\(\int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} (-12-4 x-4 x^2+4 \log (x))}{x^3} \, dx\) [5384]

Optimal result

Integrand size = 38, antiderivative size = 31 \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=\frac {4 e^{-9+\frac {2}{x}-x-\frac {x+\log (x)}{x}}-x}{x} \]

[Out]

(4*exp(2/x-9-(x+ln(x))/x-x)-x)/x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(31)=62\).

Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.23, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2326} \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=\frac {4 e^{\frac {-x^2-10 x+2}{x}} x^{-\frac {1}{x}-3} \left (x^2-\log (x)+3\right )}{\frac {-x^2-10 x-\log (x)+2}{x^2}+\frac {2 x+\frac {1}{x}+10}{x}} \]

[In]

Int[(E^((2 - 10*x - x^2 - Log[x])/x)*(-12 - 4*x - 4*x^2 + 4*Log[x]))/x^3,x]

[Out]

(4*E^((2 - 10*x - x^2)/x)*x^(-3 - x^(-1))*(3 + x^2 - Log[x]))/((10 + x^(-1) + 2*x)/x + (2 - 10*x - x^2 - Log[x
])/x^2)

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 {4 e^{\frac {2-10 x-x^2}{x}} x^{-3-\frac {1}{x}} \left (3+x^2-\log (x)\right )}{\frac {10+\frac {1}{x}+2 x}{x}+\frac {2-10 x-x^2-\log (x)}{x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=4 e^{-10+\frac {2}{x}-x} x^{-1-\frac {1}{x}} \]

[In]

Integrate[(E^((2 - 10*x - x^2 - Log[x])/x)*(-12 - 4*x - 4*x^2 + 4*Log[x]))/x^3,x]

[Out]

4*E^(-10 + 2/x - x)*x^(-1 - x^(-1))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-\frac {x^{2}+\ln \left (x \right )+10 x -2}{x}}}{x}\)	\(22\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{-\frac {x^{2}+\ln \left (x \right )+10 x -2}{x}}}{x}\)	\(22\)
norman	\(\frac {4 \,{\mathrm e}^{\frac {-\ln \left (x \right )-x^{2}-10 x +2}{x}}}{x}\)	\(25\)

[In]

int((4*ln(x)-4*x^2-4*x-12)*exp((-ln(x)-x^2-10*x+2)/x)/x^3,x,method=_RETURNVERBOSE)

[Out]

4/x*exp(-(x^2+ln(x)+10*x-2)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=\frac {4 \, e^{\left (-\frac {x^{2} + 10 \, x + \log \left (x\right ) - 2}{x}\right )}}{x} \]

[In]

integrate((4*log(x)-4*x^2-4*x-12)*exp((-log(x)-x^2-10*x+2)/x)/x^3,x, algorithm="fricas")

[Out]

4*e^(-(x^2 + 10*x + log(x) - 2)/x)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=\frac {4 e^{\frac {- x^{2} - 10 x - \log {\left (x \right )} + 2}{x}}}{x} \]

[In]

integrate((4*ln(x)-4*x**2-4*x-12)*exp((-ln(x)-x**2-10*x+2)/x)/x**3,x)

[Out]

4*exp((-x**2 - 10*x - log(x) + 2)/x)/x

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=\frac {4 \, e^{\left (-x - \frac {\log \left (x\right )}{x} + \frac {2}{x} - 10\right )}}{x} \]

[In]

integrate((4*log(x)-4*x^2-4*x-12)*exp((-log(x)-x^2-10*x+2)/x)/x^3,x, algorithm="maxima")

[Out]

4*e^(-x - log(x)/x + 2/x - 10)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=\frac {4 \, e^{\left (-\frac {x^{2} + 10 \, x + \log \left (x\right ) - 2}{x}\right )}}{x} \]

[In]

integrate((4*log(x)-4*x^2-4*x-12)*exp((-log(x)-x^2-10*x+2)/x)/x^3,x, algorithm="giac")

[Out]

4*e^(-(x^2 + 10*x + log(x) - 2)/x)/x

Mupad [B] (verification not implemented)

Time = 12.40 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {2-10 x-x^2-\log (x)}{x}} \left (-12-4 x-4 x^2+4 \log (x)\right )}{x^3} \, dx=\frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-10}\,{\mathrm {e}}^{2/x}}{x^{\frac {1}{x}+1}} \]

[In]

int(-(exp(-(10*x + log(x) + x^2 - 2)/x)*(4*x - 4*log(x) + 4*x^2 + 12))/x^3,x)

[Out]

(4*exp(-x)*exp(-10)*exp(2/x))/x^(1/x + 1)