3.11.11 \(\int \frac {e^x x-x^2+(e^x (-3+x)+3 x-x^2) \log (x)+(x^2-e^x x^2) \log (x) \log (-10 e^{3/x} x \log (x)) \log (\log (-10 e^{3/x} x \log (x)))}{(3 e^{2 x} x^2-6 e^x x^3+3 x^4) \log (x) \log (-10 e^{3/x} x \log (x))} \, dx\) [1011]

3.11.11.1 Optimal result

Integrand size = 118, antiderivative size = 27 \begin {dmath*} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx=\frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{3 \left (e^x-x\right )} \end {dmath*}

output

1/3*ln(ln(-10*x*exp(3/x)*ln(x)))/(exp(x)-x)
 

3.11.11.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx=\frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{3 \left (e^x-x\right )} \end {dmath*}

input

Integrate[(E^x*x - x^2 + (E^x*(-3 + x) + 3*x - x^2)*Log[x] + (x^2 - E^x*x^ 
2)*Log[x]*Log[-10*E^(3/x)*x*Log[x]]*Log[Log[-10*E^(3/x)*x*Log[x]]])/((3*E^ 
(2*x)*x^2 - 6*E^x*x^3 + 3*x^4)*Log[x]*Log[-10*E^(3/x)*x*Log[x]]),x]
 

output

Log[Log[-10*E^(3/x)*x*Log[x]]]/(3*(E^x - x))
 

3.11.11.3 Rubi [F]

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.

\(\displaystyle \int \frac {-x^2+\left (-x^2+3 x+e^x (x-3)\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )+e^x x}{\left (3 x^4-6 e^x x^3+3 e^{2 x} x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-x^2+\left (-x^2+3 x+e^x (x-3)\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )+e^x x}{3 \left (e^x-x\right )^2 x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {-x^2+e^x x-\left (x^2-3 x+e^x (3-x)\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2 x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{3} \int \left (-\frac {(x-1) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}-\frac {\log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right ) x^2-\log (x) x-x+3 \log (x)}{\left (e^x-x\right ) x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (-3 \int \frac {1}{\left (e^x-x\right ) x^2 \log \left (-10 e^{3/x} x \log (x)\right )}dx+\int \frac {1}{\left (e^x-x\right ) x \log \left (-10 e^{3/x} x \log (x)\right )}dx+\int \frac {1}{\left (e^x-x\right ) x \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}dx+\int \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}dx-\int \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{e^x-x}dx-\int \frac {x \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}dx\right )\)

input

Int[(E^x*x - x^2 + (E^x*(-3 + x) + 3*x - x^2)*Log[x] + (x^2 - E^x*x^2)*Log 
[x]*Log[-10*E^(3/x)*x*Log[x]]*Log[Log[-10*E^(3/x)*x*Log[x]]])/((3*E^(2*x)* 
x^2 - 6*E^x*x^3 + 3*x^4)*Log[x]*Log[-10*E^(3/x)*x*Log[x]]),x]
 

output

$Aborted
 

3.11.11.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.11.11.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.44 (sec) , antiderivative size = 195, normalized size of antiderivative = 7.22

\[-\frac {\ln \left (\ln \left (2\right )+\ln \left (5\right )+i \pi +\ln \left (x \right )+\ln \left ({\mathrm e}^{\frac {3}{x}}\right )+\ln \left (\ln \left (x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right )+\operatorname {csgn}\left (i {\mathrm e}^{\frac {3}{x}}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right )+\operatorname {csgn}\left (i \ln \left (x \right )\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right )+\operatorname {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right )\right )}{2}+i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right )^{2} \left (\operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \left (x \right )\right )-1\right )\right )}{3 \left (x -{\mathrm e}^{x}\right )}\]

input

int(((-exp(x)*x^2+x^2)*ln(x)*ln(-10*x*exp(3/x)*ln(x))*ln(ln(-10*x*exp(3/x) 
*ln(x)))+((-3+x)*exp(x)-x^2+3*x)*ln(x)+exp(x)*x-x^2)/(3*exp(x)^2*x^2-6*exp 
(x)*x^3+3*x^4)/ln(x)/ln(-10*x*exp(3/x)*ln(x)),x)
 

output

-1/3/(x-exp(x))*ln(ln(2)+ln(5)+I*Pi+ln(x)+ln(exp(3/x))+ln(ln(x))-1/2*I*Pi* 
csgn(I*exp(3/x)*ln(x))*(-csgn(I*exp(3/x)*ln(x))+csgn(I*exp(3/x)))*(-csgn(I 
*exp(3/x)*ln(x))+csgn(I*ln(x)))-1/2*I*Pi*csgn(I*x*exp(3/x)*ln(x))*(-csgn(I 
*x*exp(3/x)*ln(x))+csgn(I*x))*(-csgn(I*x*exp(3/x)*ln(x))+csgn(I*exp(3/x)*l 
n(x)))+I*Pi*csgn(I*x*exp(3/x)*ln(x))^2*(csgn(I*x*exp(3/x)*ln(x))-1))
 

3.11.11.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \begin {dmath*} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx=-\frac {\log \left (\log \left (-10 \, x e^{\frac {3}{x}} \log \left (x\right )\right )\right )}{3 \, {\left (x - e^{x}\right )}} \end {dmath*}

input

integrate(((-exp(x)*x^2+x^2)*log(x)*log(-10*x*exp(3/x)*log(x))*log(log(-10 
*x*exp(3/x)*log(x)))+((-3+x)*exp(x)-x^2+3*x)*log(x)+exp(x)*x-x^2)/(3*exp(x 
)^2*x^2-6*exp(x)*x^3+3*x^4)/log(x)/log(-10*x*exp(3/x)*log(x)),x, algorithm 
=\
 

output

-1/3*log(log(-10*x*e^(3/x)*log(x)))/(x - e^x)
 

3.11.11.6 Sympy [A] (verification not implemented)

Time = 1.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx=- \frac {\log {\left (\log {\left (- 10 x e^{\frac {3}{x}} \log {\left (x \right )} \right )} \right )}}{3 x - 3 e^{x}} \end {dmath*}

input

integrate(((-exp(x)*x**2+x**2)*ln(x)*ln(-10*x*exp(3/x)*ln(x))*ln(ln(-10*x* 
exp(3/x)*ln(x)))+((-3+x)*exp(x)-x**2+3*x)*ln(x)+exp(x)*x-x**2)/(3*exp(x)** 
2*x**2-6*exp(x)*x**3+3*x**4)/ln(x)/ln(-10*x*exp(3/x)*ln(x)),x)
 

output

-log(log(-10*x*exp(3/x)*log(x)))/(3*x - 3*exp(x))
 

3.11.11.7 Maxima [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx=-\frac {\log \left (x {\left (\log \left (5\right ) + \log \left (2\right )\right )} + x \log \left (x\right ) + x \log \left (-\log \left (x\right )\right ) + 3\right ) - \log \left (x\right )}{3 \, {\left (x - e^{x}\right )}} \end {dmath*}

input

integrate(((-exp(x)*x^2+x^2)*log(x)*log(-10*x*exp(3/x)*log(x))*log(log(-10 
*x*exp(3/x)*log(x)))+((-3+x)*exp(x)-x^2+3*x)*log(x)+exp(x)*x-x^2)/(3*exp(x 
)^2*x^2-6*exp(x)*x^3+3*x^4)/log(x)/log(-10*x*exp(3/x)*log(x)),x, algorithm 
=\
 

output

-1/3*(log(x*(log(5) + log(2)) + x*log(x) + x*log(-log(x)) + 3) - log(x))/( 
x - e^x)
 

3.11.11.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \begin {dmath*} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx=-\frac {\log \left (x \log \left (x\right ) + x \log \left (-10 \, \log \left (x\right )\right ) + 3\right ) - \log \left (x\right )}{3 \, {\left (x - e^{x}\right )}} \end {dmath*}

input

integrate(((-exp(x)*x^2+x^2)*log(x)*log(-10*x*exp(3/x)*log(x))*log(log(-10 
*x*exp(3/x)*log(x)))+((-3+x)*exp(x)-x^2+3*x)*log(x)+exp(x)*x-x^2)/(3*exp(x 
)^2*x^2-6*exp(x)*x^3+3*x^4)/log(x)/log(-10*x*exp(3/x)*log(x)),x, algorithm 
=\
 

output

-1/3*(log(x*log(x) + x*log(-10*log(x)) + 3) - log(x))/(x - e^x)
 

3.11.11.9 Mupad [F(-1)]

Timed out. \begin {dmath*} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx=\int \frac {x\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (3\,x+{\mathrm {e}}^x\,\left (x-3\right )-x^2\right )-x^2-\ln \left (\ln \left (-10\,x\,{\mathrm {e}}^{3/x}\,\ln \left (x\right )\right )\right )\,\ln \left (-10\,x\,{\mathrm {e}}^{3/x}\,\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (x^2\,{\mathrm {e}}^x-x^2\right )}{\ln \left (-10\,x\,{\mathrm {e}}^{3/x}\,\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (3\,x^2\,{\mathrm {e}}^{2\,x}-6\,x^3\,{\mathrm {e}}^x+3\,x^4\right )} \,d x \end {dmath*}

input

int((x*exp(x) + log(x)*(3*x + exp(x)*(x - 3) - x^2) - x^2 - log(log(-10*x* 
exp(3/x)*log(x)))*log(-10*x*exp(3/x)*log(x))*log(x)*(x^2*exp(x) - x^2))/(l 
og(-10*x*exp(3/x)*log(x))*log(x)*(3*x^2*exp(2*x) - 6*x^3*exp(x) + 3*x^4)), 
x)
 

output

int((x*exp(x) + log(x)*(3*x + exp(x)*(x - 3) - x^2) - x^2 - log(log(-10*x* 
exp(3/x)*log(x)))*log(-10*x*exp(3/x)*log(x))*log(x)*(x^2*exp(x) - x^2))/(l 
og(-10*x*exp(3/x)*log(x))*log(x)*(3*x^2*exp(2*x) - 6*x^3*exp(x) + 3*x^4)), 
 x)