\(\int \frac {3-4 x+(2 x-2 x^2) \log (\frac {e^2}{-x^3+x^4})+(1-x) \log (\frac {e^2}{-x^3+x^4}) \log (e^{-2 x} \log (\frac {e^2}{-x^3+x^4}))}{(-1+x) \log (\frac {e^2}{-x^3+x^4}) \log ^2(e^{-2 x} \log (\frac {e^2}{-x^3+x^4}))} \, dx\) [7810]

Optimal result

Integrand size = 125, antiderivative size = 31 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=1-\frac {x}{\log \left (e^{-2 x} \log \left (\frac {e^2}{x^2 \left (-x+x^2\right )}\right )\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=\int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-3+4 x-\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )-(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(1-x) \log \left (\frac {e^2}{(-1+x) x^3}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{(-1+x) x^3}\right )\right )} \, dx \\ & = \int \left (\frac {3-4 x^2+2 x \log \left (\frac {1}{(-1+x) x^3}\right )-2 x^2 \log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}-\frac {1}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}\right ) \, dx \\ & = \int \frac {3-4 x^2+2 x \log \left (\frac {1}{(-1+x) x^3}\right )-2 x^2 \log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-\int \frac {1}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx \\ & = \int \left (\frac {3}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}-\frac {4 x^2}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}+\frac {2 x \log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}-\frac {2 x^2 \log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}\right ) \, dx-\int \frac {1}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx \\ & = 2 \int \frac {x \log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-2 \int \frac {x^2 \log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx+3 \int \frac {1}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-4 \int \frac {x^2}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-\int \frac {1}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx \\ & = 2 \int \left (\frac {\log \left (\frac {1}{(-1+x) x^3}\right )}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}+\frac {\log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}\right ) \, dx-2 \int \left (\frac {\log \left (\frac {1}{(-1+x) x^3}\right )}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}+\frac {\log \left (\frac {1}{(-1+x) x^3}\right )}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}+\frac {x \log \left (\frac {1}{(-1+x) x^3}\right )}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}\right ) \, dx+3 \int \frac {1}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-4 \int \left (\frac {1}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}+\frac {1}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}+\frac {x}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )}\right ) \, dx-\int \frac {1}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx \\ & = -\left (2 \int \frac {x \log \left (\frac {1}{(-1+x) x^3}\right )}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx\right )+3 \int \frac {1}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-4 \int \frac {1}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-4 \int \frac {1}{(-1+x) \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-4 \int \frac {x}{\left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right ) \log ^2\left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx-\int \frac {1}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=-\frac {x}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(-\frac {x}{\ln \left (\ln \left (\frac {{\mathrm e}^{2}}{x^{3} \left (-1+x \right )}\right ) {\mathrm e}^{-2 x}\right )}\)	\(28\)
risch	\(\text {Expression too large to display}\)	\(3079\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=-\frac {x}{\log \left (e^{\left (-2 \, x\right )} \log \left (\frac {e^{2}}{x^{4} - x^{3}}\right )\right )} \]

[In]

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

[Out]

-x/log(e^(-2*x)*log(e^2/(x^4 - x^3)))

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=- \frac {x}{\log {\left (e^{- 2 x} \log {\left (\frac {e^{2}}{x^{4} - x^{3}} \right )} \right )}} \]

[In]

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

[Out]

-x/log(exp(-2*x)*log(exp(2)/(x**4 - x**3)))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=\frac {x}{2 \, x - \log \left (-\log \left (x - 1\right ) - 3 \, \log \left (x\right ) + 2\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=\frac {x}{2 \, x - \log \left (-\log \left (x^{4} - x^{3}\right ) + 2\right )} \]

[In]

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

[Out]

x/(2*x - log(-log(x^4 - x^3) + 2))

Mupad [F(-1)]

Timed out. \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=-\int \frac {4\,x-\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\,\left (2\,x-2\,x^2\right )+\ln \left ({\mathrm {e}}^{-2\,x}\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\right )\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\,\left (x-1\right )-3}{{\ln \left ({\mathrm {e}}^{-2\,x}\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\right )}^2\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\,\left (x-1\right )} \,d x \]

[In]

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

[Out]

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