3.4.10 \(\int \frac {(x+e^{1-x} x-2 x^2) \log (x)+(-2-2 e^{1-x}) \log ^2(x)+8 \log ^3(x)+(-x-e^{1-x} x+2 x^2+(-2 x^2-e^{1-x} x^2) \log (x)+(1+e^{1-x}-2 x) \log ^2(x)+(2 x+e^{1-x} x) \log ^3(x)) \log (-x+\log ^2(x))}{((x^2+e^{1-x} x^2-2 x^3) \log (x)-4 x^2 \log ^2(x)+(-x-e^{1-x} x+2 x^2) \log ^3(x)+4 x \log ^4(x)) \log (-x+\log ^2(x))} \, dx\)

Optimal. Leaf size=32 \[ \log \left (\left (4+\frac {-1-e^{1-x}+2 x}{\log (x)}\right ) \log \left (-x+\log ^2(x)\right )\right ) \]

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Rubi [F]  time = 14.84, 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 (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((x + E^(1 - x)*x - 2*x^2)*Log[x] + (-2 - 2*E^(1 - x))*Log[x]^2 + 8*Log[x]^3 + (-x - E^(1 - x)*x + 2*x^2 +
(-2*x^2 - E^(1 - x)*x^2)*Log[x] + (1 + E^(1 - x) - 2*x)*Log[x]^2 + (2*x + E^(1 - x)*x)*Log[x]^3)*Log[-x + Log
[x]^2])/(((x^2 + E^(1 - x)*x^2 - 2*x^3)*Log[x] - 4*x^2*Log[x]^2 + (-x - E^(1 - x)*x + 2*x^2)*Log[x]^3 + 4*x*Lo
g[x]^4)*Log[-x + Log[x]^2]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )\right )}{x \log (x) \left (e+e^x-2 e^x x-4 e^x \log (x)\right ) \left (x-\log ^2(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx\\ &=\int \left (\frac {e^x \left (4+x+2 x^2+4 x \log (x)\right )}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )}-\frac {-x \log (x)+2 \log ^2(x)+x \log \left (-x+\log ^2(x)\right )+x^2 \log (x) \log \left (-x+\log ^2(x)\right )-\log ^2(x) \log \left (-x+\log ^2(x)\right )-x \log ^3(x) \log \left (-x+\log ^2(x)\right )}{x \log (x) \left (x-\log ^2(x)\right ) \log \left (-x+\log ^2(x)\right )}\right ) \, dx\\ &=\int \frac {e^x \left (4+x+2 x^2+4 x \log (x)\right )}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )} \, dx-\int \frac {-x \log (x)+2 \log ^2(x)+x \log \left (-x+\log ^2(x)\right )+x^2 \log (x) \log \left (-x+\log ^2(x)\right )-\log ^2(x) \log \left (-x+\log ^2(x)\right )-x \log ^3(x) \log \left (-x+\log ^2(x)\right )}{x \log (x) \left (x-\log ^2(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx\\ &=\int \left (\frac {e^x}{-e-e^x+2 e^x x+4 e^x \log (x)}+\frac {4 e^x}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )}+\frac {2 e^x x}{-e-e^x+2 e^x x+4 e^x \log (x)}+\frac {4 e^x \log (x)}{-e-e^x+2 e^x x+4 e^x \log (x)}\right ) \, dx-\int \left (\frac {1+x \log (x)}{x \log (x)}+\frac {-x+2 \log (x)}{x \left (x-\log ^2(x)\right ) \log \left (-x+\log ^2(x)\right )}\right ) \, dx\\ &=2 \int \frac {e^x x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+4 \int \frac {e^x}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )} \, dx+4 \int \frac {e^x \log (x)}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+\int \frac {e^x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx-\int \frac {1+x \log (x)}{x \log (x)} \, dx-\int \frac {-x+2 \log (x)}{x \left (x-\log ^2(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx\\ &=\log \left (\log \left (-x+\log ^2(x)\right )\right )+2 \int \frac {e^x x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+4 \int \frac {e^x}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )} \, dx+4 \int \frac {e^x \log (x)}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx-\int \left (1+\frac {1}{x \log (x)}\right ) \, dx+\int \frac {e^x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx\\ &=-x+\log \left (\log \left (-x+\log ^2(x)\right )\right )+2 \int \frac {e^x x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+4 \int \frac {e^x}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )} \, dx+4 \int \frac {e^x \log (x)}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx-\int \frac {1}{x \log (x)} \, dx+\int \frac {e^x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx\\ &=-x+\log \left (\log \left (-x+\log ^2(x)\right )\right )+2 \int \frac {e^x x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+4 \int \frac {e^x}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )} \, dx+4 \int \frac {e^x \log (x)}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+\int \frac {e^x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-x-\log (\log (x))+\log \left (\log \left (-x+\log ^2(x)\right )\right )+2 \int \frac {e^x x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+4 \int \frac {e^x}{x \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )} \, dx+4 \int \frac {e^x \log (x)}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx+\int \frac {e^x}{-e-e^x+2 e^x x+4 e^x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 42, normalized size = 1.31 \begin {gather*} -x-\log (\log (x))+\log \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )+\log \left (\log \left (-x+\log ^2(x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((x + E^(1 - x)*x - 2*x^2)*Log[x] + (-2 - 2*E^(1 - x))*Log[x]^2 + 8*Log[x]^3 + (-x - E^(1 - x)*x + 2
*x^2 + (-2*x^2 - E^(1 - x)*x^2)*Log[x] + (1 + E^(1 - x) - 2*x)*Log[x]^2 + (2*x + E^(1 - x)*x)*Log[x]^3)*Log[-x
+ Log[x]^2])/(((x^2 + E^(1 - x)*x^2 - 2*x^3)*Log[x] - 4*x^2*Log[x]^2 + (-x - E^(1 - x)*x + 2*x^2)*Log[x]^3 +
4*x*Log[x]^4)*Log[-x + Log[x]^2]),x]

[Out]

-x - Log[Log[x]] + Log[-E - E^x + 2*E^x*x + 4*E^x*Log[x]] + Log[Log[-x + Log[x]^2]]

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fricas [A]  time = 1.20, size = 34, normalized size = 1.06 \begin {gather*} \log \left (2 \, x - e^{\left (-x + 1\right )} + 4 \, \log \relax (x) - 1\right ) + \log \left (\log \left (\log \relax (x)^{2} - x\right )\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(-x+1)+2*x)*log(x)^3+(exp(-x+1)+1-2*x)*log(x)^2+(-x^2*exp(-x+1)-2*x^2)*log(x)-x*exp(-x+1)+2*
x^2-x)*log(log(x)^2-x)+8*log(x)^3+(-2*exp(-x+1)-2)*log(x)^2+(x*exp(-x+1)-2*x^2+x)*log(x))/(4*x*log(x)^4+(-x*ex
p(-x+1)+2*x^2-x)*log(x)^3-4*x^2*log(x)^2+(x^2*exp(-x+1)-2*x^3+x^2)*log(x))/log(log(x)^2-x),x, algorithm="frica
s")

[Out]

log(2*x - e^(-x + 1) + 4*log(x) - 1) + log(log(log(x)^2 - x)) - log(log(x))

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giac [A]  time = 0.92, size = 34, normalized size = 1.06 \begin {gather*} \log \left (2 \, x - e^{\left (-x + 1\right )} + 4 \, \log \relax (x) - 1\right ) + \log \left (\log \left (\log \relax (x)^{2} - x\right )\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(-x+1)+2*x)*log(x)^3+(exp(-x+1)+1-2*x)*log(x)^2+(-x^2*exp(-x+1)-2*x^2)*log(x)-x*exp(-x+1)+2*
x^2-x)*log(log(x)^2-x)+8*log(x)^3+(-2*exp(-x+1)-2)*log(x)^2+(x*exp(-x+1)-2*x^2+x)*log(x))/(4*x*log(x)^4+(-x*ex
p(-x+1)+2*x^2-x)*log(x)^3-4*x^2*log(x)^2+(x^2*exp(-x+1)-2*x^3+x^2)*log(x))/log(log(x)^2-x),x, algorithm="giac"
)

[Out]

log(2*x - e^(-x + 1) + 4*log(x) - 1) + log(log(log(x)^2 - x)) - log(log(x))

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maple [A]  time = 0.05, size = 33, normalized size = 1.03

method	result	size
risch	\(\ln \left (\frac {x}{2}-\frac {{\mathrm e}^{1-x}}{4}+\ln \relax (x )-\frac {1}{4}\right )-\ln \left (\ln \relax (x )\right )+\ln \left (\ln \left (\ln \relax (x )^{2}-x \right )\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x*exp(1-x)+2*x)*ln(x)^3+(exp(1-x)+1-2*x)*ln(x)^2+(-x^2*exp(1-x)-2*x^2)*ln(x)-x*exp(1-x)+2*x^2-x)*ln(ln(
x)^2-x)+8*ln(x)^3+(-2*exp(1-x)-2)*ln(x)^2+(x*exp(1-x)-2*x^2+x)*ln(x))/(4*x*ln(x)^4+(-x*exp(1-x)+2*x^2-x)*ln(x)
^3-4*x^2*ln(x)^2+(x^2*exp(1-x)-2*x^3+x^2)*ln(x))/ln(ln(x)^2-x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.55, size = 57, normalized size = 1.78 \begin {gather*} -x + \log \left (\frac {1}{2} \, x + \log \relax (x) - \frac {1}{4}\right ) + \log \left (\frac {{\left (2 \, x + 4 \, \log \relax (x) - 1\right )} e^{x} - e}{2 \, x + 4 \, \log \relax (x) - 1}\right ) + \log \left (\log \left (\log \relax (x)^{2} - x\right )\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(-x+1)+2*x)*log(x)^3+(exp(-x+1)+1-2*x)*log(x)^2+(-x^2*exp(-x+1)-2*x^2)*log(x)-x*exp(-x+1)+2*
x^2-x)*log(log(x)^2-x)+8*log(x)^3+(-2*exp(-x+1)-2)*log(x)^2+(x*exp(-x+1)-2*x^2+x)*log(x))/(4*x*log(x)^4+(-x*ex
p(-x+1)+2*x^2-x)*log(x)^3-4*x^2*log(x)^2+(x^2*exp(-x+1)-2*x^3+x^2)*log(x))/log(log(x)^2-x),x, algorithm="maxim
a")

[Out]

-x + log(1/2*x + log(x) - 1/4) + log(((2*x + 4*log(x) - 1)*e^x - e)/(2*x + 4*log(x) - 1)) + log(log(log(x)^2 -
x)) - log(log(x))

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mupad [B]  time = 1.01, size = 80, normalized size = 2.50 \begin {gather*} \ln \left (\frac {2\,x-{\mathrm {e}}^{1-x}+4\,\ln \relax (x)-1}{x}\right )+\ln \left (\frac {2\,x+x\,{\mathrm {e}}^{1-x}+4}{x}\right )+\ln \left (\ln \left ({\ln \relax (x)}^2-x\right )\right )-\ln \left (\frac {4\,\ln \relax (x)+2\,x\,\ln \relax (x)+x\,{\mathrm {e}}^{1-x}\,\ln \relax (x)}{x}\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(x)^2 - x)*(x - log(x)^3*(2*x + x*exp(1 - x)) + x*exp(1 - x) - log(x)^2*(exp(1 - x) - 2*x + 1) +
log(x)*(x^2*exp(1 - x) + 2*x^2) - 2*x^2) - 8*log(x)^3 - log(x)*(x + x*exp(1 - x) - 2*x^2) + log(x)^2*(2*exp(1
- x) + 2))/(log(log(x)^2 - x)*(4*x*log(x)^4 - log(x)^3*(x + x*exp(1 - x) - 2*x^2) - 4*x^2*log(x)^2 + log(x)*(x
^2*exp(1 - x) + x^2 - 2*x^3))),x)

[Out]

log((2*x - exp(1 - x) + 4*log(x) - 1)/x) + log((2*x + x*exp(1 - x) + 4)/x) + log(log(log(x)^2 - x)) - log((4*l
og(x) + 2*x*log(x) + x*exp(1 - x)*log(x))/x) + log(x)

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sympy [A]  time = 0.78, size = 31, normalized size = 0.97 \begin {gather*} \log {\left (- 2 x + e^{1 - x} - 4 \log {\relax (x )} + 1 \right )} - \log {\left (\log {\relax (x )} \right )} + \log {\left (\log {\left (- x + \log {\relax (x )}^{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(-x+1)+2*x)*ln(x)**3+(exp(-x+1)+1-2*x)*ln(x)**2+(-x**2*exp(-x+1)-2*x**2)*ln(x)-x*exp(-x+1)+2
*x**2-x)*ln(ln(x)**2-x)+8*ln(x)**3+(-2*exp(-x+1)-2)*ln(x)**2+(x*exp(-x+1)-2*x**2+x)*ln(x))/(4*x*ln(x)**4+(-x*e
xp(-x+1)+2*x**2-x)*ln(x)**3-4*x**2*ln(x)**2+(x**2*exp(-x+1)-2*x**3+x**2)*ln(x))/ln(ln(x)**2-x),x)

[Out]

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

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