3.28.4 \(\int \frac {2 e^{\frac {1}{5} (-10-2 \log (x))}}{(-20 x+5 e^{\frac {1}{5} (-10-2 \log (x))} x) \log (4-e^{\frac {1}{5} (-10-2 \log (x))})} \, dx\)

Optimal. Leaf size=23 \[ \log (3)-\log \left (\log \left (4-e^{\frac {2}{5} (-5-\log (x))}\right )\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 16, normalized size of antiderivative = 0.70, number of steps used = 4, number of rules used = 3, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 2274, 6684} \begin {gather*} -\log \left (\log \left (4-\frac {1}{e^2 x^{2/5}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^((-10 - 2*Log[x])/5))/((-20*x + 5*E^((-10 - 2*Log[x])/5)*x)*Log[4 - E^((-10 - 2*Log[x])/5)]),x]

[Out]

-Log[Log[4 - 1/(E^2*x^(2/5))]]

Rule 12

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

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 6684

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int \frac {e^{\frac {1}{5} (-10-2 \log (x))}}{\left (-20 x+5 e^{\frac {1}{5} (-10-2 \log (x))} x\right ) \log \left (4-e^{\frac {1}{5} (-10-2 \log (x))}\right )} \, dx\\ &=2 \int \frac {1}{e^2 x^{2/5} \left (-20 x+5 e^{\frac {1}{5} (-10-2 \log (x))} x\right ) \log \left (4-e^{\frac {1}{5} (-10-2 \log (x))}\right )} \, dx\\ &=\frac {2 \int \frac {1}{x^{2/5} \left (-20 x+5 e^{\frac {1}{5} (-10-2 \log (x))} x\right ) \log \left (4-e^{\frac {1}{5} (-10-2 \log (x))}\right )} \, dx}{e^2}\\ &=-\log \left (\log \left (4-\frac {1}{e^2 x^{2/5}}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 16, normalized size = 0.70 \begin {gather*} -\log \left (\log \left (4-\frac {1}{e^2 x^{2/5}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^((-10 - 2*Log[x])/5))/((-20*x + 5*E^((-10 - 2*Log[x])/5)*x)*Log[4 - E^((-10 - 2*Log[x])/5)]),x]

[Out]

-Log[Log[4 - 1/(E^2*x^(2/5))]]

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fricas [A]  time = 0.55, size = 21, normalized size = 0.91 \begin {gather*} -\log \left (\log \left (\frac {{\left (4 \, x e^{2} - x^{\frac {3}{5}}\right )} e^{\left (-2\right )}}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(-2/5*log(x)-2)/(5*x*exp(-2/5*log(x)-2)-20*x)/log(-exp(-2/5*log(x)-2)+4),x, algorithm="fricas")

[Out]

-log(log((4*x*e^2 - x^(3/5))*e^(-2)/x))

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giac [A]  time = 0.26, size = 15, normalized size = 0.65 \begin {gather*} -\log \left (\log \left (-e^{\left (-\frac {2}{5} \, \log \relax (x) - 2\right )} + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(-2/5*log(x)-2)/(5*x*exp(-2/5*log(x)-2)-20*x)/log(-exp(-2/5*log(x)-2)+4),x, algorithm="giac")

[Out]

-log(log(-e^(-2/5*log(x) - 2) + 4))

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {2 \,{\mathrm e}^{-\frac {2 \ln \relax (x )}{5}-2}}{\left (5 x \,{\mathrm e}^{-\frac {2 \ln \relax (x )}{5}-2}-20 x \right ) \ln \left (-{\mathrm e}^{-\frac {2 \ln \relax (x )}{5}-2}+4\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(-2/5*ln(x)-2)/(5*x*exp(-2/5*ln(x)-2)-20*x)/ln(-exp(-2/5*ln(x)-2)+4),x)

[Out]

int(2*exp(-2/5*ln(x)-2)/(5*x*exp(-2/5*ln(x)-2)-20*x)/ln(-exp(-2/5*ln(x)-2)+4),x)

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maxima [A]  time = 0.57, size = 29, normalized size = 1.26 \begin {gather*} -\log \left (\log \left (2 \, x^{\frac {1}{5}} e + 1\right ) + \log \left (2 \, x^{\frac {1}{5}} e - 1\right ) - \frac {2}{5} \, \log \relax (x) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(-2/5*log(x)-2)/(5*x*exp(-2/5*log(x)-2)-20*x)/log(-exp(-2/5*log(x)-2)+4),x, algorithm="maxima")

[Out]

-log(log(2*x^(1/5)*e + 1) + log(2*x^(1/5)*e - 1) - 2/5*log(x) - 2)

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mupad [B]  time = 2.73, size = 13, normalized size = 0.57 \begin {gather*} -\ln \left (\ln \left (4-\frac {{\mathrm {e}}^{-2}}{x^{2/5}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(- (2*log(x))/5 - 2))/(log(4 - exp(- (2*log(x))/5 - 2))*(20*x - 5*x*exp(- (2*log(x))/5 - 2))),x)

[Out]

-log(log(4 - exp(-2)/x^(2/5)))

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sympy [A]  time = 143.00, size = 15, normalized size = 0.65 \begin {gather*} - \log {\left (\log {\left (4 - \frac {1}{x^{\frac {2}{5}} e^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(-2/5*ln(x)-2)/(5*x*exp(-2/5*ln(x)-2)-20*x)/ln(-exp(-2/5*ln(x)-2)+4),x)

[Out]

-log(log(4 - exp(-2)/x**(2/5)))

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