\(\int \frac {e^{-1/x} (-12 x^2+12 x \log (3)+(12 x+12 x^2-12 \log (3)) \log (x))}{x^2 \log ^2(x)} \, dx\) [4915]

Optimal result

Integrand size = 42, antiderivative size = 28 \[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=-81+e^3+3 \left (-2+\frac {4 e^{-1/x} (x-\log (3))}{\log (x)}\right ) \]

[Out]

12/exp(1/x)/ln(x)*(-ln(3)+x)-87+exp(3)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6873, 12, 2326} \[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=\frac {12 e^{-1/x} (x \log (x)-\log (3) \log (x))}{\log ^2(x)} \]

[In]

Int[(-12*x^2 + 12*x*Log[3] + (12*x + 12*x^2 - 12*Log[3])*Log[x])/(E^x^(-1)*x^2*Log[x]^2),x]

[Out]

(12*(x*Log[x] - Log[3]*Log[x]))/(E^x^(-1)*Log[x]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {12 e^{-1/x} \left (-x^2+x \log (3)+x \log (x)+x^2 \log (x)-\log (3) \log (x)\right )}{x^2 \log ^2(x)} \, dx \\ & = 12 \int \frac {e^{-1/x} \left (-x^2+x \log (3)+x \log (x)+x^2 \log (x)-\log (3) \log (x)\right )}{x^2 \log ^2(x)} \, dx \\ & = \frac {12 e^{-1/x} (x \log (x)-\log (3) \log (x))}{\log ^2(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=\frac {12 e^{-1/x} (x-\log (3))}{\log (x)} \]

[In]

Integrate[(-12*x^2 + 12*x*Log[3] + (12*x + 12*x^2 - 12*Log[3])*Log[x])/(E^x^(-1)*x^2*Log[x]^2),x]

[Out]

(12*(x - Log[3]))/(E^x^(-1)*Log[x])

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68

method	result	size
risch	\(-\frac {12 \left (\ln \left (3\right )-x \right ) {\mathrm e}^{-\frac {1}{x}}}{\ln \left (x \right )}\)	\(19\)
derivativedivides	\(\frac {\left (12-\frac {12 \ln \left (3\right )}{x}\right ) x \,{\mathrm e}^{-\frac {1}{x}}}{\ln \left (x \right )}\)	\(22\)
norman	\(\frac {\left (12 x^{2}-12 x \ln \left (3\right )\right ) {\mathrm e}^{-\frac {1}{x}}}{x \ln \left (x \right )}\)	\(26\)
parallelrisch	\(-\frac {\left (12 x \ln \left (3\right )-12 x^{2}\right ) {\mathrm e}^{-\frac {1}{x}}}{x \ln \left (x \right )}\)	\(27\)

[In]

int(((-12*ln(3)+12*x^2+12*x)*ln(x)+12*x*ln(3)-12*x^2)/x^2/exp(1/x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-12*(ln(3)-x)*exp(-1/x)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=\frac {12 \, {\left (x - \log \left (3\right )\right )} e^{\left (-\frac {1}{x}\right )}}{\log \left (x\right )} \]

[In]

integrate(((-12*log(3)+12*x^2+12*x)*log(x)+12*x*log(3)-12*x^2)/x^2/exp(1/x)/log(x)^2,x, algorithm="fricas")

[Out]

12*(x - log(3))*e^(-1/x)/log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=\frac {\left (12 x - 12 \log {\left (3 \right )}\right ) e^{- \frac {1}{x}}}{\log {\left (x \right )}} \]

[In]

integrate(((-12*ln(3)+12*x**2+12*x)*ln(x)+12*x*ln(3)-12*x**2)/x**2/exp(1/x)/ln(x)**2,x)

[Out]

(12*x - 12*log(3))*exp(-1/x)/log(x)

Maxima [F]

\[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=\int { -\frac {12 \, {\left (x^{2} - x \log \left (3\right ) - {\left (x^{2} + x - \log \left (3\right )\right )} \log \left (x\right )\right )} e^{\left (-\frac {1}{x}\right )}}{x^{2} \log \left (x\right )^{2}} \,d x } \]

[In]

integrate(((-12*log(3)+12*x^2+12*x)*log(x)+12*x*log(3)-12*x^2)/x^2/exp(1/x)/log(x)^2,x, algorithm="maxima")

[Out]

-12*integrate((x^2 - x*log(3) - (x^2 + x - log(3))*log(x))*e^(-1/x)/(x^2*log(x)^2), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=\frac {12 \, {\left (x e^{\left (-\frac {1}{x}\right )} - e^{\left (-\frac {1}{x}\right )} \log \left (3\right )\right )}}{\log \left (x\right )} \]

[In]

integrate(((-12*log(3)+12*x^2+12*x)*log(x)+12*x*log(3)-12*x^2)/x^2/exp(1/x)/log(x)^2,x, algorithm="giac")

[Out]

12*(x*e^(-1/x) - e^(-1/x)*log(3))/log(x)

Mupad [B] (verification not implemented)

Time = 10.83 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-1/x} \left (-12 x^2+12 x \log (3)+\left (12 x+12 x^2-12 \log (3)\right ) \log (x)\right )}{x^2 \log ^2(x)} \, dx=\frac {12\,{\mathrm {e}}^{-\frac {1}{x}}\,\left (x-\ln \left (3\right )\right )}{\ln \left (x\right )} \]

[In]

int((exp(-1/x)*(12*x*log(3) + log(x)*(12*x - 12*log(3) + 12*x^2) - 12*x^2))/(x^2*log(x)^2),x)

[Out]

(12*exp(-1/x)*(x - log(3)))/log(x)