Integrand size = 68, antiderivative size = 26 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx=\log ^4\left (2 x+\frac {1}{2} e^e \left (-3+\frac {1}{5} (-x+\log (x))\right )\right ) \] Output:
ln(2*x+1/2*exp(exp(1))*(1/5*ln(x)-1/5*x-3))^4
Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx=\log ^4\left (\frac {1}{10} \left (20 x-e^e (15+x)+e^e \log (x)\right )\right ) \] Input:
Integrate[((E^E*(4 - 4*x) + 80*x)*Log[(E^E*(-15 - x) + 20*x + E^E*Log[x])/ 10]^3)/(20*x^2 + E^E*(-15*x - x^2) + E^E*x*Log[x]),x]
Output:
Log[(20*x - E^E*(15 + x) + E^E*Log[x])/10]^4
Time = 0.41 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {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.
\(\displaystyle \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-x-15)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-x^2-15 x\right )+e^e x \log (x)} \, dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \log ^4\left (\frac {1}{10} \left (20 x-e^e (x+15)+e^e \log (x)\right )\right )\) |
Input:
Int[((E^E*(4 - 4*x) + 80*x)*Log[(E^E*(-15 - x) + 20*x + E^E*Log[x])/10]^3) /(20*x^2 + E^E*(-15*x - x^2) + E^E*x*Log[x]),x]
Output:
Log[(20*x - E^E*(15 + x) + E^E*Log[x])/10]^4
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]
Time = 0.49 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\ln \left (\frac {{\mathrm e}^{{\mathrm e}} \ln \left (x \right )}{10}+\frac {\left (-x -15\right ) {\mathrm e}^{{\mathrm e}}}{10}+2 x \right )^{4}\) | \(25\) |
default | \(-4 \ln \left (10\right )^{3} \ln \left (x \,{\mathrm e}^{{\mathrm e}}-{\mathrm e}^{{\mathrm e}} \ln \left (x \right )-20 x +15 \,{\mathrm e}^{{\mathrm e}}\right )+\ln \left ({\mathrm e}^{{\mathrm e}} \ln \left (x \right )-x \,{\mathrm e}^{{\mathrm e}}-15 \,{\mathrm e}^{{\mathrm e}}+20 x \right )^{4}+6 \ln \left (10\right )^{2} \ln \left ({\mathrm e}^{{\mathrm e}} \ln \left (x \right )-x \,{\mathrm e}^{{\mathrm e}}-15 \,{\mathrm e}^{{\mathrm e}}+20 x \right )^{2}-4 \ln \left (10\right ) \ln \left ({\mathrm e}^{{\mathrm e}} \ln \left (x \right )-x \,{\mathrm e}^{{\mathrm e}}-15 \,{\mathrm e}^{{\mathrm e}}+20 x \right )^{3}\) | \(112\) |
Input:
int(((4-4*x)*exp(exp(1))+80*x)*ln(1/10*exp(exp(1))*ln(x)+1/10*(-x-15)*exp( exp(1))+2*x)^3/(x*exp(exp(1))*ln(x)+(-x^2-15*x)*exp(exp(1))+20*x^2),x,meth od=_RETURNVERBOSE)
Output:
ln(1/10*exp(exp(1))*ln(x)+1/10*(-x-15)*exp(exp(1))+2*x)^4
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx=\log \left (-\frac {1}{10} \, {\left (x + 15\right )} e^{e} + \frac {1}{10} \, e^{e} \log \left (x\right ) + 2 \, x\right )^{4} \] Input:
integrate(((4-4*x)*exp(exp(1))+80*x)*log(1/10*exp(exp(1))*log(x)+1/10*(-x- 15)*exp(exp(1))+2*x)^3/(x*exp(exp(1))*log(x)+(-x^2-15*x)*exp(exp(1))+20*x^ 2),x, algorithm="fricas")
Output:
log(-1/10*(x + 15)*e^e + 1/10*e^e*log(x) + 2*x)^4
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx=\log {\left (2 x + \left (- \frac {x}{10} - \frac {3}{2}\right ) e^{e} + \frac {e^{e} \log {\left (x \right )}}{10} \right )}^{4} \] Input:
integrate(((4-4*x)*exp(exp(1))+80*x)*ln(1/10*exp(exp(1))*ln(x)+1/10*(-x-15 )*exp(exp(1))+2*x)**3/(x*exp(exp(1))*ln(x)+(-x**2-15*x)*exp(exp(1))+20*x** 2),x)
Output:
log(2*x + (-x/10 - 3/2)*exp(E) + exp(E)*log(x)/10)**4
Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (19) = 38\).
Time = 0.17 (sec) , antiderivative size = 381, normalized size of antiderivative = 14.65 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx=-12 \, {\left (\log \left (5\right ) + \log \left (2\right )\right )} e \log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right )^{2} + 4 \, e \log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right )^{3} - \log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right )^{4} + 4 \, \log \left (-{\left (x {\left (e^{e} - 20\right )} - e^{e} \log \left (x\right ) + 15 \, e^{e}\right )} e^{\left (-e\right )}\right ) \log \left (-\frac {1}{10} \, {\left (x + 15\right )} e^{e} + \frac {1}{10} \, e^{e} \log \left (x\right ) + 2 \, x\right )^{3} - 6 \, {\left (\log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right )^{2} - 2 \, \log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right ) \log \left (-\frac {1}{10} \, {\left (x + 15\right )} e^{e} + \frac {1}{10} \, e^{e} \log \left (x\right ) + 2 \, x\right ) + 2 \, \log \left (-{\left (x {\left (e^{e} - 20\right )} - e^{e} \log \left (x\right ) + 15 \, e^{e}\right )} e^{\left (-e\right )}\right ) \log \left (-\frac {1}{10} \, {\left (x + 15\right )} e^{e} + \frac {1}{10} \, e^{e} \log \left (x\right ) + 2 \, x\right )\right )} \log \left (-\frac {1}{10} \, {\left (x + 15\right )} e^{e} + \frac {1}{10} \, e^{e} \log \left (x\right ) + 2 \, x\right )^{2} + 4 \, {\left (6 \, {\left (\log \left (5\right ) + \log \left (2\right )\right )} e \log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right ) - 3 \, e \log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right )^{2} + \log \left (-x {\left (e^{e} - 20\right )} + e^{e} \log \left (x\right ) - 15 \, e^{e}\right )^{3}\right )} \log \left (-\frac {1}{10} \, {\left (x + 15\right )} e^{e} + \frac {1}{10} \, e^{e} \log \left (x\right ) + 2 \, x\right ) \] Input:
integrate(((4-4*x)*exp(exp(1))+80*x)*log(1/10*exp(exp(1))*log(x)+1/10*(-x- 15)*exp(exp(1))+2*x)^3/(x*exp(exp(1))*log(x)+(-x^2-15*x)*exp(exp(1))+20*x^ 2),x, algorithm="maxima")
Output:
-12*(log(5) + log(2))*e*log(-x*(e^e - 20) + e^e*log(x) - 15*e^e)^2 + 4*e*l og(-x*(e^e - 20) + e^e*log(x) - 15*e^e)^3 - log(-x*(e^e - 20) + e^e*log(x) - 15*e^e)^4 + 4*log(-(x*(e^e - 20) - e^e*log(x) + 15*e^e)*e^(-e))*log(-1/ 10*(x + 15)*e^e + 1/10*e^e*log(x) + 2*x)^3 - 6*(log(-x*(e^e - 20) + e^e*lo g(x) - 15*e^e)^2 - 2*log(-x*(e^e - 20) + e^e*log(x) - 15*e^e)*log(-1/10*(x + 15)*e^e + 1/10*e^e*log(x) + 2*x) + 2*log(-(x*(e^e - 20) - e^e*log(x) + 15*e^e)*e^(-e))*log(-1/10*(x + 15)*e^e + 1/10*e^e*log(x) + 2*x))*log(-1/10 *(x + 15)*e^e + 1/10*e^e*log(x) + 2*x)^2 + 4*(6*(log(5) + log(2))*e*log(-x *(e^e - 20) + e^e*log(x) - 15*e^e) - 3*e*log(-x*(e^e - 20) + e^e*log(x) - 15*e^e)^2 + log(-x*(e^e - 20) + e^e*log(x) - 15*e^e)^3)*log(-1/10*(x + 15) *e^e + 1/10*e^e*log(x) + 2*x)
Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (19) = 38\).
Time = 0.43 (sec) , antiderivative size = 111, normalized size of antiderivative = 4.27 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx=-4 \, \log \left (10\right )^{3} \log \left (-x e^{e} + e^{e} \log \left (x\right ) + 20 \, x - 15 \, e^{e}\right ) + 6 \, \log \left (10\right )^{2} \log \left (-x e^{e} + e^{e} \log \left (x\right ) + 20 \, x - 15 \, e^{e}\right )^{2} - 4 \, \log \left (10\right ) \log \left (-x e^{e} + e^{e} \log \left (x\right ) + 20 \, x - 15 \, e^{e}\right )^{3} + \log \left (-x e^{e} + e^{e} \log \left (x\right ) + 20 \, x - 15 \, e^{e}\right )^{4} \] Input:
integrate(((4-4*x)*exp(exp(1))+80*x)*log(1/10*exp(exp(1))*log(x)+1/10*(-x- 15)*exp(exp(1))+2*x)^3/(x*exp(exp(1))*log(x)+(-x^2-15*x)*exp(exp(1))+20*x^ 2),x, algorithm="giac")
Output:
-4*log(10)^3*log(-x*e^e + e^e*log(x) + 20*x - 15*e^e) + 6*log(10)^2*log(-x *e^e + e^e*log(x) + 20*x - 15*e^e)^2 - 4*log(10)*log(-x*e^e + e^e*log(x) + 20*x - 15*e^e)^3 + log(-x*e^e + e^e*log(x) + 20*x - 15*e^e)^4
Time = 3.49 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx={\ln \left (2\,x-\frac {{\mathrm {e}}^{\mathrm {e}}\,\left (x+15\right )}{10}+\frac {{\mathrm {e}}^{\mathrm {e}}\,\ln \left (x\right )}{10}\right )}^4 \] Input:
int((log(2*x - (exp(exp(1))*(x + 15))/10 + (exp(exp(1))*log(x))/10)^3*(80* x - exp(exp(1))*(4*x - 4)))/(20*x^2 - exp(exp(1))*(15*x + x^2) + x*exp(exp (1))*log(x)),x)
Output:
log(2*x - (exp(exp(1))*(x + 15))/10 + (exp(exp(1))*log(x))/10)^4
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (e^e (4-4 x)+80 x\right ) \log ^3\left (\frac {1}{10} \left (e^e (-15-x)+20 x+e^e \log (x)\right )\right )}{20 x^2+e^e \left (-15 x-x^2\right )+e^e x \log (x)} \, dx=\mathrm {log}\left (\frac {e^{e} \mathrm {log}\left (x \right )}{10}-\frac {e^{e} x}{10}-\frac {3 e^{e}}{2}+2 x \right )^{4} \] Input:
int(((4-4*x)*exp(exp(1))+80*x)*log(1/10*exp(exp(1))*log(x)+1/10*(-x-15)*ex p(exp(1))+2*x)^3/(x*exp(exp(1))*log(x)+(-x^2-15*x)*exp(exp(1))+20*x^2),x)
Output:
log((e**e*log(x) - e**e*x - 15*e**e + 20*x)/10)**4