3.39.37 \(\int \frac {1+(-1+(-2+e) e^{-2-2 x+e x}) \log ^2(\log (3))}{x+(e^{-2-2 x+e x}-x) \log ^2(\log (3))} \, dx\)

Optimal. Leaf size=25 \[ \log \left (-x+\left (-e^{-2-2 x+e x}+x\right ) \log ^2(\log (3))\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 46, normalized size of antiderivative = 1.84, number of steps used = 1, number of rules used = 1, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6684} \begin {gather*} \log \left (-e^{-2 x-2} \left (e^{2 x+2} x-e^{2 x+2} x \log ^2(\log (3))+e^{e x} \log ^2(\log (3))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + (-1 + (-2 + E)*E^(-2 - 2*x + E*x))*Log[Log[3]]^2)/(x + (E^(-2 - 2*x + E*x) - x)*Log[Log[3]]^2),x]

[Out]

Log[-(E^(-2 - 2*x)*(E^(2 + 2*x)*x + E^(E*x)*Log[Log[3]]^2 - E^(2 + 2*x)*x*Log[Log[3]]^2))]

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} &=\log \left (-e^{-2-2 x} \left (e^{2+2 x} x+e^{e x} \log ^2(\log (3))-e^{2+2 x} x \log ^2(\log (3))\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 22, normalized size = 0.88 \begin {gather*} \log \left (x+\left (e^{-2+(-2+e) x}-x\right ) \log ^2(\log (3))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + (-1 + (-2 + E)*E^(-2 - 2*x + E*x))*Log[Log[3]]^2)/(x + (E^(-2 - 2*x + E*x) - x)*Log[Log[3]]^2),
x]

[Out]

Log[x + (E^(-2 + (-2 + E)*x) - x)*Log[Log[3]]^2]

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fricas [A]  time = 0.80, size = 24, normalized size = 0.96 \begin {gather*} \log \left (-{\left (x - e^{\left (x e - 2 \, x - 2\right )}\right )} \log \left (\log \relax (3)\right )^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(1)-2)*exp(x*exp(1)-2*x-2)-1)*log(log(3))^2+1)/((exp(x*exp(1)-2*x-2)-x)*log(log(3))^2+x),x, al
gorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 32, normalized size = 1.28 \begin {gather*} \log \left (x e^{2} \log \left (\log \relax (3)\right )^{2} - e^{\left (x e - 2 \, x\right )} \log \left (\log \relax (3)\right )^{2} - x e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(1)-2)*exp(x*exp(1)-2*x-2)-1)*log(log(3))^2+1)/((exp(x*exp(1)-2*x-2)-x)*log(log(3))^2+x),x, al
gorithm="giac")

[Out]

log(x*e^2*log(log(3))^2 - e^(x*e - 2*x)*log(log(3))^2 - x*e^2)

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maple [A]  time = 0.05, size = 24, normalized size = 0.96

method	result	size
derivativedivides	\(\ln \left (\left ({\mathrm e}^{x \,{\mathrm e}-2 x -2}-x \right ) \ln \left (\ln \relax (3)\right )^{2}+x \right )\)	\(24\)
default	\(\ln \left (\left ({\mathrm e}^{x \,{\mathrm e}-2 x -2}-x \right ) \ln \left (\ln \relax (3)\right )^{2}+x \right )\)	\(24\)
norman	\(\ln \left (\ln \left (\ln \relax (3)\right )^{2} x -\ln \left (\ln \relax (3)\right )^{2} {\mathrm e}^{x \,{\mathrm e}-2 x -2}-x \right )\)	\(30\)
risch	\(2+\ln \left ({\mathrm e}^{x \,{\mathrm e}-2 x -2}-\frac {x \left (\ln \left (\ln \relax (3)\right )^{2}-1\right )}{\ln \left (\ln \relax (3)\right )^{2}}\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(1)-2)*exp(x*exp(1)-2*x-2)-1)*ln(ln(3))^2+1)/((exp(x*exp(1)-2*x-2)-x)*ln(ln(3))^2+x),x,method=_RETUR
NVERBOSE)

[Out]

ln((exp(x*exp(1)-2*x-2)-x)*ln(ln(3))^2+x)

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maxima [A]  time = 0.55, size = 40, normalized size = 1.60 \begin {gather*} -2 \, x + \log \left (-\frac {{\left (\log \left (\log \relax (3)\right )^{2} - 1\right )} x e^{\left (2 \, x + 2\right )} - e^{\left (x e\right )} \log \left (\log \relax (3)\right )^{2}}{\log \left (\log \relax (3)\right )^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(1)-2)*exp(x*exp(1)-2*x-2)-1)*log(log(3))^2+1)/((exp(x*exp(1)-2*x-2)-x)*log(log(3))^2+x),x, al
gorithm="maxima")

[Out]

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

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mupad [B]  time = 2.48, size = 28, normalized size = 1.12 \begin {gather*} \ln \left (x-x\,{\ln \left (\ln \relax (3)\right )}^2+{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{x\,\mathrm {e}}\,{\ln \left (\ln \relax (3)\right )}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(3))^2*(exp(x*exp(1) - 2*x - 2)*(exp(1) - 2) - 1) + 1)/(x - log(log(3))^2*(x - exp(x*exp(1) - 2*x
- 2))),x)

[Out]

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

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sympy [A]  time = 0.22, size = 29, normalized size = 1.16 \begin {gather*} \log {\left (\frac {- x \log {\left (\log {\relax (3 )} \right )}^{2} + x}{\log {\left (\log {\relax (3 )} \right )}^{2}} + e^{- 2 x + e x - 2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(1)-2)*exp(x*exp(1)-2*x-2)-1)*ln(ln(3))**2+1)/((exp(x*exp(1)-2*x-2)-x)*ln(ln(3))**2+x),x)

[Out]

log((-x*log(log(3))**2 + x)/log(log(3))**2 + exp(-2*x + E*x - 2))

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