Integrand size = 89, antiderivative size = 30 \[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=e^{-e^{e+x}+x+\frac {e^2}{x \log ^2(x)}}+(4-x)^2 \] Output:
(4-x)^2+exp(x-exp(x+exp(1))+exp(2)/ln(x)^2/x)
Time = 1.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=e^{-e^{e+x}+x+\frac {e^2}{x \log ^2(x)}}-8 x+x^2 \] Input:
Integrate[((-8*x^2 + 2*x^3)*Log[x]^3 + E^((E^2 + (-(E^(E + x)*x) + x^2)*Lo g[x]^2)/(x*Log[x]^2))*(-2*E^2 - E^2*Log[x] + (x^2 - E^(E + x)*x^2)*Log[x]^ 3))/(x^2*Log[x]^3),x]
Output:
E^(-E^(E + x) + x + E^2/(x*Log[x]^2)) - 8*x + x^2
Time = 2.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {\left (\left (x^2-e^{x+e} x^2\right ) \log ^3(x)-e^2 \log (x)-2 e^2\right ) \exp \left (\frac {\left (x^2-e^{x+e} x\right ) \log ^2(x)+e^2}{x \log ^2(x)}\right )+\left (2 x^3-8 x^2\right ) \log ^3(x)}{x^2 \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 (x-4)-\frac {e^{x-e^{x+e}+\frac {e^2}{x \log ^2(x)}} \left (e^{x+e} x^2 \log ^3(x)-x^2 \log ^3(x)+e^2 \log (x)+2 e^2\right )}{x^2 \log ^3(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle (x-4)^2+e^{x-e^{x+e}+\frac {e^2}{x \log ^2(x)}}\) |
Input:
Int[((-8*x^2 + 2*x^3)*Log[x]^3 + E^((E^2 + (-(E^(E + x)*x) + x^2)*Log[x]^2 )/(x*Log[x]^2))*(-2*E^2 - E^2*Log[x] + (x^2 - E^(E + x)*x^2)*Log[x]^3))/(x ^2*Log[x]^3),x]
Output:
E^(-E^(E + x) + x + E^2/(x*Log[x]^2)) + (-4 + x)^2
Time = 91.69 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23
method | result | size |
parallelrisch | \(x^{2}-8 x +{\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{x +{\mathrm e}}+x^{2}\right ) \ln \left (x \right )^{2}+{\mathrm e}^{2}}{x \ln \left (x \right )^{2}}}\) | \(37\) |
risch | \(x^{2}-8 x +{\mathrm e}^{\frac {-{\mathrm e}^{x +{\mathrm e}} \ln \left (x \right )^{2} x +x^{2} \ln \left (x \right )^{2}+{\mathrm e}^{2}}{x \ln \left (x \right )^{2}}}\) | \(40\) |
Input:
int((((-x^2*exp(x+exp(1))+x^2)*ln(x)^3-exp(2)*ln(x)-2*exp(2))*exp(((-x*exp (x+exp(1))+x^2)*ln(x)^2+exp(2))/x/ln(x)^2)+(2*x^3-8*x^2)*ln(x)^3)/x^2/ln(x )^3,x,method=_RETURNVERBOSE)
Output:
x^2-8*x+exp(((-x*exp(x+exp(1))+x^2)*ln(x)^2+exp(2))/x/ln(x)^2)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=x^{2} - 8 \, x + e^{\left (\frac {{\left (x^{2} - x e^{\left (x + e\right )}\right )} \log \left (x\right )^{2} + e^{2}}{x \log \left (x\right )^{2}}\right )} \] Input:
integrate((((-x^2*exp(x+exp(1))+x^2)*log(x)^3-exp(2)*log(x)-2*exp(2))*exp( ((-x*exp(x+exp(1))+x^2)*log(x)^2+exp(2))/x/log(x)^2)+(2*x^3-8*x^2)*log(x)^ 3)/x^2/log(x)^3,x, algorithm="fricas")
Output:
x^2 - 8*x + e^(((x^2 - x*e^(x + e))*log(x)^2 + e^2)/(x*log(x)^2))
Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=x^{2} - 8 x + e^{\frac {\left (x^{2} - x e^{x + e}\right ) \log {\left (x \right )}^{2} + e^{2}}{x \log {\left (x \right )}^{2}}} \] Input:
integrate((((-x**2*exp(x+exp(1))+x**2)*ln(x)**3-exp(2)*ln(x)-2*exp(2))*exp (((-x*exp(x+exp(1))+x**2)*ln(x)**2+exp(2))/x/ln(x)**2)+(2*x**3-8*x**2)*ln( x)**3)/x**2/ln(x)**3,x)
Output:
x**2 - 8*x + exp(((x**2 - x*exp(x + E))*log(x)**2 + exp(2))/(x*log(x)**2))
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=x^{2} - 8 \, x + e^{\left (x + \frac {e^{2}}{x \log \left (x\right )^{2}} - e^{\left (x + e\right )}\right )} \] Input:
integrate((((-x^2*exp(x+exp(1))+x^2)*log(x)^3-exp(2)*log(x)-2*exp(2))*exp( ((-x*exp(x+exp(1))+x^2)*log(x)^2+exp(2))/x/log(x)^2)+(2*x^3-8*x^2)*log(x)^ 3)/x^2/log(x)^3,x, algorithm="maxima")
Output:
x^2 - 8*x + e^(x + e^2/(x*log(x)^2) - e^(x + e))
\[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=\int { \frac {2 \, {\left (x^{3} - 4 \, x^{2}\right )} \log \left (x\right )^{3} - {\left ({\left (x^{2} e^{\left (x + e\right )} - x^{2}\right )} \log \left (x\right )^{3} + e^{2} \log \left (x\right ) + 2 \, e^{2}\right )} e^{\left (\frac {{\left (x^{2} - x e^{\left (x + e\right )}\right )} \log \left (x\right )^{2} + e^{2}}{x \log \left (x\right )^{2}}\right )}}{x^{2} \log \left (x\right )^{3}} \,d x } \] Input:
integrate((((-x^2*exp(x+exp(1))+x^2)*log(x)^3-exp(2)*log(x)-2*exp(2))*exp( ((-x*exp(x+exp(1))+x^2)*log(x)^2+exp(2))/x/log(x)^2)+(2*x^3-8*x^2)*log(x)^ 3)/x^2/log(x)^3,x, algorithm="giac")
Output:
undef
Time = 2.79 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=x^2-8\,x+{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x\,{\ln \left (x\right )}^2}}\,{\mathrm {e}}^{-{\mathrm {e}}^{\mathrm {e}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^x \] Input:
int(-(exp((exp(2) - log(x)^2*(x*exp(x + exp(1)) - x^2))/(x*log(x)^2))*(2*e xp(2) + exp(2)*log(x) - log(x)^3*(x^2 - x^2*exp(x + exp(1)))) + log(x)^3*( 8*x^2 - 2*x^3))/(x^2*log(x)^3),x)
Output:
x^2 - 8*x + exp(exp(2)/(x*log(x)^2))*exp(-exp(exp(1))*exp(x))*exp(x)
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {\left (-8 x^2+2 x^3\right ) \log ^3(x)+e^{\frac {e^2+\left (-e^{e+x} x+x^2\right ) \log ^2(x)}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)+\left (x^2-e^{e+x} x^2\right ) \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx=\frac {e^{e^{e +x}} x^{2}-8 e^{e^{e +x}} x +e^{\frac {\mathrm {log}\left (x \right )^{2} x^{2}+e^{2}}{\mathrm {log}\left (x \right )^{2} x}}}{e^{e^{e +x}}} \] Input:
int((((-x^2*exp(x+exp(1))+x^2)*log(x)^3-exp(2)*log(x)-2*exp(2))*exp(((-x*e xp(x+exp(1))+x^2)*log(x)^2+exp(2))/x/log(x)^2)+(2*x^3-8*x^2)*log(x)^3)/x^2 /log(x)^3,x)
Output:
(e**(e**(e + x))*x**2 - 8*e**(e**(e + x))*x + e**((log(x)**2*x**2 + e**2)/ (log(x)**2*x)))/e**(e**(e + x))