ) \]
Integrate[(12*E^(4/Log[x]) - 6*x*Log[x]^2)/(E^(8/Log[x])*x*Log[x]^2 + E^(4 /Log[x])*(3*x - 2*E*x + 4*x^2)*Log[x]^2 + (E^2*x + 6*x^2 + 4*x^3 + E*(-3*x - 4*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.
\(\displaystyle \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{\left (4 x^2-2 e x+3 x\right ) e^{\frac {4}{\log (x)}} \log ^2(x)+\left (4 x^3+6 x^2+e \left (-4 x^2-3 x\right )+e^2 x\right ) \log ^2(x)+x e^{\frac {8}{\log (x)}} \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {6 \left (x \log ^2(x)-2 e^{\frac {4}{\log (x)}}\right )}{x \left (-4 x^2-6 \left (1-\frac {2 e}{3}\right ) x-4 x e^{\frac {4}{\log (x)}}+2 \left (1-\frac {3}{2 e}\right ) e^{\frac {4}{\log (x)}+1}-e^{\frac {8}{\log (x)}}+3 \left (1-\frac {e}{3}\right ) e\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 6 \int -\frac {2 e^{\frac {4}{\log (x)}}-x \log ^2(x)}{x \left (-4 x^2-4 e^{\frac {4}{\log (x)}} x-2 (3-2 e) x+\left (2-\frac {3}{e}\right ) e^{1+\frac {4}{\log (x)}}-e^{\frac {8}{\log (x)}}+(3-e) e\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -6 \int \frac {2 e^{\frac {4}{\log (x)}}-x \log ^2(x)}{x \left (-4 x^2-4 e^{\frac {4}{\log (x)}} x-2 (3-2 e) x+\left (2-\frac {3}{e}\right ) e^{1+\frac {4}{\log (x)}}-e^{\frac {8}{\log (x)}}+(3-e) e\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -6 \int \left (\frac {-x \log ^2(x)-4 x-6 \left (1-\frac {e}{3}\right )}{3 x \left (2 x+e^{\frac {4}{\log (x)}}+3 \left (1-\frac {e}{3}\right )\right ) \log ^2(x)}+\frac {x \log ^2(x)+4 x-2 e}{3 x \left (2 x+e^{\frac {4}{\log (x)}}-e\right ) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -6 \left (\frac {4}{3} \int \frac {1}{\left (-2 x-e^{\frac {4}{\log (x)}}-3 \left (1-\frac {e}{3}\right )\right ) \log ^2(x)}dx+\frac {2}{3} e \int \frac {1}{\left (-2 x-e^{\frac {4}{\log (x)}}+e\right ) x \log ^2(x)}dx-\frac {2}{3} (3-e) \int \frac {1}{x \left (2 x+e^{\frac {4}{\log (x)}}+3 \left (1-\frac {e}{3}\right )\right ) \log ^2(x)}dx+\frac {4}{3} \int \frac {1}{\left (2 x+e^{\frac {4}{\log (x)}}-e\right ) \log ^2(x)}dx+\frac {1}{3} \int \frac {1}{-2 x-e^{\frac {4}{\log (x)}}-3 \left (1-\frac {e}{3}\right )}dx+\frac {1}{3} \int \frac {1}{2 x+e^{\frac {4}{\log (x)}}-e}dx\right )\) |
Int[(12*E^(4/Log[x]) - 6*x*Log[x]^2)/(E^(8/Log[x])*x*Log[x]^2 + E^(4/Log[x ])*(3*x - 2*E*x + 4*x^2)*Log[x]^2 + (E^2*x + 6*x^2 + 4*x^3 + E*(-3*x - 4*x ^2))*Log[x]^2),x]
3.28.96.3.1 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]]
Time = 2.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54
method | result | size |
risch | \(\ln \left ({\mathrm e}^{\frac {4}{\ln \left (x \right )}}+3+2 x -{\mathrm e}\right )-\ln \left (2 x -{\mathrm e}+{\mathrm e}^{\frac {4}{\ln \left (x \right )}}\right )\) | \(37\) |
parallelrisch | \(-\ln \left (\frac {{\mathrm e}^{\frac {4}{\ln \left (x \right )}}}{2}+x -\frac {{\mathrm e}}{2}\right )+\ln \left (\frac {{\mathrm e}^{\frac {4}{\ln \left (x \right )}}}{2}+\frac {3}{2}+x -\frac {{\mathrm e}}{2}\right )\) | \(37\) |
int((12*exp(4/ln(x))-6*x*ln(x)^2)/(x*ln(x)^2*exp(4/ln(x))^2+(-2*x*exp(1)+4 *x^2+3*x)*ln(x)^2*exp(4/ln(x))+(x*exp(1)^2+(-4*x^2-3*x)*exp(1)+4*x^3+6*x^2 )*ln(x)^2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx=\log \left (2 \, x - e + e^{\frac {4}{\log \left (x\right )}} + 3\right ) - \log \left (2 \, x - e + e^{\frac {4}{\log \left (x\right )}}\right ) \]
integrate((12*exp(4/log(x))-6*x*log(x)^2)/(x*log(x)^2*exp(4/log(x))^2+(-2* x*exp(1)+4*x^2+3*x)*log(x)^2*exp(4/log(x))+(x*exp(1)^2+(-4*x^2-3*x)*exp(1) +4*x^3+6*x^2)*log(x)^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx=- \log {\left (2 x + e^{\frac {4}{\log {\left (x \right )}}} - e \right )} + \log {\left (2 x + e^{\frac {4}{\log {\left (x \right )}}} - e + 3 \right )} \]
integrate((12*exp(4/ln(x))-6*x*ln(x)**2)/(x*ln(x)**2*exp(4/ln(x))**2+(-2*x *exp(1)+4*x**2+3*x)*ln(x)**2*exp(4/ln(x))+(x*exp(1)**2+(-4*x**2-3*x)*exp(1 )+4*x**3+6*x**2)*ln(x)**2),x)
Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx=\log \left (2 \, x - e + e^{\frac {4}{\log \left (x\right )}} + 3\right ) - \log \left (2 \, x - e + e^{\frac {4}{\log \left (x\right )}}\right ) \]
integrate((12*exp(4/log(x))-6*x*log(x)^2)/(x*log(x)^2*exp(4/log(x))^2+(-2* x*exp(1)+4*x^2+3*x)*log(x)^2*exp(4/log(x))+(x*exp(1)^2+(-4*x^2-3*x)*exp(1) +4*x^3+6*x^2)*log(x)^2),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx=\log \left (2 \, x - e + e^{\frac {4}{\log \left (x\right )}} + 3\right ) - \log \left (2 \, x - e + e^{\frac {4}{\log \left (x\right )}}\right ) \]
integrate((12*exp(4/log(x))-6*x*log(x)^2)/(x*log(x)^2*exp(4/log(x))^2+(-2* x*exp(1)+4*x^2+3*x)*log(x)^2*exp(4/log(x))+(x*exp(1)^2+(-4*x^2-3*x)*exp(1) +4*x^3+6*x^2)*log(x)^2),x, algorithm=\
Timed out. \[ \int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx=\int \frac {12\,{\mathrm {e}}^{\frac {4}{\ln \left (x\right )}}-6\,x\,{\ln \left (x\right )}^2}{{\ln \left (x\right )}^2\,\left (x\,{\mathrm {e}}^2-\mathrm {e}\,\left (4\,x^2+3\,x\right )+6\,x^2+4\,x^3\right )+x\,{\mathrm {e}}^{\frac {8}{\ln \left (x\right )}}\,{\ln \left (x\right )}^2+{\mathrm {e}}^{\frac {4}{\ln \left (x\right )}}\,{\ln \left (x\right )}^2\,\left (3\,x-2\,x\,\mathrm {e}+4\,x^2\right )} \,d x \]
int((12*exp(4/log(x)) - 6*x*log(x)^2)/(log(x)^2*(x*exp(2) - exp(1)*(3*x + 4*x^2) + 6*x^2 + 4*x^3) + x*exp(8/log(x))*log(x)^2 + exp(4/log(x))*log(x)^ 2*(3*x - 2*x*exp(1) + 4*x^2)),x)