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

Optimal. Leaf size=31 \[ 1-\frac {x}{\log \left (e^{-2 x} \log \left (\frac {e^2}{x^2 \left (-x+x^2\right )}\right )\right )} \]

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Rubi [F]  time = 4.51, 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 {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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 24, normalized size = 0.77 \begin {gather*} -\frac {x}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[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)])

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fricas [A]  time = 0.77, size = 26, normalized size = 0.84 \begin {gather*} -\frac {x}{\log \left (e^{\left (-2 \, x\right )} \log \left (\frac {e^{2}}{x^{4} - x^{3}}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*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)/(x-1)/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)))

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giac [A]  time = 0.48, size = 25, normalized size = 0.81 \begin {gather*} \frac {x}{2 \, x - \log \left (-\log \left (x^{4} - x^{3}\right ) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*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)/(x-1)/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))

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maple [C]  time = 0.86, size = 3055, normalized size = 98.55

method	result	size
risch	\(\text {Expression too large to display}\)	\(3055\)

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-1)/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]

-2*I*x/(Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3
)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I
/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*
csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I))*csgn(I*exp(-2*x))*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*
csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*
csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1
))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I)*exp(-2*x))+Pi*csgn(I
*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*cs
gn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))+Pi*cs
gn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I
*ln(x)+2*I*ln(x-1)-4*I))*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x
^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I
/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^
3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I)*exp(-2*x))^2-Pi*csgn(I*exp(-2*x))*csgn(I*(Pi*csgn(I*x)^
2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*
csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(
I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x
-1)-4*I)*exp(-2*x))^2-Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x
^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I
/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^
3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I)*exp(-2*x))^3+Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi
*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*c
sgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*
csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I)*exp(-2*x)
)*csgn(exp(-2*x)*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)
-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/
x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*c
sgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I))^2+Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+
Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-
Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/
(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I)*exp(-2*x))*csgn(exp(-2*x)*(Pi*csgn
(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)
^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)
*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*
I*ln(x-1)-4*I))-Pi*csgn(exp(-2*x)*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I
*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn
(I/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/
x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I))^3-Pi*csgn(exp(-2*x)*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi
*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*c
sgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))+Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2+Pi*
csgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I/x^3/(x-1))^3+Pi*csgn(I*x^3)^3+6*I*ln(x)+2*I*ln(x-1)-4*I))^2+Pi-2*I
*ln(2)+2*I*ln(4*I-Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)+Pi*csgn(I/x^3)*csgn(I/(x-1))*csgn(I/x^3/(x-1))-Pi*csgn(
I*x^2)^3-Pi*csgn(I*x^3)^3+Pi*csgn(I/x^3/(x-1))^3-2*I*ln(x-1)-6*I*ln(x)-Pi*csgn(I/x^3)*csgn(I/x^3/(x-1))^2-Pi*c
sgn(I/(x-1))*csgn(I/x^3/(x-1))^2-Pi*csgn(I*x)^2*csgn(I*x^2)+2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x
^3)^2+Pi*csgn(I*x^2)*csgn(I*x^3)^2)-2*I*ln(exp(2*x)))

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maxima [A]  time = 0.41, size = 23, normalized size = 0.74 \begin {gather*} \frac {x}{2 \, x - \log \left (-\log \left (x - 1\right ) - 3 \, \log \relax (x) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*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)/(x-1)/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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.77, size = 20, normalized size = 0.65 \begin {gather*} - \frac {x}{\log {\left (e^{- 2 x} \log {\left (\frac {e^{2}}{x^{4} - x^{3}} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*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)/(x-1)/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)))

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