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

Optimal. Leaf size=18 \[ \frac {1}{4+e^{\frac {1}{5} \log ^2(2-x)}} \]

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Rubi [A]  time = 1.10, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {12, 6688, 6742, 6686} \begin {gather*} \frac {1}{e^{\frac {1}{5} \log ^2(2-x)}+4} \end {gather*}

Antiderivative was successfully verified.

[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 6686

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{4+e^{\frac {1}{5} \log ^2(2-x)}} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.62, size = 15, normalized size = 0.83 \begin {gather*} \frac {1}{e^{\left (\frac {1}{5} \, \log \left (-x + 2\right )^{2}\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.32, size = 15, normalized size = 0.83 \begin {gather*} \frac {1}{e^{\left (\frac {1}{5} \, \log \left (-x + 2\right )^{2}\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 16, normalized size = 0.89

method	result	size
risch	\(\frac {1}{{\mathrm e}^{\frac {\ln \left (2-x \right )^{2}}{5}}+4}\)	\(16\)
norman	\(-\frac {{\mathrm e}^{\frac {\ln \left (2-x \right )^{2}}{5}}}{4 \left ({\mathrm e}^{\frac {\ln \left (2-x \right )^{2}}{5}}+4\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.65, size = 15, normalized size = 0.83 \begin {gather*} \frac {1}{e^{\left (\frac {1}{5} \, \log \left (-x + 2\right )^{2}\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.57, size = 15, normalized size = 0.83 \begin {gather*} \frac {1}{{\mathrm {e}}^{\frac {{\ln \left (2-x\right )}^2}{5}}+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.26, size = 12, normalized size = 0.67 \begin {gather*} \frac {1}{e^{\frac {\log {\left (2 - x \right )}^{2}}{5}} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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