\(\int \frac {-2+e^{-4+x+x^2} (-2 x-4 x^2)}{-5 e^{-4+x+x^2} x-5 x \log (x)+(2 e^{-4+x+x^2} x+2 x \log (x)) \log (4 e^{-4+x+x^2}+4 \log (x))} \, dx\) [654]

Optimal result

Integrand size = 74, antiderivative size = 23 \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=5-\log \left (-\frac {5}{2}+\log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right ) \]

[Out]

5-ln(ln(4*ln(x)+4*exp(x^2+x-4))-5/2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6873, 12, 6816} \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=-\log \left (5-2 \log \left (4 \left (e^{x^2+x-4}+\log (x)\right )\right )\right ) \]

[In]

Int[(-2 + E^(-4 + x + x^2)*(-2*x - 4*x^2))/(-5*E^(-4 + x + x^2)*x - 5*x*Log[x] + (2*E^(-4 + x + x^2)*x + 2*x*L
og[x])*Log[4*E^(-4 + x + x^2) + 4*Log[x]]),x]

[Out]

-Log[5 - 2*Log[4*(E^(-4 + x + x^2) + Log[x])]]

Rule 12

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

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^4 \left (2-e^{-4+x+x^2} \left (-2 x-4 x^2\right )\right )}{x \left (e^{x+x^2}+e^4 \log (x)\right ) \left (5-2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right )} \, dx \\ & = e^4 \int \frac {2-e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{x \left (e^{x+x^2}+e^4 \log (x)\right ) \left (5-2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right )} \, dx \\ & = -\log \left (5-2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=-\log \left (-5+2 \log \left (4 \left (e^{-4+x+x^2}+\log (x)\right )\right )\right ) \]

[In]

Integrate[(-2 + E^(-4 + x + x^2)*(-2*x - 4*x^2))/(-5*E^(-4 + x + x^2)*x - 5*x*Log[x] + (2*E^(-4 + x + x^2)*x +
2*x*Log[x])*Log[4*E^(-4 + x + x^2) + 4*Log[x]]),x]

[Out]

-Log[-5 + 2*Log[4*(E^(-4 + x + x^2) + Log[x])]]

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\ln \left (\ln \left (4 \ln \left (x \right )+4 \,{\mathrm e}^{x^{2}+x -4}\right )-\frac {5}{2}\right )\)	\(21\)
parallelrisch	\(-\ln \left (\ln \left (4 \ln \left (x \right )+4 \,{\mathrm e}^{x^{2}+x -4}\right )-\frac {5}{2}\right )\)	\(21\)

[In]

int(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*ln(x)+2*x*exp(x^2+x-4))*ln(4*ln(x)+4*exp(x^2+x-4))-5*x*ln(x)-5*x*exp(x
^2+x-4)),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(4*ln(x)+4*exp(x^2+x-4))-5/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=-\log \left (2 \, \log \left (4 \, e^{\left (x^{2} + x - 4\right )} + 4 \, \log \left (x\right )\right ) - 5\right ) \]

[In]

integrate(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*log(x)+2*x*exp(x^2+x-4))*log(4*log(x)+4*exp(x^2+x-4))-5*x*log(x)
-5*x*exp(x^2+x-4)),x, algorithm="fricas")

[Out]

-log(2*log(4*e^(x^2 + x - 4) + 4*log(x)) - 5)

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=- \log {\left (\log {\left (4 e^{x^{2} + x - 4} + 4 \log {\left (x \right )} \right )} - \frac {5}{2} \right )} \]

[In]

integrate(((-4*x**2-2*x)*exp(x**2+x-4)-2)/((2*x*ln(x)+2*x*exp(x**2+x-4))*ln(4*ln(x)+4*exp(x**2+x-4))-5*x*ln(x)
-5*x*exp(x**2+x-4)),x)

[Out]

-log(log(4*exp(x**2 + x - 4) + 4*log(x)) - 5/2)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=-\log \left (2 \, \log \left (2\right ) + \log \left (e^{4} \log \left (x\right ) + e^{\left (x^{2} + x\right )}\right ) - \frac {13}{2}\right ) \]

[In]

integrate(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*log(x)+2*x*exp(x^2+x-4))*log(4*log(x)+4*exp(x^2+x-4))-5*x*log(x)
-5*x*exp(x^2+x-4)),x, algorithm="maxima")

[Out]

-log(2*log(2) + log(e^4*log(x) + e^(x^2 + x)) - 13/2)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=-\log \left (2 \, \log \left (4 \, e^{4} \log \left (x\right ) + 4 \, e^{\left (x^{2} + x\right )}\right ) - 13\right ) \]

[In]

integrate(((-4*x^2-2*x)*exp(x^2+x-4)-2)/((2*x*log(x)+2*x*exp(x^2+x-4))*log(4*log(x)+4*exp(x^2+x-4))-5*x*log(x)
-5*x*exp(x^2+x-4)),x, algorithm="giac")

[Out]

-log(2*log(4*e^4*log(x) + 4*e^(x^2 + x)) - 13)

Mupad [B] (verification not implemented)

Time = 8.56 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-2+e^{-4+x+x^2} \left (-2 x-4 x^2\right )}{-5 e^{-4+x+x^2} x-5 x \log (x)+\left (2 e^{-4+x+x^2} x+2 x \log (x)\right ) \log \left (4 e^{-4+x+x^2}+4 \log (x)\right )} \, dx=-\ln \left (\ln \left (4\,\ln \left (x\right )+4\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^x\right )-\frac {5}{2}\right ) \]

[In]

int((exp(x + x^2 - 4)*(2*x + 4*x^2) + 2)/(5*x*log(x) - log(4*exp(x + x^2 - 4) + 4*log(x))*(2*x*log(x) + 2*x*ex
p(x + x^2 - 4)) + 5*x*exp(x + x^2 - 4)),x)

[Out]

-log(log(4*log(x) + 4*exp(x^2)*exp(-4)*exp(x)) - 5/2)