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

Optimal. Leaf size=35 \[ 2 x-x^2-\frac {\log (x+\log (x-(2-x) x))}{\log \left (4 e^{e^x}\right )} \]

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Rubi [F]  time = 1.29, 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 (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 30, normalized size = 0.86 \begin {gather*} 2 x-x^2-\frac {\log (x+\log ((-1+x) x))}{\log \left (4 e^{e^x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 44, normalized size = 1.26 \begin {gather*} -\frac {{\left (x^{2} - 2 \, x\right )} e^{x} + 2 \, {\left (x^{2} - 2 \, x\right )} \log \relax (2) + \log \left (x + \log \left (x^{2} - x\right )\right )}{e^{x} + 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^3+4*x^2-2*x)*log(x^2-x)-2*x^4+4*x^3-2*x^2)*log(4*exp(exp(x)))^2+(-x^2-x+1)*log(4*exp(exp(x))
)+((x^2-x)*exp(x)*log(x^2-x)+(x^3-x^2)*exp(x))*log(log(x^2-x)+x))/((x^2-x)*log(x^2-x)+x^3-x^2)/log(4*exp(exp(x
)))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.51, size = 96, normalized size = 2.74 \begin {gather*} -\frac {x^{2} e^{x} + 2 \, x^{2} \log \relax (2) - 2 \, x e^{x} - 4 \, x \log \relax (2) + 4 \, e^{x} \log \left (e^{x} + 2 \, \log \relax (2)\right ) + 8 \, \log \relax (2) \log \left (e^{x} + 2 \, \log \relax (2)\right ) - 4 \, e^{x} \log \left (-e^{x} - 2 \, \log \relax (2)\right ) - 8 \, \log \relax (2) \log \left (-e^{x} - 2 \, \log \relax (2)\right ) + \log \left (x + \log \left (x - 1\right ) + \log \relax (x)\right )}{e^{x} + 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^2*e^x + 2*x^2*log(2) - 2*x*e^x - 4*x*log(2) + 4*e^x*log(e^x + 2*log(2)) + 8*log(2)*log(e^x + 2*log(2)) - 4
*e^x*log(-e^x - 2*log(2)) - 8*log(2)*log(-e^x - 2*log(2)) + log(x + log(x - 1) + log(x)))/(e^x + 2*log(2))

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maple [C]  time = 1.34, size = 83, normalized size = 2.37

method	result	size
risch	\(-x^{2}+2 x -\frac {2 i \ln \left (\ln \left (x -1\right )+\ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x \left (x -1\right )\right ) \left (-\mathrm {csgn}\left (i x \left (x -1\right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \left (x -1\right )\right )+\mathrm {csgn}\left (i \left (x -1\right )\right )\right )}{2}+x \right )}{4 i \ln \relax (2)+2 i \ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.96, size = 43, normalized size = 1.23 \begin {gather*} -\frac {2 \, x^{2} \log \relax (2) + {\left (x^{2} - 2 \, x\right )} e^{x} - 4 \, x \log \relax (2) + \log \left (x + \log \left (x - 1\right ) + \log \relax (x)\right )}{e^{x} + 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^3+4*x^2-2*x)*log(x^2-x)-2*x^4+4*x^3-2*x^2)*log(4*exp(exp(x)))^2+(-x^2-x+1)*log(4*exp(exp(x))
)+((x^2-x)*exp(x)*log(x^2-x)+(x^3-x^2)*exp(x))*log(log(x^2-x)+x))/((x^2-x)*log(x^2-x)+x^3-x^2)/log(4*exp(exp(x
)))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.23, size = 29, normalized size = 0.83 \begin {gather*} 2\,x-\frac {\ln \left (x+\ln \left (x^2-x\right )\right )}{\ln \relax (4)+{\mathrm {e}}^x}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.94, size = 24, normalized size = 0.69 \begin {gather*} - x^{2} + 2 x - \frac {\log {\left (x + \log {\left (x^{2} - x \right )} \right )}}{e^{x} + 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**3+4*x**2-2*x)*ln(x**2-x)-2*x**4+4*x**3-2*x**2)*ln(4*exp(exp(x)))**2+(-x**2-x+1)*ln(4*exp(ex
p(x)))+((x**2-x)*exp(x)*ln(x**2-x)+(x**3-x**2)*exp(x))*ln(ln(x**2-x)+x))/((x**2-x)*ln(x**2-x)+x**3-x**2)/ln(4*
exp(exp(x)))**2,x)

[Out]

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

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