\(\int \frac {e^x x-x^2+(e^x (-3+x)+3 x-x^2) \log (x)+(x^2-e^x x^2) \log (x) \log (-10 e^{3/x} x \log (x)) \log (\log (-10 e^{3/x} x \log (x)))}{(3 e^{2 x} x^2-6 e^x x^3+3 x^4) \log (x) \log (-10 e^{3/x} x \log (x))} \, dx\) [10011]

Optimal result

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

[Out]

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

Rubi [F]

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

[In]

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[Lo
g[-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]

[Out]

-Defer[Int][1/((E^x - x)*x^2*Log[-10*E^(3/x)*x*Log[x]]), x] + Defer[Int][1/((E^x - x)*x*Log[-10*E^(3/x)*x*Log[
x]]), x]/3 + Defer[Int][1/((E^x - x)*x*Log[x]*Log[-10*E^(3/x)*x*Log[x]]), x]/3 + Defer[Int][Log[Log[-10*E^(3/x
)*x*Log[x]]]/(E^x - x)^2, x]/3 - Defer[Int][Log[Log[-10*E^(3/x)*x*Log[x]]]/(E^x - x), x]/3 - Defer[Int][(x*Log
[Log[-10*E^(3/x)*x*Log[x]]])/(E^x - x)^2, x]/3

Rubi steps \begin{align*} \text {integral}& = \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 )}{3 \left (e^x-x\right )^2 x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx \\ & = \frac {1}{3} \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 (e^x-x\right )^2 x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx \\ & = \frac {1}{3} \int \left (-\frac {(-1+x) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}-\frac {-x+3 \log (x)-x \log (x)+x^2 \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 ) x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}\right ) \, dx \\ & = -\left (\frac {1}{3} \int \frac {(-1+x) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2} \, dx\right )-\frac {1}{3} \int \frac {-x+3 \log (x)-x \log (x)+x^2 \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 ) x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx \\ & = -\left (\frac {1}{3} \int \left (\frac {3}{\left (e^x-x\right ) x^2 \log \left (-10 e^{3/x} x \log (x)\right )}-\frac {1}{\left (e^x-x\right ) x \log \left (-10 e^{3/x} x \log (x)\right )}-\frac {1}{\left (e^x-x\right ) x \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}+\frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{e^x-x}\right ) \, dx\right )-\frac {1}{3} \int \left (-\frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}+\frac {x \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}\right ) \, dx \\ & = \frac {1}{3} \int \frac {1}{\left (e^x-x\right ) x \log \left (-10 e^{3/x} x \log (x)\right )} \, dx+\frac {1}{3} \int \frac {1}{\left (e^x-x\right ) x \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2} \, dx-\frac {1}{3} \int \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{e^x-x} \, dx-\frac {1}{3} \int \frac {x \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2} \, dx-\int \frac {1}{\left (e^x-x\right ) x^2 \log \left (-10 e^{3/x} x \log (x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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]

[Out]

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

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

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

[In]

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)

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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, algorit
hm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

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

[In]

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)

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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, algorit
hm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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, algorit
hm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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