Integrand size = 150, antiderivative size = 27 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{e^2}-\frac {\log (4)}{-4+e^{\left (e^4+x\right )^2}+x-\log (\log (x))} \] Output:
exp(-2)-2*ln(2)/(exp((x+exp(4))^2)+x-4-ln(ln(x)))
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {\log (4)}{-4+e^{\left (e^4+x\right )^2}+x-\log (\log (x))} \] Input:
Integrate[(-Log[4] + (x*Log[4] + E^(E^8 + 2*E^4*x + x^2)*(2*E^4*x + 2*x^2) *Log[4])*Log[x])/((16*x + E^(2*E^8 + 4*E^4*x + 2*x^2)*x - 8*x^2 + x^3 + E^ (E^8 + 2*E^4*x + x^2)*(-8*x + 2*x^2))*Log[x] + (8*x - 2*E^(E^8 + 2*E^4*x + x^2)*x - 2*x^2)*Log[x]*Log[Log[x]] + x*Log[x]*Log[Log[x]]^2),x]
Output:
-(Log[4]/(-4 + E^(E^4 + x)^2 + x - Log[Log[x]]))
Time = 1.86 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {\left (e^{x^2+2 e^4 x+e^8} \left (2 x^2+2 e^4 x\right ) \log (4)+x \log (4)\right ) \log (x)-\log (4)}{\left (-2 x^2-2 e^{x^2+2 e^4 x+e^8} x+8 x\right ) \log (x) \log (\log (x))+\left (x^3-8 x^2+e^{2 x^2+4 e^4 x+2 e^8} x+e^{x^2+2 e^4 x+e^8} \left (2 x^2-8 x\right )+16 x\right ) \log (x)+x \log (x) \log ^2(\log (x))} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log (4) \left (x \left (2 e^{x^2+2 e^4 x+e^8+4}+2 e^{\left (x+e^4\right )^2} x+1\right ) \log (x)-1\right )}{x \log (x) \left (-x-e^{\left (x+e^4\right )^2}+\log (\log (x))+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (4) \int -\frac {1-x \left (2 e^{\left (x+e^4\right )^2} x+2 e^{x^2+2 e^4 x+e^8+4}+1\right ) \log (x)}{x \log (x) \left (-x-e^{\left (x+e^4\right )^2}+\log (\log (x))+4\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\log (4) \int \frac {1-x \left (2 e^{\left (x+e^4\right )^2} x+2 e^{x^2+2 e^4 x+e^8+4}+1\right ) \log (x)}{x \log (x) \left (-x-e^{\left (x+e^4\right )^2}+\log (\log (x))+4\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {\log (4)}{-x-e^{\left (x+e^4\right )^2}+\log (\log (x))+4}\) |
Input:
Int[(-Log[4] + (x*Log[4] + E^(E^8 + 2*E^4*x + x^2)*(2*E^4*x + 2*x^2)*Log[4 ])*Log[x])/((16*x + E^(2*E^8 + 4*E^4*x + 2*x^2)*x - 8*x^2 + x^3 + E^(E^8 + 2*E^4*x + x^2)*(-8*x + 2*x^2))*Log[x] + (8*x - 2*E^(E^8 + 2*E^4*x + x^2)* x - 2*x^2)*Log[x]*Log[Log[x]] + x*Log[x]*Log[Log[x]]^2),x]
Output:
Log[4]/(4 - E^(E^4 + x)^2 - x + Log[Log[x]])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 5.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {2 \ln \left (2\right )}{x -4+{\mathrm e}^{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}}-\ln \left (\ln \left (x \right )\right )}\) | \(27\) |
parallelrisch | \(-\frac {2 \ln \left (2\right )}{x -4+{\mathrm e}^{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}}-\ln \left (\ln \left (x \right )\right )}\) | \(29\) |
Input:
int(((2*(2*x*exp(4)+2*x^2)*ln(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*ln(2))*l n(x)-2*ln(2))/(x*ln(x)*ln(ln(x))^2+(-2*x*exp(exp(4)^2+2*x*exp(4)+x^2)-2*x^ 2+8*x)*ln(x)*ln(ln(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2+(2*x^2-8*x)*exp(e xp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*ln(x)),x,method=_RETURNVERBOSE)
Output:
-2*ln(2)/(x-4+exp(exp(8)+2*x*exp(4)+x^2)-ln(ln(x)))
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2 \, \log \left (2\right )}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \left (x\right )\right ) - 4} \] Input:
integrate(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*l og(2))*log(x)-2*log(2))/(x*log(x)*log(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp (4)+x^2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2+( 2*x^2-8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x, algorit hm="fricas")
Output:
-2*log(2)/(x + e^(x^2 + 2*x*e^4 + e^8) - log(log(x)) - 4)
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=- \frac {2 \log {\left (2 \right )}}{x + e^{x^{2} + 2 x e^{4} + e^{8}} - \log {\left (\log {\left (x \right )} \right )} - 4} \] Input:
integrate(((2*(2*x*exp(4)+2*x**2)*ln(2)*exp(exp(4)**2+2*x*exp(4)+x**2)+2*x *ln(2))*ln(x)-2*ln(2))/(x*ln(x)*ln(ln(x))**2+(-2*x*exp(exp(4)**2+2*x*exp(4 )+x**2)-2*x**2+8*x)*ln(x)*ln(ln(x))+(x*exp(exp(4)**2+2*x*exp(4)+x**2)**2+( 2*x**2-8*x)*exp(exp(4)**2+2*x*exp(4)+x**2)+x**3-8*x**2+16*x)*ln(x)),x)
Output:
-2*log(2)/(x + exp(x**2 + 2*x*exp(4) + exp(8)) - log(log(x)) - 4)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2 \, \log \left (2\right )}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \left (x\right )\right ) - 4} \] Input:
integrate(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*l og(2))*log(x)-2*log(2))/(x*log(x)*log(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp (4)+x^2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2+( 2*x^2-8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x, algorit hm="maxima")
Output:
-2*log(2)/(x + e^(x^2 + 2*x*e^4 + e^8) - log(log(x)) - 4)
Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2 \, \log \left (2\right )}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \left (x\right )\right ) - 4} \] Input:
integrate(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*l og(2))*log(x)-2*log(2))/(x*log(x)*log(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp (4)+x^2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2+( 2*x^2-8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x, algorit hm="giac")
Output:
-2*log(2)/(x + e^(x^2 + 2*x*e^4 + e^8) - log(log(x)) - 4)
Time = 2.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=-\frac {2\,\ln \left (2\right )}{x-\ln \left (\ln \left (x\right )\right )+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^4}\,{\mathrm {e}}^{{\mathrm {e}}^8}-4} \] Input:
int(-(2*log(2) - log(x)*(2*x*log(2) + 2*exp(exp(8) + 2*x*exp(4) + x^2)*log (2)*(2*x*exp(4) + 2*x^2)))/(log(x)*(16*x - exp(exp(8) + 2*x*exp(4) + x^2)* (8*x - 2*x^2) + x*exp(2*exp(8) + 4*x*exp(4) + 2*x^2) - 8*x^2 + x^3) - log( log(x))*log(x)*(2*x*exp(exp(8) + 2*x*exp(4) + x^2) - 8*x + 2*x^2) + x*log( log(x))^2*log(x)),x)
Output:
-(2*log(2))/(x - log(log(x)) + exp(x^2)*exp(2*x*exp(4))*exp(exp(8)) - 4)
\[ \int \frac {-\log (4)+\left (x \log (4)+e^{e^8+2 e^4 x+x^2} \left (2 e^4 x+2 x^2\right ) \log (4)\right ) \log (x)}{\left (16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} \left (-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2\right ) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx=\int \frac {\left (2 \left (2 x \,{\mathrm e}^{4}+2 x^{2}\right ) \mathrm {log}\left (2\right ) {\mathrm e}^{\left ({\mathrm e}^{4}\right )^{2}+2 x \,{\mathrm e}^{4}+x^{2}}+2 \,\mathrm {log}\left (2\right ) x \right ) \mathrm {log}\left (x \right )-2 \,\mathrm {log}\left (2\right )}{x \,\mathrm {log}\left (x \right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2}+\left (-2 x \,{\mathrm e}^{\left ({\mathrm e}^{4}\right )^{2}+2 x \,{\mathrm e}^{4}+x^{2}}-2 x^{2}+8 x \right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right )+\left (x \left ({\mathrm e}^{\left ({\mathrm e}^{4}\right )^{2}+2 x \,{\mathrm e}^{4}+x^{2}}\right )^{2}+\left (2 x^{2}-8 x \right ) {\mathrm e}^{\left ({\mathrm e}^{4}\right )^{2}+2 x \,{\mathrm e}^{4}+x^{2}}+x^{3}-8 x^{2}+16 x \right ) \mathrm {log}\left (x \right )}d x \] Input:
int(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*log(2)) *log(x)-2*log(2))/(x*log(x)*log(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp(4)+x^ 2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2+(2*x^2- 8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x)
Output:
int(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*log(2)) *log(x)-2*log(2))/(x*log(x)*log(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp(4)+x^ 2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2+(2*x^2- 8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x)