\(\int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+(-40 x-2 x^2) \log (x)+x^2 \log ^2(x)} \, dx\) [7758]

Optimal result

Integrand size = 59, antiderivative size = 19 \[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\frac {e-\log (3)}{20+x-x \log (x)}} \]

[Out]

exp((-ln(3)+exp(1))/(20+x-x*ln(x)))

Rubi [F]

\[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx \]

[In]

Int[(E^((-E + Log[3])/(-20 - x + x*Log[x]))*(E - Log[3])*Log[x])/(400 + 40*x + x^2 + (-40*x - 2*x^2)*Log[x] +
x^2*Log[x]^2),x]

[Out]

(E - Log[3])*Defer[Int][(3^(-20 - x + x*Log[x])^(-1)*E^(E/(20 + x - x*Log[x])))/(-20 - x + x*Log[x])^2, x] + 2
0*(E - Log[3])*Defer[Int][(3^(-20 - x + x*Log[x])^(-1)*E^(E/(20 + x - x*Log[x])))/(x*(-20 - x + x*Log[x])^2),
x] + (E - Log[3])*Defer[Int][(3^(-20 - x + x*Log[x])^(-1)*E^(E/(20 + x - x*Log[x])))/(x*(-20 - x + x*Log[x])),
x]

Rubi steps \begin{align*} \text {integral}& = (e-\log (3)) \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx \\ & = (e-\log (3)) \int \frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}} \log (x)}{(20+x-x \log (x))^2} \, dx \\ & = (e-\log (3)) \int \left (\frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}} (20+x)}{x (-20-x+x \log (x))^2}+\frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{x (-20-x+x \log (x))}\right ) \, dx \\ & = (e-\log (3)) \int \frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}} (20+x)}{x (-20-x+x \log (x))^2} \, dx+(e-\log (3)) \int \frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{x (-20-x+x \log (x))} \, dx \\ & = (e-\log (3)) \int \frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{x (-20-x+x \log (x))} \, dx+(e-\log (3)) \int \left (\frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{(-20-x+x \log (x))^2}+\frac {20\ 3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{x (-20-x+x \log (x))^2}\right ) \, dx \\ & = (e-\log (3)) \int \frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{(-20-x+x \log (x))^2} \, dx+(e-\log (3)) \int \frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{x (-20-x+x \log (x))} \, dx+(20 (e-\log (3))) \int \frac {3^{\frac {1}{-20-x+x \log (x)}} e^{\frac {e}{20+x-x \log (x)}}}{x (-20-x+x \log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\frac {e-\log (3)}{20+x-x \log (x)}} \]

[In]

Integrate[(E^((-E + Log[3])/(-20 - x + x*Log[x]))*(E - Log[3])*Log[x])/(400 + 40*x + x^2 + (-40*x - 2*x^2)*Log
[x] + x^2*Log[x]^2),x]

[Out]

E^((E - Log[3])/(20 + x - x*Log[x]))

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16

method	result	size
risch	\({\mathrm e}^{-\frac {-\ln \left (3\right )+{\mathrm e}}{x \ln \left (x \right )-x -20}}\)	\(22\)
parallelrisch	\({\mathrm e}^{-\frac {-\ln \left (3\right )+{\mathrm e}}{x \ln \left (x \right )-x -20}}\)	\(22\)
norman	\(\frac {x \ln \left (x \right ) {\mathrm e}^{\frac {\ln \left (3\right )-{\mathrm e}}{x \ln \left (x \right )-x -20}}-x \,{\mathrm e}^{\frac {\ln \left (3\right )-{\mathrm e}}{x \ln \left (x \right )-x -20}}-20 \,{\mathrm e}^{\frac {\ln \left (3\right )-{\mathrm e}}{x \ln \left (x \right )-x -20}}}{x \ln \left (x \right )-x -20}\)	\(83\)

[In]

int((-ln(3)+exp(1))*ln(x)*exp((ln(3)-exp(1))/(x*ln(x)-x-20))/(x^2*ln(x)^2+(-2*x^2-40*x)*ln(x)+x^2+40*x+400),x,
method=_RETURNVERBOSE)

[Out]

exp(-(-ln(3)+exp(1))/(x*ln(x)-x-20))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (-\frac {e - \log \left (3\right )}{x \log \left (x\right ) - x - 20}\right )} \]

[In]

integrate((-log(3)+exp(1))*log(x)*exp((log(3)-exp(1))/(x*log(x)-x-20))/(x^2*log(x)^2+(-2*x^2-40*x)*log(x)+x^2+
40*x+400),x, algorithm="fricas")

[Out]

e^(-(e - log(3))/(x*log(x) - x - 20))

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((-ln(3)+exp(1))*ln(x)*exp((ln(3)-exp(1))/(x*ln(x)-x-20))/(x**2*ln(x)**2+(-2*x**2-40*x)*ln(x)+x**2+40
*x+400),x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((-log(3)+exp(1))*log(x)*exp((log(3)-exp(1))/(x*log(x)-x-20))/(x^2*log(x)^2+(-2*x^2-40*x)*log(x)+x^2+
40*x+400),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (-\frac {e}{x \log \left (x\right ) - x - 20} + \frac {\log \left (3\right )}{x \log \left (x\right ) - x - 20}\right )} \]

[In]

integrate((-log(3)+exp(1))*log(x)*exp((log(3)-exp(1))/(x*log(x)-x-20))/(x^2*log(x)^2+(-2*x^2-40*x)*log(x)+x^2+
40*x+400),x, algorithm="giac")

[Out]

e^(-e/(x*log(x) - x - 20) + log(3)/(x*log(x) - x - 20))

Mupad [B] (verification not implemented)

Time = 14.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\frac {-e+\log (3)}{-20-x+x \log (x)}} (e-\log (3)) \log (x)}{400+40 x+x^2+\left (-40 x-2 x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{\frac {\mathrm {e}}{x-x\,\ln \left (x\right )+20}}}{3^{\frac {1}{x-x\,\ln \left (x\right )+20}}} \]

[In]

int((exp((exp(1) - log(3))/(x - x*log(x) + 20))*log(x)*(exp(1) - log(3)))/(40*x + x^2*log(x)^2 - log(x)*(40*x
+ 2*x^2) + x^2 + 400),x)

[Out]

exp(exp(1)/(x - x*log(x) + 20))/3^(1/(x - x*log(x) + 20))