3.36.98 \(\int \frac {(-e^x+x) \log (e^x-x)+(2 e^x-2 x) \log ^2(e^x-x)+(-2+2 e^x) \log (\frac {8-16 \log (e^x-x)}{\log (e^x-x)})}{(-e^x+x) \log (e^x-x)+(2 e^x-2 x) \log ^2(e^x-x)} \, dx\)

Optimal. Leaf size=21 \[ x+\log ^2\left (4 \left (-4+\frac {2}{\log \left (e^x-x\right )}\right )\right ) \]

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Rubi [F]  time = 3.44, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2 + (-2 + 2*E^x)*Log[(8 - 16*Log[E^x - x])/Log[E^x -
x]])/((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2),x]

[Out]

x + 2*Defer[Int][Log[8*(-2 + Log[E^x - x]^(-1))]/(Log[E^x - x]*(-1 + 2*Log[E^x - x])), x] - 2*Defer[Int][Log[8
*(-2 + Log[E^x - x]^(-1))]/((E^x - x)*Log[E^x - x]*(-1 + 2*Log[E^x - x])), x] + 2*Defer[Int][(x*Log[8*(-2 + Lo
g[E^x - x]^(-1))])/((E^x - x)*Log[E^x - x]*(-1 + 2*Log[E^x - x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\left (\left (-e^x+x\right ) \log \left (e^x-x\right )\right )-\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )-\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (e^x-x\right ) \left (1-2 \log \left (e^x-x\right )\right ) \log \left (e^x-x\right )} \, dx\\ &=\int \left (\frac {2 (-1+x) \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}+\frac {-\log \left (e^x-x\right )+2 \log ^2\left (e^x-x\right )+2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx\\ &=2 \int \frac {(-1+x) \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx+\int \frac {-\log \left (e^x-x\right )+2 \log ^2\left (e^x-x\right )+2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx\\ &=2 \int \left (-\frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}+\frac {x \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx+\int \left (1+\frac {2 \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )}\right ) \, dx\\ &=x+2 \int \frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx-2 \int \frac {\log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx+2 \int \frac {x \log \left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (-1+2 \log \left (e^x-x\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 19, normalized size = 0.90 \begin {gather*} x+\log ^2\left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2 + (-2 + 2*E^x)*Log[(8 - 16*Log[E^x - x])/Log
[E^x - x]])/((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2),x]

[Out]

x + Log[8*(-2 + Log[E^x - x]^(-1))]^2

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fricas [A]  time = 1.01, size = 27, normalized size = 1.29 \begin {gather*} \log \left (-\frac {8 \, {\left (2 \, \log \left (-x + e^{x}\right ) - 1\right )}}{\log \left (-x + e^{x}\right )}\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log
(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm="fricas")

[Out]

log(-8*(2*log(-x + e^x) - 1)/log(-x + e^x))^2 + x

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giac [B]  time = 0.29, size = 76, normalized size = 3.62 \begin {gather*} -\log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} + 2 \, \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) - 2 \, \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log
(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm="giac")

[Out]

-log(2*log(-x + e^x) - 1)^2 + 2*log(2*log(-x + e^x) - 1)*log(-16*log(-x + e^x) + 8) - 2*log(-16*log(-x + e^x)
+ 8)*log(log(-x + e^x)) + log(log(-x + e^x))^2 + x

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maple [C]  time = 0.24, size = 567, normalized size = 27.00

method	result	size
risch	\(\ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )^{2}-2 \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )+\ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right )^{2}+i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right ) \pi \,\mathrm {csgn}\left (\frac {i}{\ln \left ({\mathrm e}^{x}-x \right )}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{2}-i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \pi \,\mathrm {csgn}\left (i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{2}-2 i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right ) \pi \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{2}-2 i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \pi +i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \pi \,\mathrm {csgn}\left (i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (\frac {i}{\ln \left ({\mathrm e}^{x}-x \right )}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )-i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \pi \,\mathrm {csgn}\left (\frac {i}{\ln \left ({\mathrm e}^{x}-x \right )}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{2}+2 i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \pi \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{2}+i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right ) \pi \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{3}-i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right ) \pi \,\mathrm {csgn}\left (i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (\frac {i}{\ln \left ({\mathrm e}^{x}-x \right )}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )+2 i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right ) \pi -i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \pi \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{3}+i \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right ) \pi \,\mathrm {csgn}\left (i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right )}{\ln \left ({\mathrm e}^{x}-x \right )}\right )^{2}-8 \ln \left (\ln \left ({\mathrm e}^{x}-x \right )\right ) \ln \relax (2)+8 \ln \left (\ln \left ({\mathrm e}^{x}-x \right )-\frac {1}{2}\right ) \ln \relax (2)+x\)	\(567\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)-2)*ln((-16*ln(exp(x)-x)+8)/ln(exp(x)-x))+(2*exp(x)-2*x)*ln(exp(x)-x)^2+(x-exp(x))*ln(exp(x)-x))
/((2*exp(x)-2*x)*ln(exp(x)-x)^2+(x-exp(x))*ln(exp(x)-x)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp(x)-x)-1/2)^2-2*ln(ln(exp(x)-x))*ln(ln(exp(x)-x)-1/2)+ln(ln(exp(x)-x))^2+I*ln(ln(exp(x)-x)-1/2)*Pi*cs
gn(I/ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2-I*ln(ln(exp(x)-x))*Pi*csgn(I*(ln(exp(x)-x)-1/2))*
csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2-2*I*ln(ln(exp(x)-x)-1/2)*Pi*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^
2-2*I*ln(ln(exp(x)-x))*Pi+I*ln(ln(exp(x)-x))*Pi*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(
x)-x)*(ln(exp(x)-x)-1/2))-I*ln(ln(exp(x)-x))*Pi*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2
+2*I*ln(ln(exp(x)-x))*Pi*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2+I*ln(ln(exp(x)-x)-1/2)*Pi*csgn(I/ln(exp(x)-
x)*(ln(exp(x)-x)-1/2))^3-I*ln(ln(exp(x)-x)-1/2)*Pi*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x))*csgn(I/ln(e
xp(x)-x)*(ln(exp(x)-x)-1/2))+2*I*ln(ln(exp(x)-x)-1/2)*Pi-I*ln(ln(exp(x)-x))*Pi*csgn(I/ln(exp(x)-x)*(ln(exp(x)-
x)-1/2))^3+I*ln(ln(exp(x)-x)-1/2)*Pi*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2-8*ln
(ln(exp(x)-x))*ln(2)+8*ln(ln(exp(x)-x)-1/2)*ln(2)+x

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maxima [C]  time = 1.09, size = 74, normalized size = 3.52 \begin {gather*} -2 \, {\left (-i \, \pi - 3 \, \log \relax (2) + \log \left (\log \left (-x + e^{x}\right )\right )\right )} \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) + \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} - 2 \, {\left (i \, \pi + 3 \, \log \relax (2)\right )} \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log
(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.39, size = 18, normalized size = 0.86 \begin {gather*} {\ln \left (\frac {8}{\ln \left ({\mathrm {e}}^x-x\right )}-16\right )}^2+x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(-(16*log(exp(x) - x) - 8)/log(exp(x) - x))*(2*exp(x) - 2) - log(exp(x) - x)^2*(2*x - 2*exp(x)) + log
(exp(x) - x)*(x - exp(x)))/(log(exp(x) - x)^2*(2*x - 2*exp(x)) - log(exp(x) - x)*(x - exp(x))),x)

[Out]

x + log(8/log(exp(x) - x) - 16)^2

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sympy [A]  time = 1.59, size = 20, normalized size = 0.95 \begin {gather*} x + \log {\left (\frac {8 - 16 \log {\left (- x + e^{x} \right )}}{\log {\left (- x + e^{x} \right )}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)-2)*ln((-16*ln(exp(x)-x)+8)/ln(exp(x)-x))+(2*exp(x)-2*x)*ln(exp(x)-x)**2+(x-exp(x))*ln(exp
(x)-x))/((2*exp(x)-2*x)*ln(exp(x)-x)**2+(x-exp(x))*ln(exp(x)-x)),x)

[Out]

x + log((8 - 16*log(-x + exp(x)))/log(-x + exp(x)))**2

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