\(\int \frac {e^{-29+e^x} (-1+e^x x \log (3 x))}{x \log ^2(3 x)} \, dx\) [8114]

Optimal result

Integrand size = 28, antiderivative size = 14 \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {e^{-29+e^x}}{\log (3 x)} \]

[Out]

exp(exp(x))/ln(3*x)/exp(29)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2326} \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {e^{e^x-29}}{\log (3 x)} \]

[In]

Int[(E^(-29 + E^x)*(-1 + E^x*x*Log[3*x]))/(x*Log[3*x]^2),x]

[Out]

E^(-29 + E^x)/Log[3*x]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{-29+e^x}}{\log (3 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {e^{-29+e^x}}{\log (3 x)} \]

[In]

Integrate[(E^(-29 + E^x)*(-1 + E^x*x*Log[3*x]))/(x*Log[3*x]^2),x]

[Out]

E^(-29 + E^x)/Log[3*x]

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{x}-29}}{\ln \left (3 x \right )}\)	\(13\)
parallelrisch	\(\frac {{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{-29}}{\ln \left (3 x \right )}\)	\(15\)

[In]

int((x*exp(x)*ln(3*x)-1)*exp(exp(x))/x/exp(29)/ln(3*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(3*x)*exp(exp(x)-29)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {e^{\left (e^{x} - 29\right )}}{\log \left (3 \, x\right )} \]

[In]

integrate((x*exp(x)*log(3*x)-1)*exp(exp(x))/x/exp(29)/log(3*x)^2,x, algorithm="fricas")

[Out]

e^(e^x - 29)/log(3*x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {e^{e^{x}}}{e^{29} \log {\left (3 x \right )}} \]

[In]

integrate((x*exp(x)*ln(3*x)-1)*exp(exp(x))/x/exp(29)/ln(3*x)**2,x)

[Out]

exp(-29)*exp(exp(x))/log(3*x)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {e^{\left (e^{x}\right )}}{e^{29} \log \left (3\right ) + e^{29} \log \left (x\right )} \]

[In]

integrate((x*exp(x)*log(3*x)-1)*exp(exp(x))/x/exp(29)/log(3*x)^2,x, algorithm="maxima")

[Out]

e^(e^x)/(e^29*log(3) + e^29*log(x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {e^{\left (e^{x} - 29\right )}}{\log \left (3 \, x\right )} \]

[In]

integrate((x*exp(x)*log(3*x)-1)*exp(exp(x))/x/exp(29)/log(3*x)^2,x, algorithm="giac")

[Out]

e^(e^x - 29)/log(3*x)

Mupad [B] (verification not implemented)

Time = 12.62 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-29+e^x} \left (-1+e^x x \log (3 x)\right )}{x \log ^2(3 x)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-29}}{\ln \left (3\,x\right )} \]

[In]

int((exp(exp(x))*exp(-29)*(x*log(3*x)*exp(x) - 1))/(x*log(3*x)^2),x)

[Out]

(exp(exp(x))*exp(-29))/log(3*x)