\(\int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx\) [2807]

Optimal result

Integrand size = 69, antiderivative size = 18 \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{4+e^{\frac {1}{5} \log ^2(2-x)}} \]

[Out]

1/(exp(1/5*ln(2-x)^2)+4)

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {12, 6820, 6874, 6818} \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{e^{\frac {1}{5} \log ^2(2-x)}+4} \]

[In]

Int[(-2*E^(Log[2 - x]^2/5)*Log[2 - x])/(-160 + 80*x + E^((2*Log[2 - x]^2)/5)*(-10 + 5*x) + E^(Log[2 - x]^2/5)*
(-80 + 40*x)),x]

[Out]

(4 + E^(Log[2 - x]^2/5))^(-1)

Rule 12

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

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = -\left (2 \int \frac {e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx\right ) \\ & = -\left (2 \int \frac {e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{5 \left (4+e^{\frac {1}{5} \log ^2(2-x)}\right )^2 (-2+x)} \, dx\right ) \\ & = -\left (\frac {2}{5} \int \frac {e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{\left (4+e^{\frac {1}{5} \log ^2(2-x)}\right )^2 (-2+x)} \, dx\right ) \\ & = \frac {1}{4+e^{\frac {1}{5} \log ^2(2-x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{4+e^{\frac {1}{5} \log ^2(2-x)}} \]

[In]

Integrate[(-2*E^(Log[2 - x]^2/5)*Log[2 - x])/(-160 + 80*x + E^((2*Log[2 - x]^2)/5)*(-10 + 5*x) + E^(Log[2 - x]
^2/5)*(-80 + 40*x)),x]

[Out]

(4 + E^(Log[2 - x]^2/5))^(-1)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
norman	\(\frac {1}{{\mathrm e}^{\frac {\ln \left (2-x \right )^{2}}{5}}+4}\)	\(16\)
risch	\(\frac {1}{{\mathrm e}^{\frac {\ln \left (2-x \right )^{2}}{5}}+4}\)	\(16\)
parallelrisch	\(\frac {1}{{\mathrm e}^{\frac {\ln \left (2-x \right )^{2}}{5}}+4}\)	\(16\)

[In]

int(-2*ln(2-x)*exp(1/5*ln(2-x)^2)/((5*x-10)*exp(1/5*ln(2-x)^2)^2+(40*x-80)*exp(1/5*ln(2-x)^2)+80*x-160),x,meth
od=_RETURNVERBOSE)

[Out]

1/(exp(1/5*ln(2-x)^2)+4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{e^{\left (\frac {1}{5} \, \log \left (-x + 2\right )^{2}\right )} + 4} \]

[In]

integrate(-2*log(2-x)*exp(1/5*log(2-x)^2)/((5*x-10)*exp(1/5*log(2-x)^2)^2+(40*x-80)*exp(1/5*log(2-x)^2)+80*x-1
60),x, algorithm="fricas")

[Out]

1/(e^(1/5*log(-x + 2)^2) + 4)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{e^{\frac {\log {\left (2 - x \right )}^{2}}{5}} + 4} \]

[In]

integrate(-2*ln(2-x)*exp(1/5*ln(2-x)**2)/((5*x-10)*exp(1/5*ln(2-x)**2)**2+(40*x-80)*exp(1/5*ln(2-x)**2)+80*x-1
60),x)

[Out]

1/(exp(log(2 - x)**2/5) + 4)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{e^{\left (\frac {1}{5} \, \log \left (-x + 2\right )^{2}\right )} + 4} \]

[In]

integrate(-2*log(2-x)*exp(1/5*log(2-x)^2)/((5*x-10)*exp(1/5*log(2-x)^2)^2+(40*x-80)*exp(1/5*log(2-x)^2)+80*x-1
60),x, algorithm="maxima")

[Out]

1/(e^(1/5*log(-x + 2)^2) + 4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{e^{\left (\frac {1}{5} \, \log \left (-x + 2\right )^{2}\right )} + 4} \]

[In]

integrate(-2*log(2-x)*exp(1/5*log(2-x)^2)/((5*x-10)*exp(1/5*log(2-x)^2)^2+(40*x-80)*exp(1/5*log(2-x)^2)+80*x-1
60),x, algorithm="giac")

[Out]

1/(e^(1/5*log(-x + 2)^2) + 4)

Mupad [B] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {2 e^{\frac {1}{5} \log ^2(2-x)} \log (2-x)}{-160+80 x+e^{\frac {2}{5} \log ^2(2-x)} (-10+5 x)+e^{\frac {1}{5} \log ^2(2-x)} (-80+40 x)} \, dx=\frac {1}{{\mathrm {e}}^{\frac {{\ln \left (2-x\right )}^2}{5}}+4} \]

[In]

int(-(2*exp(log(2 - x)^2/5)*log(2 - x))/(80*x + exp((2*log(2 - x)^2)/5)*(5*x - 10) + exp(log(2 - x)^2/5)*(40*x
- 80) - 160),x)

[Out]

1/(exp(log(2 - x)^2/5) + 4)