\(\int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} (-x+x \log (x)+(-2+x) \log ^2(x))}{x^3 \log ^2(x)} \, dx\) [7165]

Optimal result

Integrand size = 39, antiderivative size = 15 \[ \int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} \left (-x+x \log (x)+(-2+x) \log ^2(x)\right )}{x^3 \log ^2(x)} \, dx=\frac {e^{3+x+\frac {x}{\log (x)}}}{x^2} \]

[Out]

exp(x+x/ln(x)+3)/x^2

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} \left (-x+x \log (x)+(-2+x) \log ^2(x)\right )}{x^3 \log ^2(x)} \, dx=\frac {e^{3+x+\frac {x}{\log (x)}}}{x^2} \]

[In]

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

[Out]

E^(3 + x + x/Log[x])/x^2

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{\frac {\left (3+x \right ) \ln \left (x \right )+x}{\ln \left (x \right )}}}{x^{2}}\)	\(19\)
risch	\(\frac {{\mathrm e}^{\frac {x \ln \left (x \right )+3 \ln \left (x \right )+x}{\ln \left (x \right )}}}{x^{2}}\)	\(21\)

[In]

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

[Out]

1/x^2*exp(((3+x)*ln(x)+x)/ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} \left (-x+x \log (x)+(-2+x) \log ^2(x)\right )}{x^3 \log ^2(x)} \, dx=\frac {e^{\left (\frac {{\left (x + 3\right )} \log \left (x\right ) + x}{\log \left (x\right )}\right )}}{x^{2}} \]

[In]

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

[Out]

e^(((x + 3)*log(x) + x)/log(x))/x^2

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} \left (-x+x \log (x)+(-2+x) \log ^2(x)\right )}{x^3 \log ^2(x)} \, dx=\frac {e^{\frac {x + \left (x + 3\right ) \log {\left (x \right )}}{\log {\left (x \right )}}}}{x^{2}} \]

[In]

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

[Out]

exp((x + (x + 3)*log(x))/log(x))/x**2

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} \left (-x+x \log (x)+(-2+x) \log ^2(x)\right )}{x^3 \log ^2(x)} \, dx=\frac {e^{\left (x + \frac {x}{\log \left (x\right )} + 3\right )}}{x^{2}} \]

[In]

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

[Out]

e^(x + x/log(x) + 3)/x^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} \left (-x+x \log (x)+(-2+x) \log ^2(x)\right )}{x^3 \log ^2(x)} \, dx=\frac {e^{\left (\frac {x \log \left (x\right ) + x + 3 \, \log \left (x\right )}{\log \left (x\right )}\right )}}{x^{2}} \]

[In]

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

[Out]

e^((x*log(x) + x + 3*log(x))/log(x))/x^2

Mupad [B] (verification not implemented)

Time = 12.87 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x+(3+x) \log (x)}{\log (x)}} \left (-x+x \log (x)+(-2+x) \log ^2(x)\right )}{x^3 \log ^2(x)} \, dx=\frac {{\mathrm {e}}^3\,{\mathrm {e}}^{\frac {x}{\ln \left (x\right )}}\,{\mathrm {e}}^x}{x^2} \]

[In]

int((exp((x + log(x)*(x + 3))/log(x))*(log(x)^2*(x - 2) - x + x*log(x)))/(x^3*log(x)^2),x)

[Out]

(exp(3)*exp(x/log(x))*exp(x))/x^2