3.63.3 \(\int \frac {(-12 x^2+12 x^2 \log (1-x)) \log (5-4 x+(-2+2 x) \log (1-x))+(30 x-24 x^2+(-12 x+12 x^2) \log (1-x)) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx\)

Optimal. Leaf size=30 \[ 3 x^2 \log ^2(2-x-(1-x) (-3+2 \log (1-x))) \]

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Rubi [F]  time = 0.76, 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 (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-12*x^2 + 12*x^2*Log[1 - x])*Log[5 - 4*x + (-2 + 2*x)*Log[1 - x]] + (30*x - 24*x^2 + (-12*x + 12*x^2)*Lo
g[1 - x])*Log[5 - 4*x + (-2 + 2*x)*Log[1 - x]]^2)/(5 - 4*x + (-2 + 2*x)*Log[1 - x]),x]

[Out]

-12*Defer[Int][(x^2*Log[5 - 4*x + 2*(-1 + x)*Log[1 - x]])/(5 - 4*x - 2*Log[1 - x] + 2*x*Log[1 - x]), x] + 12*D
efer[Int][(x^2*Log[1 - x]*Log[5 - 4*x + 2*(-1 + x)*Log[1 - x]])/(5 - 4*x - 2*Log[1 - x] + 2*x*Log[1 - x]), x]
+ 6*Defer[Int][x*Log[5 - 4*x + 2*(-1 + x)*Log[1 - x]]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {12 x^2 (-1+\log (1-x)) \log (5-4 x+2 (-1+x) \log (1-x))}{5-4 x-2 \log (1-x)+2 x \log (1-x)}+6 x \log ^2(5-4 x+2 (-1+x) \log (1-x))\right ) \, dx\\ &=6 \int x \log ^2(5-4 x+2 (-1+x) \log (1-x)) \, dx+12 \int \frac {x^2 (-1+\log (1-x)) \log (5-4 x+2 (-1+x) \log (1-x))}{5-4 x-2 \log (1-x)+2 x \log (1-x)} \, dx\\ &=6 \int x \log ^2(5-4 x+2 (-1+x) \log (1-x)) \, dx+12 \int \left (-\frac {x^2 \log (5-4 x+2 (-1+x) \log (1-x))}{5-4 x-2 \log (1-x)+2 x \log (1-x)}+\frac {x^2 \log (1-x) \log (5-4 x+2 (-1+x) \log (1-x))}{5-4 x-2 \log (1-x)+2 x \log (1-x)}\right ) \, dx\\ &=6 \int x \log ^2(5-4 x+2 (-1+x) \log (1-x)) \, dx-12 \int \frac {x^2 \log (5-4 x+2 (-1+x) \log (1-x))}{5-4 x-2 \log (1-x)+2 x \log (1-x)} \, dx+12 \int \frac {x^2 \log (1-x) \log (5-4 x+2 (-1+x) \log (1-x))}{5-4 x-2 \log (1-x)+2 x \log (1-x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 0.80 \begin {gather*} 3 x^2 \log ^2(5-4 x+2 (-1+x) \log (1-x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-12*x^2 + 12*x^2*Log[1 - x])*Log[5 - 4*x + (-2 + 2*x)*Log[1 - x]] + (30*x - 24*x^2 + (-12*x + 12*x
^2)*Log[1 - x])*Log[5 - 4*x + (-2 + 2*x)*Log[1 - x]]^2)/(5 - 4*x + (-2 + 2*x)*Log[1 - x]),x]

[Out]

3*x^2*Log[5 - 4*x + 2*(-1 + x)*Log[1 - x]]^2

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fricas [A]  time = 0.72, size = 24, normalized size = 0.80 \begin {gather*} 3 \, x^{2} \log \left (2 \, {\left (x - 1\right )} \log \left (-x + 1\right ) - 4 \, x + 5\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2-12*x)*log(-x+1)-24*x^2+30*x)*log((2*x-2)*log(-x+1)-4*x+5)^2+(12*x^2*log(-x+1)-12*x^2)*log(
(2*x-2)*log(-x+1)-4*x+5))/((2*x-2)*log(-x+1)-4*x+5),x, algorithm="fricas")

[Out]

3*x^2*log(2*(x - 1)*log(-x + 1) - 4*x + 5)^2

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giac [A]  time = 0.38, size = 30, normalized size = 1.00 \begin {gather*} 3 \, x^{2} \log \left (2 \, x \log \left (-x + 1\right ) - 4 \, x - 2 \, \log \left (-x + 1\right ) + 5\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2-12*x)*log(-x+1)-24*x^2+30*x)*log((2*x-2)*log(-x+1)-4*x+5)^2+(12*x^2*log(-x+1)-12*x^2)*log(
(2*x-2)*log(-x+1)-4*x+5))/((2*x-2)*log(-x+1)-4*x+5),x, algorithm="giac")

[Out]

3*x^2*log(2*x*log(-x + 1) - 4*x - 2*log(-x + 1) + 5)^2

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maple [A]  time = 0.05, size = 26, normalized size = 0.87

method	result	size
risch	\(3 x^{2} \ln \left (\left (2 x -2\right ) \ln \left (1-x \right )-4 x +5\right )^{2}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*x^2-12*x)*ln(1-x)-24*x^2+30*x)*ln((2*x-2)*ln(1-x)-4*x+5)^2+(12*x^2*ln(1-x)-12*x^2)*ln((2*x-2)*ln(1-x
)-4*x+5))/((2*x-2)*ln(1-x)-4*x+5),x,method=_RETURNVERBOSE)

[Out]

3*x^2*ln((2*x-2)*ln(1-x)-4*x+5)^2

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maxima [A]  time = 0.40, size = 29, normalized size = 0.97 \begin {gather*} 3 \, x^{2} \log \left (2 \, x {\left (\log \left (-x + 1\right ) - 2\right )} - 2 \, \log \left (-x + 1\right ) + 5\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2-12*x)*log(-x+1)-24*x^2+30*x)*log((2*x-2)*log(-x+1)-4*x+5)^2+(12*x^2*log(-x+1)-12*x^2)*log(
(2*x-2)*log(-x+1)-4*x+5))/((2*x-2)*log(-x+1)-4*x+5),x, algorithm="maxima")

[Out]

3*x^2*log(2*x*(log(-x + 1) - 2) - 2*log(-x + 1) + 5)^2

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mupad [B]  time = 4.53, size = 25, normalized size = 0.83 \begin {gather*} 3\,x^2\,{\ln \left (\ln \left (1-x\right )\,\left (2\,x-2\right )-4\,x+5\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(1 - x)*(2*x - 2) - 4*x + 5)^2*(log(1 - x)*(12*x - 12*x^2) - 30*x + 24*x^2) + log(log(1 - x)*(2*x
- 2) - 4*x + 5)*(12*x^2 - 12*x^2*log(1 - x)))/(log(1 - x)*(2*x - 2) - 4*x + 5),x)

[Out]

3*x^2*log(log(1 - x)*(2*x - 2) - 4*x + 5)^2

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sympy [A]  time = 0.44, size = 22, normalized size = 0.73 \begin {gather*} 3 x^{2} \log {\left (- 4 x + \left (2 x - 2\right ) \log {\left (1 - x \right )} + 5 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x**2-12*x)*ln(-x+1)-24*x**2+30*x)*ln((2*x-2)*ln(-x+1)-4*x+5)**2+(12*x**2*ln(-x+1)-12*x**2)*ln(
(2*x-2)*ln(-x+1)-4*x+5))/((2*x-2)*ln(-x+1)-4*x+5),x)

[Out]

3*x**2*log(-4*x + (2*x - 2)*log(1 - x) + 5)**2

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