Integrand size = 128, antiderivative size = 28 \[ \int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx=\frac {2 \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x \log (x)} \]
Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx=\frac {2 \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x \log (x)} \]
Integrate[(-4*x*Log[x] + 4*E^(2*x + 4*E^(2*x)*x^2)*x^2*(8 + 8*x)*Log[x] + (2*x + E^(4*E^(2*x)*x^2)*(-2 - 2*Log[x]) + 2*x*Log[x])*Log[E^(8*E^(2*x)*x^ 2) - 2*E^(4*E^(2*x)*x^2)*x + x^2])/(E^(4*E^(2*x)*x^2)*x^2*Log[x]^2 - x^3*L og[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.
\(\displaystyle \int \frac {4 e^{4 e^{2 x} x^2+2 x} (8 x+8) x^2 \log (x)+\left (e^{4 e^{2 x} x^2} (-2 \log (x)-2)+2 x+2 x \log (x)\right ) \log \left (x^2-2 e^{4 e^{2 x} x^2} x+e^{8 e^{2 x} x^2}\right )-4 x \log (x)}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 e^{4 e^{2 x} x^2+2 x} (8 x+8) x^2 \log (x)+\left (e^{4 e^{2 x} x^2} (-2 \log (x)-2)+2 x+2 x \log (x)\right ) \log \left (x^2-2 e^{4 e^{2 x} x^2} x+e^{8 e^{2 x} x^2}\right )-4 x \log (x)}{\left (e^{4 e^{2 x} x^2}-x\right ) x^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (16 e^{2 x} x^3 \log (x)+16 e^{2 x} x^2 \log (x)-\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )-\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) \log (x)\right )}{x^2 \log ^2(x)}+\frac {4 \left (8 e^{2 x} x^3+8 e^{2 x} x^2-1\right )}{\left (e^{4 e^{2 x} x^2}-x\right ) x \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x^2 \log ^2(x)}dx-4 \int \frac {1}{\left (e^{4 e^{2 x} x^2}-x\right ) x \log (x)}dx+32 \int \frac {e^{2 x} x}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)}dx+32 \int \frac {e^{2 x} x^2}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)}dx-2 \int \frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x^2 \log (x)}dx+32 \int \frac {e^{2 x}}{\log (x)}dx+32 \int \frac {e^{2 x} x}{\log (x)}dx\) |
Int[(-4*x*Log[x] + 4*E^(2*x + 4*E^(2*x)*x^2)*x^2*(8 + 8*x)*Log[x] + (2*x + E^(4*E^(2*x)*x^2)*(-2 - 2*Log[x]) + 2*x*Log[x])*Log[E^(8*E^(2*x)*x^2) - 2 *E^(4*E^(2*x)*x^2)*x + x^2])/(E^(4*E^(2*x)*x^2)*x^2*Log[x]^2 - x^3*Log[x]^ 2),x]
3.24.98.3.1 Defintions of rubi rules used
Time = 15.38 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43
method | result | size |
parallelrisch | \(\frac {2 \ln \left ({\mathrm e}^{8 \,{\mathrm e}^{2 x} x^{2}}-2 x \,{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x^{2}\right )}{x \ln \left (x \right )}\) | \(40\) |
risch | \(\frac {4 \ln \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )}{x \ln \left (x \right )}-\frac {i \pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )^{2}\right ) \left ({\operatorname {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )\right )}^{2}-2 \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )^{2}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )\right )+{\operatorname {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )^{2}\right )}^{2}\right )}{x \ln \left (x \right )}\) | \(140\) |
int((((-2*ln(x)-2)*exp(exp(ln(2*x)+x)^2)+2*x*ln(x)+2*x)*ln(exp(exp(ln(2*x) +x)^2)^2-2*x*exp(exp(ln(2*x)+x)^2)+x^2)+(8*x+8)*ln(x)*exp(ln(2*x)+x)^2*exp (exp(ln(2*x)+x)^2)-4*x*ln(x))/(x^2*ln(x)^2*exp(exp(ln(2*x)+x)^2)-x^3*ln(x) ^2),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (28) = 56\).
Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.54 \[ \int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx=\frac {2 \, \log \left ({\left (x^{2} e^{\left (4 \, x + 4 \, \log \left (2\right ) + 4 \, \log \left (x\right )\right )} - 2 \, x e^{\left (4 \, x + e^{\left (2 \, x + 2 \, \log \left (2\right ) + 2 \, \log \left (x\right )\right )} + 4 \, \log \left (2\right ) + 4 \, \log \left (x\right )\right )} + e^{\left (4 \, x + 2 \, e^{\left (2 \, x + 2 \, \log \left (2\right ) + 2 \, \log \left (x\right )\right )} + 4 \, \log \left (2\right ) + 4 \, \log \left (x\right )\right )}\right )} e^{\left (-4 \, x - 4 \, \log \left (2\right ) - 4 \, \log \left (x\right )\right )}\right )}{x \log \left (x\right )} \]
integrate((((-2*log(x)-2)*exp(exp(log(2*x)+x)^2)+2*x*log(x)+2*x)*log(exp(e xp(log(2*x)+x)^2)^2-2*x*exp(exp(log(2*x)+x)^2)+x^2)+(8*x+8)*log(x)*exp(log (2*x)+x)^2*exp(exp(log(2*x)+x)^2)-4*x*log(x))/(x^2*log(x)^2*exp(exp(log(2* x)+x)^2)-x^3*log(x)^2),x, algorithm=\
2*log((x^2*e^(4*x + 4*log(2) + 4*log(x)) - 2*x*e^(4*x + e^(2*x + 2*log(2) + 2*log(x)) + 4*log(2) + 4*log(x)) + e^(4*x + 2*e^(2*x + 2*log(2) + 2*log( x)) + 4*log(2) + 4*log(x)))*e^(-4*x - 4*log(2) - 4*log(x)))/(x*log(x))
Exception generated. \[ \int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
integrate((((-2*ln(x)-2)*exp(exp(ln(2*x)+x)**2)+2*x*ln(x)+2*x)*ln(exp(exp( ln(2*x)+x)**2)**2-2*x*exp(exp(ln(2*x)+x)**2)+x**2)+(8*x+8)*ln(x)*exp(ln(2* x)+x)**2*exp(exp(ln(2*x)+x)**2)-4*x*ln(x))/(x**2*ln(x)**2*exp(exp(ln(2*x)+ x)**2)-x**3*ln(x)**2),x)
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx=\frac {4 \, \log \left (x - e^{\left (4 \, x^{2} e^{\left (2 \, x\right )}\right )}\right )}{x \log \left (x\right )} \]
integrate((((-2*log(x)-2)*exp(exp(log(2*x)+x)^2)+2*x*log(x)+2*x)*log(exp(e xp(log(2*x)+x)^2)^2-2*x*exp(exp(log(2*x)+x)^2)+x^2)+(8*x+8)*log(x)*exp(log (2*x)+x)^2*exp(exp(log(2*x)+x)^2)-4*x*log(x))/(x^2*log(x)^2*exp(exp(log(2* x)+x)^2)-x^3*log(x)^2),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx=\frac {2 \, \log \left (x^{2} - 2 \, x e^{\left (4 \, x^{2} e^{\left (2 \, x\right )}\right )} + e^{\left (8 \, x^{2} e^{\left (2 \, x\right )}\right )}\right )}{x \log \left (x\right )} \]
integrate((((-2*log(x)-2)*exp(exp(log(2*x)+x)^2)+2*x*log(x)+2*x)*log(exp(e xp(log(2*x)+x)^2)^2-2*x*exp(exp(log(2*x)+x)^2)+x^2)+(8*x+8)*log(x)*exp(log (2*x)+x)^2*exp(exp(log(2*x)+x)^2)-4*x*log(x))/(x^2*log(x)^2*exp(exp(log(2* x)+x)^2)-x^3*log(x)^2),x, algorithm=\
Time = 9.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx=\frac {2\,\ln \left ({\mathrm {e}}^{8\,x^2\,{\mathrm {e}}^{2\,x}}-2\,x\,{\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^{2\,x}}+x^2\right )}{x\,\ln \left (x\right )} \]