3.100.53 \(\int \frac {e^{-\frac {x^2+(16 x+4 x^2) \log ^2(x)+(64+36 x+4 x^2) \log ^4(x)}{4 \log ^4(x)}} (e^5 (-2 x^2+2 x^3) \log (x-x^2)+e^5 (x^2-x^3) \log (x) \log (x-x^2)+e^5 (-16 x+12 x^2+4 x^3) \log ^2(x) \log (x-x^2)+e^5 (8 x-4 x^2-4 x^3) \log ^3(x) \log (x-x^2)+\log ^5(x) (e^5 (-2+4 x)+e^5 (18 x-14 x^2-4 x^3) \log (x-x^2)))}{(-2 x+2 x^2) \log ^5(x)} \, dx\)

Optimal. Leaf size=32 \[ e^{5-x-\left (4+x+\frac {x}{2 \log ^2(x)}\right )^2} \log \left (x-x^2\right ) \]

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Rubi [F]  time = 25.00, 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 {\exp \left (-\frac {x^2+\left (16 x+4 x^2\right ) \log ^2(x)+\left (64+36 x+4 x^2\right ) \log ^4(x)}{4 \log ^4(x)}\right ) \left (e^5 \left (-2 x^2+2 x^3\right ) \log \left (x-x^2\right )+e^5 \left (x^2-x^3\right ) \log (x) \log \left (x-x^2\right )+e^5 \left (-16 x+12 x^2+4 x^3\right ) \log ^2(x) \log \left (x-x^2\right )+e^5 \left (8 x-4 x^2-4 x^3\right ) \log ^3(x) \log \left (x-x^2\right )+\log ^5(x) \left (e^5 (-2+4 x)+e^5 \left (18 x-14 x^2-4 x^3\right ) \log \left (x-x^2\right )\right )\right )}{\left (-2 x+2 x^2\right ) \log ^5(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^5*(-2*x^2 + 2*x^3)*Log[x - x^2] + E^5*(x^2 - x^3)*Log[x]*Log[x - x^2] + E^5*(-16*x + 12*x^2 + 4*x^3)*Lo
g[x]^2*Log[x - x^2] + E^5*(8*x - 4*x^2 - 4*x^3)*Log[x]^3*Log[x - x^2] + Log[x]^5*(E^5*(-2 + 4*x) + E^5*(18*x -
14*x^2 - 4*x^3)*Log[x - x^2]))/(E^((x^2 + (16*x + 4*x^2)*Log[x]^2 + (64 + 36*x + 4*x^2)*Log[x]^4)/(4*Log[x]^4
))*(-2*x + 2*x^2)*Log[x]^5),x]

[Out]

Defer[Int][E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)/(-1 + x), x] + Defer[Int][E^(-11 - 9*
x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)/x, x] - 9*Defer[Int][E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4)
- (x*(4 + x))/Log[x]^2)*Log[(1 - x)*x], x] - 2*Defer[Int][E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/
Log[x]^2)*x*Log[(1 - x)*x], x] + Defer[Int][(E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)*x*L
og[(1 - x)*x])/Log[x]^5, x] - Defer[Int][(E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)*x*Log[
(1 - x)*x])/Log[x]^4, x]/2 + 8*Defer[Int][(E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)*Log[(
1 - x)*x])/Log[x]^3, x] + 2*Defer[Int][(E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)*x*Log[(1
- x)*x])/Log[x]^3, x] - 4*Defer[Int][(E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)*Log[(1 -
x)*x])/Log[x]^2, x] - 2*Defer[Int][(E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)*x*Log[(1 - x
)*x])/Log[x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {x^2+\left (16 x+4 x^2\right ) \log ^2(x)+\left (64+36 x+4 x^2\right ) \log ^4(x)}{4 \log ^4(x)}\right ) \left (e^5 \left (-2 x^2+2 x^3\right ) \log \left (x-x^2\right )+e^5 \left (x^2-x^3\right ) \log (x) \log \left (x-x^2\right )+e^5 \left (-16 x+12 x^2+4 x^3\right ) \log ^2(x) \log \left (x-x^2\right )+e^5 \left (8 x-4 x^2-4 x^3\right ) \log ^3(x) \log \left (x-x^2\right )+\log ^5(x) \left (e^5 (-2+4 x)+e^5 \left (18 x-14 x^2-4 x^3\right ) \log \left (x-x^2\right )\right )\right )}{x (-2+2 x) \log ^5(x)} \, dx\\ &=\int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \left (-2 (-1+x) x^2 \log (-((-1+x) x))+(-1+x) x^2 \log (x) \log (-((-1+x) x))-4 x \left (-4+3 x+x^2\right ) \log ^2(x) \log (-((-1+x) x))+4 x \left (-2+x+x^2\right ) \log ^3(x) \log (-((-1+x) x))+2 \log ^5(x) \left (1-2 x+x \left (-9+7 x+2 x^2\right ) \log (-((-1+x) x))\right )\right )}{2 (1-x) x \log ^5(x)} \, dx\\ &=\frac {1}{2} \int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \left (-2 (-1+x) x^2 \log (-((-1+x) x))+(-1+x) x^2 \log (x) \log (-((-1+x) x))-4 x \left (-4+3 x+x^2\right ) \log ^2(x) \log (-((-1+x) x))+4 x \left (-2+x+x^2\right ) \log ^3(x) \log (-((-1+x) x))+2 \log ^5(x) \left (1-2 x+x \left (-9+7 x+2 x^2\right ) \log (-((-1+x) x))\right )\right )}{(1-x) x \log ^5(x)} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) (-1+2 x)}{(-1+x) x}+\frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \left (2 x-x \log (x)+16 \log ^2(x)+4 x \log ^2(x)-8 \log ^3(x)-4 x \log ^3(x)-18 \log ^5(x)-4 x \log ^5(x)\right ) \log ((1-x) x)}{\log ^5(x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \left (2 x-x \log (x)+16 \log ^2(x)+4 x \log ^2(x)-8 \log ^3(x)-4 x \log ^3(x)-18 \log ^5(x)-4 x \log ^5(x)\right ) \log ((1-x) x)}{\log ^5(x)} \, dx+\int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) (-1+2 x)}{(-1+x) x} \, dx\\ &=\frac {1}{2} \int \left (-18 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \log ((1-x) x)-4 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)+\frac {2 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^5(x)}-\frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^4(x)}+\frac {16 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \log ((1-x) x)}{\log ^3(x)}+\frac {4 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^3(x)}-\frac {8 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \log ((1-x) x)}{\log ^2(x)}-\frac {4 \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^2(x)}\right ) \, dx+\int \left (\frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right )}{-1+x}+\frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right )}{x}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^4(x)} \, dx\right )-2 \int \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x) \, dx+2 \int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^3(x)} \, dx-2 \int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^2(x)} \, dx-4 \int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \log ((1-x) x)}{\log ^2(x)} \, dx+8 \int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \log ((1-x) x)}{\log ^3(x)} \, dx-9 \int \exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) \log ((1-x) x) \, dx+\int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right )}{-1+x} \, dx+\int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right )}{x} \, dx+\int \frac {\exp \left (-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}\right ) x \log ((1-x) x)}{\log ^5(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 41, normalized size = 1.28 \begin {gather*} e^{-11-9 x-x^2-\frac {x^2}{4 \log ^4(x)}-\frac {x (4+x)}{\log ^2(x)}} \log (-((-1+x) x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-2*x^2 + 2*x^3)*Log[x - x^2] + E^5*(x^2 - x^3)*Log[x]*Log[x - x^2] + E^5*(-16*x + 12*x^2 + 4*x
^3)*Log[x]^2*Log[x - x^2] + E^5*(8*x - 4*x^2 - 4*x^3)*Log[x]^3*Log[x - x^2] + Log[x]^5*(E^5*(-2 + 4*x) + E^5*(
18*x - 14*x^2 - 4*x^3)*Log[x - x^2]))/(E^((x^2 + (16*x + 4*x^2)*Log[x]^2 + (64 + 36*x + 4*x^2)*Log[x]^4)/(4*Lo
g[x]^4))*(-2*x + 2*x^2)*Log[x]^5),x]

[Out]

E^(-11 - 9*x - x^2 - x^2/(4*Log[x]^4) - (x*(4 + x))/Log[x]^2)*Log[-((-1 + x)*x)]

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fricas [A]  time = 0.65, size = 49, normalized size = 1.53 \begin {gather*} e^{\left (-\frac {4 \, {\left (x^{2} + 9 \, x + 16\right )} \log \relax (x)^{4} + 4 \, {\left (x^{2} + 4 \, x\right )} \log \relax (x)^{2} + x^{2}}{4 \, \log \relax (x)^{4}} + 5\right )} \log \left (-x^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-14*x^2+18*x)*exp(5)*log(-x^2+x)+(4*x-2)*exp(5))*log(x)^5+(-4*x^3-4*x^2+8*x)*exp(5)*log(-x^
2+x)*log(x)^3+(4*x^3+12*x^2-16*x)*exp(5)*log(-x^2+x)*log(x)^2+(-x^3+x^2)*exp(5)*log(-x^2+x)*log(x)+(2*x^3-2*x^
2)*exp(5)*log(-x^2+x))/(2*x^2-2*x)/log(x)^5/exp(1/4*((4*x^2+36*x+64)*log(x)^4+(4*x^2+16*x)*log(x)^2+x^2)/log(x
)^4),x, algorithm="fricas")

[Out]

e^(-1/4*(4*(x^2 + 9*x + 16)*log(x)^4 + 4*(x^2 + 4*x)*log(x)^2 + x^2)/log(x)^4 + 5)*log(-x^2 + x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-14*x^2+18*x)*exp(5)*log(-x^2+x)+(4*x-2)*exp(5))*log(x)^5+(-4*x^3-4*x^2+8*x)*exp(5)*log(-x^
2+x)*log(x)^3+(4*x^3+12*x^2-16*x)*exp(5)*log(-x^2+x)*log(x)^2+(-x^3+x^2)*exp(5)*log(-x^2+x)*log(x)+(2*x^3-2*x^
2)*exp(5)*log(-x^2+x))/(2*x^2-2*x)/log(x)^5/exp(1/4*((4*x^2+36*x+64)*log(x)^4+(4*x^2+16*x)*log(x)^2+x^2)/log(x
)^4),x, algorithm="giac")

[Out]

undef

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maple [C]  time = 0.36, size = 172, normalized size = 5.38

method	result	size
risch	\(\left (-i \pi \,{\mathrm e}^{5} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}-\frac {i \pi \,{\mathrm e}^{5} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )}{2}+\frac {i \pi \,{\mathrm e}^{5} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}}{2}+\frac {i \pi \,{\mathrm e}^{5} \mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (i x \left (x -1\right )\right )^{2}}{2}+\frac {i \pi \,{\mathrm e}^{5} \mathrm {csgn}\left (i x \left (x -1\right )\right )^{3}}{2}+i \pi \,{\mathrm e}^{5}+{\mathrm e}^{5} \ln \relax (x )+{\mathrm e}^{5} \ln \left (x -1\right )\right ) {\mathrm e}^{-\frac {4 x^{2} \ln \relax (x )^{4}+36 x \ln \relax (x )^{4}+64 \ln \relax (x )^{4}+4 x^{2} \ln \relax (x )^{2}+16 x \ln \relax (x )^{2}+x^{2}}{4 \ln \relax (x )^{4}}}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^3-14*x^2+18*x)*exp(5)*ln(-x^2+x)+(4*x-2)*exp(5))*ln(x)^5+(-4*x^3-4*x^2+8*x)*exp(5)*ln(-x^2+x)*ln(x
)^3+(4*x^3+12*x^2-16*x)*exp(5)*ln(-x^2+x)*ln(x)^2+(-x^3+x^2)*exp(5)*ln(-x^2+x)*ln(x)+(2*x^3-2*x^2)*exp(5)*ln(-
x^2+x))/(2*x^2-2*x)/ln(x)^5/exp(1/4*((4*x^2+36*x+64)*ln(x)^4+(4*x^2+16*x)*ln(x)^2+x^2)/ln(x)^4),x,method=_RETU
RNVERBOSE)

[Out]

(-I*Pi*exp(5)*csgn(I*x*(x-1))^2-1/2*I*Pi*exp(5)*csgn(I*x)*csgn(I*(x-1))*csgn(I*x*(x-1))+1/2*I*Pi*exp(5)*csgn(I
*x)*csgn(I*x*(x-1))^2+1/2*I*Pi*exp(5)*csgn(I*(x-1))*csgn(I*x*(x-1))^2+1/2*I*Pi*exp(5)*csgn(I*x*(x-1))^3+I*Pi*e
xp(5)+exp(5)*ln(x)+exp(5)*ln(x-1))*exp(-1/4*(4*x^2*ln(x)^4+36*x*ln(x)^4+64*ln(x)^4+4*x^2*ln(x)^2+16*x*ln(x)^2+
x^2)/ln(x)^4)

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maxima [A]  time = 0.65, size = 46, normalized size = 1.44 \begin {gather*} {\left (\log \relax (x) + \log \left (-x + 1\right )\right )} e^{\left (-x^{2} - 9 \, x - \frac {x^{2}}{\log \relax (x)^{2}} - \frac {4 \, x}{\log \relax (x)^{2}} - \frac {x^{2}}{4 \, \log \relax (x)^{4}} - 11\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3-14*x^2+18*x)*exp(5)*log(-x^2+x)+(4*x-2)*exp(5))*log(x)^5+(-4*x^3-4*x^2+8*x)*exp(5)*log(-x^
2+x)*log(x)^3+(4*x^3+12*x^2-16*x)*exp(5)*log(-x^2+x)*log(x)^2+(-x^3+x^2)*exp(5)*log(-x^2+x)*log(x)+(2*x^3-2*x^
2)*exp(5)*log(-x^2+x))/(2*x^2-2*x)/log(x)^5/exp(1/4*((4*x^2+36*x+64)*log(x)^4+(4*x^2+16*x)*log(x)^2+x^2)/log(x
)^4),x, algorithm="maxima")

[Out]

(log(x) + log(-x + 1))*e^(-x^2 - 9*x - x^2/log(x)^2 - 4*x/log(x)^2 - 1/4*x^2/log(x)^4 - 11)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{-\frac {\frac {{\ln \relax (x)}^2\,\left (4\,x^2+16\,x\right )}{4}+\frac {{\ln \relax (x)}^4\,\left (4\,x^2+36\,x+64\right )}{4}+\frac {x^2}{4}}{{\ln \relax (x)}^4}}\,\left (\left ({\mathrm {e}}^5\,\left (4\,x-2\right )-{\mathrm {e}}^5\,\ln \left (x-x^2\right )\,\left (4\,x^3+14\,x^2-18\,x\right )\right )\,{\ln \relax (x)}^5-{\mathrm {e}}^5\,\ln \left (x-x^2\right )\,\left (4\,x^3+4\,x^2-8\,x\right )\,{\ln \relax (x)}^3+{\mathrm {e}}^5\,\ln \left (x-x^2\right )\,\left (4\,x^3+12\,x^2-16\,x\right )\,{\ln \relax (x)}^2+{\mathrm {e}}^5\,\ln \left (x-x^2\right )\,\left (x^2-x^3\right )\,\ln \relax (x)-{\mathrm {e}}^5\,\ln \left (x-x^2\right )\,\left (2\,x^2-2\,x^3\right )\right )}{{\ln \relax (x)}^5\,\left (2\,x-2\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-((log(x)^2*(16*x + 4*x^2))/4 + (log(x)^4*(36*x + 4*x^2 + 64))/4 + x^2/4)/log(x)^4)*(log(x)^5*(exp(5
)*(4*x - 2) - exp(5)*log(x - x^2)*(14*x^2 - 18*x + 4*x^3)) - exp(5)*log(x - x^2)*(2*x^2 - 2*x^3) - exp(5)*log(
x - x^2)*log(x)^3*(4*x^2 - 8*x + 4*x^3) + exp(5)*log(x - x^2)*log(x)^2*(12*x^2 - 16*x + 4*x^3) + exp(5)*log(x
- x^2)*log(x)*(x^2 - x^3)))/(log(x)^5*(2*x - 2*x^2)),x)

[Out]

int(-(exp(-((log(x)^2*(16*x + 4*x^2))/4 + (log(x)^4*(36*x + 4*x^2 + 64))/4 + x^2/4)/log(x)^4)*(log(x)^5*(exp(5
)*(4*x - 2) - exp(5)*log(x - x^2)*(14*x^2 - 18*x + 4*x^3)) - exp(5)*log(x - x^2)*(2*x^2 - 2*x^3) - exp(5)*log(
x - x^2)*log(x)^3*(4*x^2 - 8*x + 4*x^3) + exp(5)*log(x - x^2)*log(x)^2*(12*x^2 - 16*x + 4*x^3) + exp(5)*log(x
- x^2)*log(x)*(x^2 - x^3)))/(log(x)^5*(2*x - 2*x^2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**3-14*x**2+18*x)*exp(5)*ln(-x**2+x)+(4*x-2)*exp(5))*ln(x)**5+(-4*x**3-4*x**2+8*x)*exp(5)*ln(
-x**2+x)*ln(x)**3+(4*x**3+12*x**2-16*x)*exp(5)*ln(-x**2+x)*ln(x)**2+(-x**3+x**2)*exp(5)*ln(-x**2+x)*ln(x)+(2*x
**3-2*x**2)*exp(5)*ln(-x**2+x))/(2*x**2-2*x)/ln(x)**5/exp(1/4*((4*x**2+36*x+64)*ln(x)**4+(4*x**2+16*x)*ln(x)**
2+x**2)/ln(x)**4),x)

[Out]

Timed out

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