3.90.91 \(\int \frac {2 \log (\frac {1-\log (2 x)}{e})}{4 x-4 x \log (2 x)+(-x+x \log (2 x)) \log ^2(\frac {1-\log (2 x)}{e})} \, dx\)

Optimal. Leaf size=20 \[ \log \left (4-\log ^2\left (\frac {1-\log (2 x)}{e}\right )\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 6696, 207, 6684} \begin {gather*} \log \left (4-\log ^2\left (\frac {1-\log (2 x)}{e}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*Log[(1 - Log[2*x])/E])/(4*x - 4*x*Log[2*x] + (-x + x*Log[2*x])*Log[(1 - Log[2*x])/E]^2),x]

[Out]

Log[4 - Log[(1 - Log[2*x])/E]^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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6684

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

Rule 6696

Int[(u_.)*((a_.) + (b_.)*(y_)^(n_))^(p_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Dist[q, Subst[In
t[(a + b*x^n)^p, x], x, y], x] /;  !FalseQ[q]] /; FreeQ[{a, b, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.90 \begin {gather*} \log \left (4-(-1+\log (1-\log (2 x)))^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*Log[(1 - Log[2*x])/E])/(4*x - 4*x*Log[2*x] + (-x + x*Log[2*x])*Log[(1 - Log[2*x])/E]^2),x]

[Out]

Log[4 - (-1 + Log[1 - Log[2*x]])^2]

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fricas [A]  time = 0.46, size = 29, normalized size = 1.45 \begin {gather*} \log \left (\log \left (-{\left (\log \left (2 \, x\right ) - 1\right )} e^{\left (-1\right )}\right ) + 2\right ) + \log \left (\log \left (-{\left (\log \left (2 \, x\right ) - 1\right )} e^{\left (-1\right )}\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log((1-log(2*x))/exp(1))/((x*log(2*x)-x)*log((1-log(2*x))/exp(1))^2-4*x*log(2*x)+4*x),x, algorithm
="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 25, normalized size = 1.25 \begin {gather*} \log \left (\log \left (-\log \left (2 \, x\right ) + 1\right )^{2} - 2 \, \log \left (-\log \left (2 \, x\right ) + 1\right ) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log((1-log(2*x))/exp(1))/((x*log(2*x)-x)*log((1-log(2*x))/exp(1))^2-4*x*log(2*x)+4*x),x, algorithm
="giac")

[Out]

log(log(-log(2*x) + 1)^2 - 2*log(-log(2*x) + 1) - 3)

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maple [A]  time = 0.05, size = 18, normalized size = 0.90

method	result	size
risch	\(\ln \left (\ln \left (\left (1-\ln \left (2 x \right )\right ) {\mathrm e}^{-1}\right )^{2}-4\right )\)	\(18\)
norman	\(\ln \left (\ln \left (\left (1-\ln \left (2 x \right )\right ) {\mathrm e}^{-1}\right )-2\right )+\ln \left (\ln \left (\left (1-\ln \left (2 x \right )\right ) {\mathrm e}^{-1}\right )+2\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*ln((1-ln(2*x))/exp(1))/((x*ln(2*x)-x)*ln((1-ln(2*x))/exp(1))^2-4*x*ln(2*x)+4*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln((1-ln(2*x))*exp(-1))^2-4)

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maxima [B]  time = 0.45, size = 103, normalized size = 5.15 \begin {gather*} -\frac {1}{2} \, {\left (\log \left (\log \left (-\log \relax (2) - \log \relax (x) + 1\right ) + 1\right ) - \log \left (\log \left (-\log \relax (2) - \log \relax (x) + 1\right ) - 3\right )\right )} \log \left (-{\left (\log \left (2 \, x\right ) - 1\right )} e^{\left (-1\right )}\right ) + \frac {1}{2} \, {\left (\log \left (-\log \relax (2) - \log \relax (x) + 1\right ) + 1\right )} \log \left (\log \left (-\log \relax (2) - \log \relax (x) + 1\right ) + 1\right ) - \frac {1}{2} \, {\left (\log \left (-\log \relax (2) - \log \relax (x) + 1\right ) - 3\right )} \log \left (\log \left (-\log \relax (2) - \log \relax (x) + 1\right ) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log((1-log(2*x))/exp(1))/((x*log(2*x)-x)*log((1-log(2*x))/exp(1))^2-4*x*log(2*x)+4*x),x, algorithm
="maxima")

[Out]

-1/2*(log(log(-log(2) - log(x) + 1) + 1) - log(log(-log(2) - log(x) + 1) - 3))*log(-(log(2*x) - 1)*e^(-1)) + 1
/2*(log(-log(2) - log(x) + 1) + 1)*log(log(-log(2) - log(x) + 1) + 1) - 1/2*(log(-log(2) - log(x) + 1) - 3)*lo
g(log(-log(2) - log(x) + 1) - 3)

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mupad [B]  time = 9.07, size = 16, normalized size = 0.80 \begin {gather*} \ln \left ({\ln \left (-{\mathrm {e}}^{-1}\,\left (\ln \left (2\,x\right )-1\right )\right )}^2-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(-exp(-1)*(log(2*x) - 1)))/(4*x*log(2*x) - 4*x + log(-exp(-1)*(log(2*x) - 1))^2*(x - x*log(2*x))),x
)

[Out]

log(log(-exp(-1)*(log(2*x) - 1))^2 - 4)

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sympy [A]  time = 0.40, size = 15, normalized size = 0.75 \begin {gather*} \log {\left (\log {\left (\frac {1 - \log {\left (2 x \right )}}{e} \right )}^{2} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*ln((1-ln(2*x))/exp(1))/((x*ln(2*x)-x)*ln((1-ln(2*x))/exp(1))**2-4*x*ln(2*x)+4*x),x)

[Out]

log(log((1 - log(2*x))*exp(-1))**2 - 4)

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