\(\int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+(5 e^{\frac {1}{e^2}} x+e^x (-4 x+x^2)) \log (x)}{5 x \log ^3(x)} \, dx\) [6892]

Optimal result

Integrand size = 55, antiderivative size = 24 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=-\frac {\left (e^{\frac {1}{e^2}}+\frac {e^x}{5}\right ) (5-x)}{\log ^2(x)} \]

[Out]

-(5-x)/ln(x)^2*(1/5*exp(x)+exp(exp(-2)))

Rubi [F]

\[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx \]

[In]

Int[(E^E^(-2)*(50 - 10*x) + E^x*(10 - 2*x) + (5*E^E^(-2)*x + E^x*(-4*x + x^2))*Log[x])/(5*x*Log[x]^3),x]

[Out]

(-5*E^E^(-2))/Log[x]^2 + (E^E^(-2)*x)/Log[x]^2 - (2*Defer[Int][E^x/Log[x]^3, x])/5 + 2*Defer[Int][E^x/(x*Log[x
]^3), x] - (4*Defer[Int][E^x/Log[x]^2, x])/5 + Defer[Int][(E^x*x)/Log[x]^2, x]/5

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{x \log ^3(x)} \, dx \\ & = \frac {1}{5} \int \left (\frac {5 e^{\frac {1}{e^2}} (10-2 x+x \log (x))}{x \log ^3(x)}+\frac {e^x \left (10-2 x-4 x \log (x)+x^2 \log (x)\right )}{x \log ^3(x)}\right ) \, dx \\ & = \frac {1}{5} \int \frac {e^x \left (10-2 x-4 x \log (x)+x^2 \log (x)\right )}{x \log ^3(x)} \, dx+e^{\frac {1}{e^2}} \int \frac {10-2 x+x \log (x)}{x \log ^3(x)} \, dx \\ & = \frac {1}{5} \int \left (-\frac {2 e^x (-5+x)}{x \log ^3(x)}+\frac {e^x (-4+x)}{\log ^2(x)}\right ) \, dx+e^{\frac {1}{e^2}} \int \left (-\frac {2 (-5+x)}{x \log ^3(x)}+\frac {1}{\log ^2(x)}\right ) \, dx \\ & = \frac {1}{5} \int \frac {e^x (-4+x)}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x (-5+x)}{x \log ^3(x)} \, dx+e^{\frac {1}{e^2}} \int \frac {1}{\log ^2(x)} \, dx-\left (2 e^{\frac {1}{e^2}}\right ) \int \frac {-5+x}{x \log ^3(x)} \, dx \\ & = -\frac {e^{\frac {1}{e^2}} x}{\log (x)}+\frac {1}{5} \int \left (-\frac {4 e^x}{\log ^2(x)}+\frac {e^x x}{\log ^2(x)}\right ) \, dx-\frac {2}{5} \int \left (\frac {e^x}{\log ^3(x)}-\frac {5 e^x}{x \log ^3(x)}\right ) \, dx+e^{\frac {1}{e^2}} \int \frac {1}{\log (x)} \, dx-\left (2 e^{\frac {1}{e^2}}\right ) \int \left (\frac {1}{\log ^3(x)}-\frac {5}{x \log ^3(x)}\right ) \, dx \\ & = -\frac {e^{\frac {1}{e^2}} x}{\log (x)}+e^{\frac {1}{e^2}} \text {li}(x)+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx-\left (2 e^{\frac {1}{e^2}}\right ) \int \frac {1}{\log ^3(x)} \, dx+\left (10 e^{\frac {1}{e^2}}\right ) \int \frac {1}{x \log ^3(x)} \, dx \\ & = \frac {e^{\frac {1}{e^2}} x}{\log ^2(x)}-\frac {e^{\frac {1}{e^2}} x}{\log (x)}+e^{\frac {1}{e^2}} \text {li}(x)+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx-e^{\frac {1}{e^2}} \int \frac {1}{\log ^2(x)} \, dx+\left (10 e^{\frac {1}{e^2}}\right ) \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right ) \\ & = -\frac {5 e^{\frac {1}{e^2}}}{\log ^2(x)}+\frac {e^{\frac {1}{e^2}} x}{\log ^2(x)}+e^{\frac {1}{e^2}} \text {li}(x)+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx-e^{\frac {1}{e^2}} \int \frac {1}{\log (x)} \, dx \\ & = -\frac {5 e^{\frac {1}{e^2}}}{\log ^2(x)}+\frac {e^{\frac {1}{e^2}} x}{\log ^2(x)}+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {\left (5 e^{\frac {1}{e^2}}+e^x\right ) (-5+x)}{5 \log ^2(x)} \]

[In]

Integrate[(E^E^(-2)*(50 - 10*x) + E^x*(10 - 2*x) + (5*E^E^(-2)*x + E^x*(-4*x + x^2))*Log[x])/(5*x*Log[x]^3),x]

[Out]

((5*E^E^(-2) + E^x)*(-5 + x))/(5*Log[x]^2)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {5 x \,{\mathrm e}^{{\mathrm e}^{-2}}+{\mathrm e}^{x} x -25 \,{\mathrm e}^{{\mathrm e}^{-2}}-5 \,{\mathrm e}^{x}}{5 \ln \left (x \right )^{2}}\)	\(27\)
parallelrisch	\(\frac {5 x \,{\mathrm e}^{{\mathrm e}^{-2}}+{\mathrm e}^{x} x -25 \,{\mathrm e}^{{\mathrm e}^{-2}}-5 \,{\mathrm e}^{x}}{5 \ln \left (x \right )^{2}}\)	\(31\)

[In]

int(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*ln(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/ln(x)^3,x,met
hod=_RETURNVERBOSE)

[Out]

1/5*(5*x*exp(exp(-2))+exp(x)*x-25*exp(exp(-2))-5*exp(x))/ln(x)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {{\left (x - 5\right )} e^{x} + 5 \, {\left (x - 5\right )} e^{\left (e^{\left (-2\right )}\right )}}{5 \, \log \left (x\right )^{2}} \]

[In]

integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/log(x)
^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {\left (x - 5\right ) e^{x}}{5 \log {\left (x \right )}^{2}} + \frac {x e^{e^{-2}} - 5 e^{e^{-2}}}{\log {\left (x \right )}^{2}} \]

[In]

integrate(1/5*(((x**2-4*x)*exp(x)+5*x*exp(1/exp(2)))*ln(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/ln(x)*
*3,x)

[Out]

(x - 5)*exp(x)/(5*log(x)**2) + (x*exp(exp(-2)) - 5*exp(exp(-2)))/log(x)**2

Maxima [F]

\[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\int { -\frac {2 \, {\left (x - 5\right )} e^{x} + 10 \, {\left (x - 5\right )} e^{\left (e^{\left (-2\right )}\right )} - {\left ({\left (x^{2} - 4 \, x\right )} e^{x} + 5 \, x e^{\left (e^{\left (-2\right )}\right )}\right )} \log \left (x\right )}{5 \, x \log \left (x\right )^{3}} \,d x } \]

[In]

integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/log(x)
^3,x, algorithm="maxima")

[Out]

2*e^(e^(-2))*gamma(-2, -log(x)) + e^(e^(-2))*integrate(1/log(x), x) - 1/5*(5*x*e^(e^(-2))*log(x) - (x - 5)*e^x
)/log(x)^2 - 5*e^(e^(-2))/log(x)^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {x e^{x} + 5 \, x e^{\left (e^{\left (-2\right )}\right )} - 5 \, e^{x} - 25 \, e^{\left (e^{\left (-2\right )}\right )}}{5 \, \log \left (x\right )^{2}} \]

[In]

integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/log(x)
^3,x, algorithm="giac")

[Out]

1/5*(x*e^x + 5*x*e^(e^(-2)) - 5*e^x - 25*e^(e^(-2)))/log(x)^2

Mupad [B] (verification not implemented)

Time = 12.66 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {\left (5\,{\mathrm {e}}^{{\mathrm {e}}^{-2}}+{\mathrm {e}}^x\right )\,\left (x-5\right )}{5\,{\ln \left (x\right )}^2} \]

[In]

int(-((exp(exp(-2))*(10*x - 50))/5 + (exp(x)*(2*x - 10))/5 - (log(x)*(5*x*exp(exp(-2)) - exp(x)*(4*x - x^2)))/
5)/(x*log(x)^3),x)

[Out]

((5*exp(exp(-2)) + exp(x))*(x - 5))/(5*log(x)^2)