3.52.8 \(\int \frac {(6-2 x-4 x^2) \log ^2(1-x)+(x^2+(x-x^2) \log (1-x)-4 x \log ^2(1-x)) \log (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x})+(-2+2 x) \log ^2(1-x) \log ^2(-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x})}{(-2+2 x) \log ^2(1-x)} \, dx\)

Optimal. Leaf size=35 \[ x+x \left (-4-x+\log ^2\left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )\right ) \]

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Rubi [F]  time = 3.09, 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 (6-2 x-4 x^2\right ) \log ^2(1-x)+\left (x^2+\left (x-x^2\right ) \log (1-x)-4 x \log ^2(1-x)\right ) \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )+(-2+2 x) \log ^2(1-x) \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-2+2 x) \log ^2(1-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((6 - 2*x - 4*x^2)*Log[1 - x]^2 + (x^2 + (x - x^2)*Log[1 - x] - 4*x*Log[1 - x]^2)*Log[-(1/(E^(x/(4*Log[1 -
x]))*(-1 + x)))] + (-2 + 2*x)*Log[1 - x]^2*Log[-(1/(E^(x/(4*Log[1 - x]))*(-1 + x)))]^2)/((-2 + 2*x)*Log[1 - x
]^2),x]

[Out]

-5*x - x^2 + ExpIntegralEi[2*Log[1 - x]]/2 - 2*x*Log[1/(E^(x/(4*Log[1 - x]))*(1 - x))] - 1/(2*Log[1 - x]) + (1
- x)/Log[1 - x] - (1 - x)^2/(2*Log[1 - x]) - 2*Log[1 - x] - LogIntegral[1 - x]/2 - 2*Defer[Int][Log[-(1/(E^(x
/(4*Log[1 - x]))*(-1 + x)))]/(-1 + x), x] + Defer[Int][Log[-(1/(E^(x/(4*Log[1 - x]))*(-1 + x)))]/Log[1 - x]^2,
x]/2 + Defer[Int][Log[-(1/(E^(x/(4*Log[1 - x]))*(-1 + x)))]/((-1 + x)*Log[1 - x]^2), x]/2 + Defer[Int][(x*Log
[-(1/(E^(x/(4*Log[1 - x]))*(-1 + x)))])/Log[1 - x]^2, x]/2 - Defer[Int][(x*Log[-(1/(E^(x/(4*Log[1 - x]))*(-1 +
x)))])/Log[1 - x], x]/2 + Defer[Int][Log[-(1/(E^(x/(4*Log[1 - x]))*(-1 + x)))]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-3-2 x-\frac {x \left (-x+(-1+x) \log (1-x)+4 \log ^2(1-x)\right ) \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{2 (-1+x) \log ^2(1-x)}+\log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )\right ) \, dx\\ &=-3 x-x^2-\frac {1}{2} \int \frac {x \left (-x+(-1+x) \log (1-x)+4 \log ^2(1-x)\right ) \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-\frac {1}{2} \int \left (\frac {\left (-x-\log (1-x)+x \log (1-x)+4 \log ^2(1-x)\right ) \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)}+\frac {\left (-x-\log (1-x)+x \log (1-x)+4 \log ^2(1-x)\right ) \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)}\right ) \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-\frac {1}{2} \int \frac {\left (-x-\log (1-x)+x \log (1-x)+4 \log ^2(1-x)\right ) \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {\left (-x-\log (1-x)+x \log (1-x)+4 \log ^2(1-x)\right ) \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-\frac {1}{2} \int \left (4 \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )-\frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)}-\frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)}+\frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)}\right ) \, dx-\frac {1}{2} \int \left (\frac {4 \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x}-\frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)}-\frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log (1-x)}+\frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log (1-x)}\right ) \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log (1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log (1-x)} \, dx-2 \int \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )-\frac {1}{2} \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right ) \text {li}(1-x)+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log (1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx+\frac {1}{2} \int \left (\frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)}+\frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)}\right ) \, dx-\frac {1}{2} \int \left (\frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)}+\frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log (1-x)}\right ) \, dx-\frac {1}{2} \int -\frac {1}{4} \left (-\frac {4}{-1+x}+\frac {x}{(-1+x) \log ^2(1-x)}-\frac {1}{\log (1-x)}\right ) \text {li}(1-x) \, dx+2 \int \frac {1}{4} x \left (-\frac {4}{-1+x}+\frac {x}{(-1+x) \log ^2(1-x)}-\frac {1}{\log (1-x)}\right ) \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )-\frac {1}{2} \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right ) \text {li}(1-x)+\frac {1}{8} \int \left (-\frac {4}{-1+x}+\frac {x}{(-1+x) \log ^2(1-x)}-\frac {1}{\log (1-x)}\right ) \text {li}(1-x) \, dx+\frac {1}{2} \int x \left (-\frac {4}{-1+x}+\frac {x}{(-1+x) \log ^2(1-x)}-\frac {1}{\log (1-x)}\right ) \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )+\frac {1}{8} \int \left (-\frac {4 \text {li}(1-x)}{-1+x}+\frac {x \text {li}(1-x)}{(-1+x) \log ^2(1-x)}-\frac {\text {li}(1-x)}{\log (1-x)}\right ) \, dx+\frac {1}{2} \int \left (-\frac {4 x}{-1+x}+\frac {x^2}{(-1+x) \log ^2(1-x)}-\frac {x}{\log (1-x)}\right ) \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx+\frac {1}{2} \int -\frac {1}{4} \left (-\frac {4}{-1+x}+\frac {x}{(-1+x) \log ^2(1-x)}-\frac {1}{\log (1-x)}\right ) \text {li}(1-x) \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )-\frac {1}{8} \int \left (-\frac {4}{-1+x}+\frac {x}{(-1+x) \log ^2(1-x)}-\frac {1}{\log (1-x)}\right ) \text {li}(1-x) \, dx+\frac {1}{8} \int \frac {x \text {li}(1-x)}{(-1+x) \log ^2(1-x)} \, dx-\frac {1}{8} \int \frac {\text {li}(1-x)}{\log (1-x)} \, dx+\frac {1}{2} \int \frac {x^2}{(-1+x) \log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x}{\log (1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx-\frac {1}{2} \int \frac {\text {li}(1-x)}{-1+x} \, dx-2 \int \frac {x}{-1+x} \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-3 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )+\frac {\text {li}(1-x)^2}{16}+\frac {1}{8} \int \left (\frac {\text {li}(1-x)}{\log ^2(1-x)}+\frac {\text {li}(1-x)}{(-1+x) \log ^2(1-x)}\right ) \, dx-\frac {1}{8} \int \left (-\frac {4 \text {li}(1-x)}{-1+x}+\frac {x \text {li}(1-x)}{(-1+x) \log ^2(1-x)}-\frac {\text {li}(1-x)}{\log (1-x)}\right ) \, dx-\frac {1}{2} \int \left (\frac {1}{\log (1-x)}-\frac {1-x}{\log (1-x)}\right ) \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx-\frac {1}{2} \int \frac {\text {li}(1-x)}{-1+x} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {(1-x)^2}{x \log ^2(x)} \, dx,x,1-x\right )-2 \int \left (1+\frac {1}{-1+x}\right ) \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-5 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )-2 \log (1-x)+\frac {\text {li}(1-x)^2}{16}+\frac {1}{8} \int \frac {\text {li}(1-x)}{\log ^2(1-x)} \, dx+\frac {1}{8} \int \frac {\text {li}(1-x)}{(-1+x) \log ^2(1-x)} \, dx-\frac {1}{8} \int \frac {x \text {li}(1-x)}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{8} \int \frac {\text {li}(1-x)}{\log (1-x)} \, dx-\frac {1}{2} \int \frac {1}{\log (1-x)} \, dx+\frac {1}{2} \int \frac {1-x}{\log (1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {2}{\log ^2(x)}+\frac {1}{x \log ^2(x)}+\frac {x}{\log ^2(x)}\right ) \, dx,x,1-x\right )-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ &=-5 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )-2 \log (1-x)-\frac {1}{8} \int \left (\frac {\text {li}(1-x)}{\log ^2(1-x)}+\frac {\text {li}(1-x)}{(-1+x) \log ^2(1-x)}\right ) \, dx+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {li}(-x)}{x \log ^2(-x)} \, dx,x,-1+x\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {li}(x)}{\log ^2(x)} \, dx,x,1-x\right )+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,1-x\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\log ^2(x)} \, dx,x,1-x\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1-x\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1-x\right )-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,1-x\right )\\ &=-5 x-x^2-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )+\frac {1-x}{\log (1-x)}-\frac {(1-x)^2}{2 \log (1-x)}-2 \log (1-x)+\frac {\text {li}(1-x)}{2}-\frac {1}{8} \int \frac {\text {li}(1-x)}{\log ^2(1-x)} \, dx-\frac {1}{8} \int \frac {\text {li}(1-x)}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {li}(-x)}{x \log ^2(-x)} \, dx,x,-1+x\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {li}(x)}{\log ^2(x)} \, dx,x,1-x\right )+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (1-x)\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1-x)\right )-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1-x\right )+\operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1-x\right )\\ &=-5 x-x^2-\frac {1}{2} \text {Ei}(2 \log (1-x))-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )-\frac {1}{2 \log (1-x)}+\frac {1-x}{\log (1-x)}-\frac {(1-x)^2}{2 \log (1-x)}-2 \log (1-x)-\frac {\text {li}(1-x)}{2}+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx+\operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1-x)\right )\\ &=-5 x-x^2+\frac {1}{2} \text {Ei}(2 \log (1-x))-2 x \log \left (\frac {e^{-\frac {x}{4 \log (1-x)}}}{1-x}\right )-\frac {1}{2 \log (1-x)}+\frac {1-x}{\log (1-x)}-\frac {(1-x)^2}{2 \log (1-x)}-2 \log (1-x)-\frac {\text {li}(1-x)}{2}+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{(-1+x) \log ^2(1-x)} \, dx+\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log ^2(1-x)} \, dx-\frac {1}{2} \int \frac {x \log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{\log (1-x)} \, dx-2 \int \frac {\log \left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )}{-1+x} \, dx+\int \log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 32, normalized size = 0.91 \begin {gather*} x \left (-3-x+\log ^2\left (-\frac {e^{-\frac {x}{4 \log (1-x)}}}{-1+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((6 - 2*x - 4*x^2)*Log[1 - x]^2 + (x^2 + (x - x^2)*Log[1 - x] - 4*x*Log[1 - x]^2)*Log[-(1/(E^(x/(4*L
og[1 - x]))*(-1 + x)))] + (-2 + 2*x)*Log[1 - x]^2*Log[-(1/(E^(x/(4*Log[1 - x]))*(-1 + x)))]^2)/((-2 + 2*x)*Log
[1 - x]^2),x]

[Out]

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

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fricas [A]  time = 0.44, size = 42, normalized size = 1.20 \begin {gather*} \frac {16 \, x \log \left (-x + 1\right )^{4} + x^{3} - 8 \, {\left (x^{2} + 6 \, x\right )} \log \left (-x + 1\right )^{2}}{16 \, \log \left (-x + 1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-2)*log(-x+1)^2*log(-exp(-1/4*x/log(-x+1))/(x-1))^2+(-4*x*log(-x+1)^2+(-x^2+x)*log(-x+1)+x^2)*l
og(-exp(-1/4*x/log(-x+1))/(x-1))+(-4*x^2-2*x+6)*log(-x+1)^2)/(2*x-2)/log(-x+1)^2,x, algorithm="fricas")

[Out]

1/16*(16*x*log(-x + 1)^4 + x^3 - 8*(x^2 + 6*x)*log(-x + 1)^2)/log(-x + 1)^2

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giac [C]  time = 2.62, size = 56, normalized size = 1.60 \begin {gather*} -2 i \, \pi x \log \left (x - 1\right ) + x \log \left (x - 1\right )^{2} - {\left (\pi ^{2} + 3\right )} x - \frac {1}{2} \, x^{2} - \frac {x^{3}}{16 \, {\left (\pi ^{2} + 2 i \, \pi \log \left (x - 1\right ) - \log \left (x - 1\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-2)*log(-x+1)^2*log(-exp(-1/4*x/log(-x+1))/(x-1))^2+(-4*x*log(-x+1)^2+(-x^2+x)*log(-x+1)+x^2)*l
og(-exp(-1/4*x/log(-x+1))/(x-1))+(-4*x^2-2*x+6)*log(-x+1)^2)/(2*x-2)/log(-x+1)^2,x, algorithm="giac")

[Out]

-2*I*pi*x*log(x - 1) + x*log(x - 1)^2 - (pi^2 + 3)*x - 1/2*x^2 - 1/16*x^3/(pi^2 + 2*I*pi*log(x - 1) - log(x -
1)^2)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (2 x -2\right ) \ln \left (1-x \right )^{2} \ln \left (-\frac {{\mathrm e}^{-\frac {x}{4 \ln \left (1-x \right )}}}{x -1}\right )^{2}+\left (-4 x \ln \left (1-x \right )^{2}+\left (-x^{2}+x \right ) \ln \left (1-x \right )+x^{2}\right ) \ln \left (-\frac {{\mathrm e}^{-\frac {x}{4 \ln \left (1-x \right )}}}{x -1}\right )+\left (-4 x^{2}-2 x +6\right ) \ln \left (1-x \right )^{2}}{\left (2 x -2\right ) \ln \left (1-x \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-2)*ln(1-x)^2*ln(-exp(-1/4*x/ln(1-x))/(x-1))^2+(-4*x*ln(1-x)^2+(-x^2+x)*ln(1-x)+x^2)*ln(-exp(-1/4*x/l
n(1-x))/(x-1))+(-4*x^2-2*x+6)*ln(1-x)^2)/(2*x-2)/ln(1-x)^2,x)

[Out]

int(((2*x-2)*ln(1-x)^2*ln(-exp(-1/4*x/ln(1-x))/(x-1))^2+(-4*x*ln(1-x)^2+(-x^2+x)*ln(1-x)+x^2)*ln(-exp(-1/4*x/l
n(1-x))/(x-1))+(-4*x^2-2*x+6)*ln(1-x)^2)/(2*x-2)/ln(1-x)^2,x)

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maxima [B]  time = 0.57, size = 72, normalized size = 2.06 \begin {gather*} x \log \left (-x + 1\right )^{2} + 2 \, x \log \left (-x + 1\right ) \log \left (e^{\left (\frac {x}{4 \, \log \left (-x + 1\right )}\right )}\right ) + x \log \left (e^{\left (\frac {x}{4 \, \log \left (-x + 1\right )}\right )}\right )^{2} - x^{2} - 3 \, x - 3 \, \log \left (x - 1\right ) + 3 \, \log \left (-x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-2)*log(-x+1)^2*log(-exp(-1/4*x/log(-x+1))/(x-1))^2+(-4*x*log(-x+1)^2+(-x^2+x)*log(-x+1)+x^2)*l
og(-exp(-1/4*x/log(-x+1))/(x-1))+(-4*x^2-2*x+6)*log(-x+1)^2)/(2*x-2)/log(-x+1)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.66, size = 157, normalized size = 4.49 \begin {gather*} \frac {{\ln \left (1-x\right )}^2}{2}-\frac {23\,x}{8}-\frac {\ln \left (x-1\right )\,\ln \left (\frac {1}{{\mathrm {e}}^{\frac {x}{4\,\ln \left (1-x\right )}}-x\,{\mathrm {e}}^{\frac {x}{4\,\ln \left (1-x\right )}}}\right )}{2}-\frac {\ln \left (x-1\right )\,\ln \left (1-x\right )}{2}+\frac {\ln \left (\frac {1}{{\mathrm {e}}^{\frac {x}{4\,\ln \left (1-x\right )}}-x\,{\mathrm {e}}^{\frac {x}{4\,\ln \left (1-x\right )}}}\right )\,\ln \left (1-x\right )}{2}+x\,{\ln \left (\frac {1}{{\mathrm {e}}^{\frac {x}{4\,\ln \left (1-x\right )}}-x\,{\mathrm {e}}^{\frac {x}{4\,\ln \left (1-x\right )}}}\right )}^2-x^2-\frac {x\,\ln \left (x-1\right )}{8\,\ln \left (1-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-exp(-x/(4*log(1 - x)))/(x - 1))*(log(1 - x)*(x - x^2) + x^2 - 4*x*log(1 - x)^2) - log(1 - x)^2*(2*x
+ 4*x^2 - 6) + log(-exp(-x/(4*log(1 - x)))/(x - 1))^2*log(1 - x)^2*(2*x - 2))/(log(1 - x)^2*(2*x - 2)),x)

[Out]

log(1 - x)^2/2 - (23*x)/8 - (log(x - 1)*log(1/(exp(x/(4*log(1 - x))) - x*exp(x/(4*log(1 - x))))))/2 - (log(x -
1)*log(1 - x))/2 + (log(1/(exp(x/(4*log(1 - x))) - x*exp(x/(4*log(1 - x)))))*log(1 - x))/2 + x*log(1/(exp(x/(
4*log(1 - x))) - x*exp(x/(4*log(1 - x)))))^2 - x^2 - (x*log(x - 1))/(8*log(1 - x))

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sympy [A]  time = 0.52, size = 27, normalized size = 0.77 \begin {gather*} \frac {x^{3}}{16 \log {\left (1 - x \right )}^{2}} - \frac {x^{2}}{2} + x \log {\left (1 - x \right )}^{2} - 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-2)*ln(-x+1)**2*ln(-exp(-1/4*x/ln(-x+1))/(x-1))**2+(-4*x*ln(-x+1)**2+(-x**2+x)*ln(-x+1)+x**2)*l
n(-exp(-1/4*x/ln(-x+1))/(x-1))+(-4*x**2-2*x+6)*ln(-x+1)**2)/(2*x-2)/ln(-x+1)**2,x)

[Out]

x**3/(16*log(1 - x)**2) - x**2/2 + x*log(1 - x)**2 - 3*x

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