\(\int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx\) [6635]

Optimal result

Integrand size = 19, antiderivative size = 9 \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx=e^{-35+x+\log ^2(x)} \]

[Out]

exp(ln(x)^2+x-35)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6838} \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx=e^{x+\log ^2(x)-35} \]

[In]

Int[(E^(-35 + x + Log[x]^2)*(x + 2*Log[x]))/x,x]

[Out]

E^(-35 + x + Log[x]^2)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e^{-35+x+\log ^2(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx=e^{-35+x+\log ^2(x)} \]

[In]

Integrate[(E^(-35 + x + Log[x]^2)*(x + 2*Log[x]))/x,x]

[Out]

E^(-35 + x + Log[x]^2)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\({\mathrm e}^{\ln \left (x \right )^{2}+x -35}\)	\(9\)
default	\({\mathrm e}^{\ln \left (x \right )^{2}+x -35}\)	\(9\)
norman	\({\mathrm e}^{\ln \left (x \right )^{2}+x -35}\)	\(9\)
risch	\({\mathrm e}^{\ln \left (x \right )^{2}+x -35}\)	\(9\)
parallelrisch	\({\mathrm e}^{\ln \left (x \right )^{2}+x -35}\)	\(9\)

[In]

int((2*ln(x)+x)*exp(ln(x)^2+x-35)/x,x,method=_RETURNVERBOSE)

[Out]

exp(ln(x)^2+x-35)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx=e^{\left (\log \left (x\right )^{2} + x - 35\right )} \]

[In]

integrate((2*log(x)+x)*exp(log(x)^2+x-35)/x,x, algorithm="fricas")

[Out]

e^(log(x)^2 + x - 35)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx=e^{x + \log {\left (x \right )}^{2} - 35} \]

[In]

integrate((2*ln(x)+x)*exp(ln(x)**2+x-35)/x,x)

[Out]

exp(x + log(x)**2 - 35)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx=e^{\left (\log \left (x\right )^{2} + x - 35\right )} \]

[In]

integrate((2*log(x)+x)*exp(log(x)^2+x-35)/x,x, algorithm="maxima")

[Out]

e^(log(x)^2 + x - 35)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx=e^{\left (\log \left (x\right )^{2} + x - 35\right )} \]

[In]

integrate((2*log(x)+x)*exp(log(x)^2+x-35)/x,x, algorithm="giac")

[Out]

e^(log(x)^2 + x - 35)

Mupad [B] (verification not implemented)

Time = 11.12 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-35+x+\log ^2(x)} (x+2 \log (x))}{x} \, dx={\mathrm {e}}^{-35}\,{\mathrm {e}}^{{\ln \left (x\right )}^2}\,{\mathrm {e}}^x \]

[In]

int((exp(x + log(x)^2 - 35)*(x + 2*log(x)))/x,x)

[Out]

exp(-35)*exp(log(x)^2)*exp(x)