3.16.18 \(\int \frac {(6 x-6 x^2) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+(3 x^2+(3 x-3 x^2) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)) \log (x^2)+(-e^5 x^2+e^5 (-x+x^2) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)) \log ^2(x^2)}{(-9 x+9 x^2) \log (2) \log ^2(1-x)+e^5 (6 x-6 x^2) \log (2) \log ^2(1-x) \log (x^2)+e^{10} (-x+x^2) \log (2) \log ^2(1-x) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=34 \[ 1-\log (x)+\frac {x}{\log (2) \log (1-x) \left (e^5-\frac {3}{\log \left (x^2\right )}\right )} \]

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Rubi [F]  time = 1.22, 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 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((6*x - 6*x^2)*Log[1 - x] + (9 - 9*x)*Log[2]*Log[1 - x]^2 + (3*x^2 + (3*x - 3*x^2)*Log[1 - x] + E^5*(-6 +
6*x)*Log[2]*Log[1 - x]^2)*Log[x^2] + (-(E^5*x^2) + E^5*(-x + x^2)*Log[1 - x] + E^10*(1 - x)*Log[2]*Log[1 - x]^
2)*Log[x^2]^2)/((-9*x + 9*x^2)*Log[2]*Log[1 - x]^2 + E^5*(6*x - 6*x^2)*Log[2]*Log[1 - x]^2*Log[x^2] + E^10*(-x
+ x^2)*Log[2]*Log[1 - x]^2*Log[x^2]^2),x]

[Out]

1/(E^5*Log[2]*Log[1 - x]) - (1 - x)/(E^5*Log[2]*Log[1 - x]) - Log[x] - (6*Defer[Int][1/(Log[1 - x]*(-3 + E^5*L
og[x^2])^2), x])/Log[2] - (3*Defer[Int][1/(Log[1 - x]^2*(-3 + E^5*Log[x^2])), x])/(E^5*Log[2]) - (3*Defer[Int]
[1/((-1 + x)*Log[1 - x]^2*(-3 + E^5*Log[x^2])), x])/(E^5*Log[2]) + (3*Defer[Int][1/(Log[1 - x]*(-3 + E^5*Log[x
^2])), x])/(E^5*Log[2])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{x}-\frac {x \log \left (x^2\right )}{(-1+x) \log (2) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )}+\frac {-6-3 \log \left (x^2\right )+e^5 \log ^2\left (x^2\right )}{\log (2) \log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=-\log (x)-\frac {\int \frac {x \log \left (x^2\right )}{(-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{\log (2)}+\frac {\int \frac {-6-3 \log \left (x^2\right )+e^5 \log ^2\left (x^2\right )}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2} \, dx}{\log (2)}\\ &=-\log (x)-\frac {\int \left (\frac {x}{e^5 (-1+x) \log ^2(1-x)}+\frac {3 x}{e^5 (-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )}\right ) \, dx}{\log (2)}+\frac {\int \left (\frac {1}{e^5 \log (1-x)}-\frac {6}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2}+\frac {3}{e^5 \log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )}\right ) \, dx}{\log (2)}\\ &=-\log (x)-\frac {6 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2} \, dx}{\log (2)}-\frac {\int \frac {x}{(-1+x) \log ^2(1-x)} \, dx}{e^5 \log (2)}+\frac {\int \frac {1}{\log (1-x)} \, dx}{e^5 \log (2)}-\frac {3 \int \frac {x}{(-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}+\frac {3 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}\\ &=-\log (x)-\frac {6 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2} \, dx}{\log (2)}-\frac {\operatorname {Subst}\left (\int \frac {1-x}{x \log ^2(x)} \, dx,x,1-x\right )}{e^5 \log (2)}-\frac {\operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1-x\right )}{e^5 \log (2)}+\frac {3 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}-\frac {3 \int \left (\frac {1}{\log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )}+\frac {1}{(-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )}\right ) \, dx}{e^5 \log (2)}\\ &=-\log (x)-\frac {\text {li}(1-x)}{e^5 \log (2)}-\frac {6 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2} \, dx}{\log (2)}-\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{\log ^2(x)}+\frac {1}{x \log ^2(x)}\right ) \, dx,x,1-x\right )}{e^5 \log (2)}-\frac {3 \int \frac {1}{\log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}-\frac {3 \int \frac {1}{(-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}+\frac {3 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}\\ &=-\log (x)-\frac {\text {li}(1-x)}{e^5 \log (2)}-\frac {6 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2} \, dx}{\log (2)}+\frac {\operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,1-x\right )}{e^5 \log (2)}-\frac {\operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,1-x\right )}{e^5 \log (2)}-\frac {3 \int \frac {1}{\log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}-\frac {3 \int \frac {1}{(-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}+\frac {3 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}\\ &=-\frac {1-x}{e^5 \log (2) \log (1-x)}-\log (x)-\frac {\text {li}(1-x)}{e^5 \log (2)}-\frac {6 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2} \, dx}{\log (2)}-\frac {\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (1-x)\right )}{e^5 \log (2)}+\frac {\operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1-x\right )}{e^5 \log (2)}-\frac {3 \int \frac {1}{\log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}-\frac {3 \int \frac {1}{(-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}+\frac {3 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}\\ &=\frac {1}{e^5 \log (2) \log (1-x)}-\frac {1-x}{e^5 \log (2) \log (1-x)}-\log (x)-\frac {6 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )^2} \, dx}{\log (2)}-\frac {3 \int \frac {1}{\log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}-\frac {3 \int \frac {1}{(-1+x) \log ^2(1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}+\frac {3 \int \frac {1}{\log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \, dx}{e^5 \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.25, size = 52, normalized size = 1.53 \begin {gather*} \frac {x}{e^5 \log (2) \log (1-x)}-\log (x)+\frac {3 x}{e^5 \log (2) \log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((6*x - 6*x^2)*Log[1 - x] + (9 - 9*x)*Log[2]*Log[1 - x]^2 + (3*x^2 + (3*x - 3*x^2)*Log[1 - x] + E^5*
(-6 + 6*x)*Log[2]*Log[1 - x]^2)*Log[x^2] + (-(E^5*x^2) + E^5*(-x + x^2)*Log[1 - x] + E^10*(1 - x)*Log[2]*Log[1
- x]^2)*Log[x^2]^2)/((-9*x + 9*x^2)*Log[2]*Log[1 - x]^2 + E^5*(6*x - 6*x^2)*Log[2]*Log[1 - x]^2*Log[x^2] + E^
10*(-x + x^2)*Log[2]*Log[1 - x]^2*Log[x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.65, size = 68, normalized size = 2.00 \begin {gather*} -\frac {e^{5} \log \relax (2) \log \left (x^{2}\right )^{2} \log \left (-x + 1\right ) - {\left (3 \, \log \relax (2) \log \left (-x + 1\right ) + 2 \, x\right )} \log \left (x^{2}\right )}{2 \, {\left (e^{5} \log \relax (2) \log \left (x^{2}\right ) \log \left (-x + 1\right ) - 3 \, \log \relax (2) \log \left (-x + 1\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*exp(5)^2*log(2)*log(-x+1)^2+(x^2-x)*exp(5)*log(-x+1)-x^2*exp(5))*log(x^2)^2+((6*x-6)*exp(5)
*log(2)*log(-x+1)^2+(-3*x^2+3*x)*log(-x+1)+3*x^2)*log(x^2)+(-9*x+9)*log(2)*log(-x+1)^2+(-6*x^2+6*x)*log(-x+1))
/((x^2-x)*exp(5)^2*log(2)*log(-x+1)^2*log(x^2)^2+(-6*x^2+6*x)*exp(5)*log(2)*log(-x+1)^2*log(x^2)+(9*x^2-9*x)*l
og(2)*log(-x+1)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.54, size = 167, normalized size = 4.91 \begin {gather*} -\frac {3 \, e^{5} \log \relax (2) \log \left (x^{2}\right ) \log \left (e^{5} \log \left (x^{2}\right ) - 3\right ) \log \left (-x + 1\right ) - 3 \, e^{5} \log \relax (2) \log \left (x^{2}\right ) \log \left (-e^{5} \log \left (x^{2}\right ) + 3\right ) \log \left (-x + 1\right ) + e^{10} \log \relax (2) \log \left (x^{2}\right ) \log \relax (x) \log \left (-x + 1\right ) - 3 \, e^{5} \log \relax (2) \log \relax (x) \log \left (-x + 1\right ) - x e^{5} \log \left (x^{2}\right ) - 9 \, \log \relax (2) \log \left (e^{5} \log \left (x^{2}\right ) - 3\right ) \log \left (-x + 1\right ) + 9 \, \log \relax (2) \log \left (-e^{5} \log \left (x^{2}\right ) + 3\right ) \log \left (-x + 1\right )}{e^{10} \log \relax (2) \log \left (x^{2}\right ) \log \left (-x + 1\right ) - 3 \, e^{5} \log \relax (2) \log \left (-x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*exp(5)^2*log(2)*log(-x+1)^2+(x^2-x)*exp(5)*log(-x+1)-x^2*exp(5))*log(x^2)^2+((6*x-6)*exp(5)
*log(2)*log(-x+1)^2+(-3*x^2+3*x)*log(-x+1)+3*x^2)*log(x^2)+(-9*x+9)*log(2)*log(-x+1)^2+(-6*x^2+6*x)*log(-x+1))
/((x^2-x)*exp(5)^2*log(2)*log(-x+1)^2*log(x^2)^2+(-6*x^2+6*x)*exp(5)*log(2)*log(-x+1)^2*log(x^2)+(9*x^2-9*x)*l
og(2)*log(-x+1)^2),x, algorithm="giac")

[Out]

-(3*e^5*log(2)*log(x^2)*log(e^5*log(x^2) - 3)*log(-x + 1) - 3*e^5*log(2)*log(x^2)*log(-e^5*log(x^2) + 3)*log(-
x + 1) + e^10*log(2)*log(x^2)*log(x)*log(-x + 1) - 3*e^5*log(2)*log(x)*log(-x + 1) - x*e^5*log(x^2) - 9*log(2)
*log(e^5*log(x^2) - 3)*log(-x + 1) + 9*log(2)*log(-e^5*log(x^2) + 3)*log(-x + 1))/(e^10*log(2)*log(x^2)*log(-x
+ 1) - 3*e^5*log(2)*log(-x + 1))

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maple [C]  time = 0.53, size = 132, normalized size = 3.88

method	result	size
risch	\(-\ln \relax (x )+\frac {x \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )\right )}{\left ({\mathrm e}^{5} \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \,{\mathrm e}^{5} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+{\mathrm e}^{5} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i {\mathrm e}^{5} \ln \relax (x )-6 i\right ) \ln \relax (2) \ln \left (1-x \right )}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((1-x)*exp(5)^2*ln(2)*ln(1-x)^2+(x^2-x)*exp(5)*ln(1-x)-x^2*exp(5))*ln(x^2)^2+((6*x-6)*exp(5)*ln(2)*ln(1-x
)^2+(-3*x^2+3*x)*ln(1-x)+3*x^2)*ln(x^2)+(-9*x+9)*ln(2)*ln(1-x)^2+(-6*x^2+6*x)*ln(1-x))/((x^2-x)*exp(5)^2*ln(2)
*ln(1-x)^2*ln(x^2)^2+(-6*x^2+6*x)*exp(5)*ln(2)*ln(1-x)^2*ln(x^2)+(9*x^2-9*x)*ln(2)*ln(1-x)^2),x,method=_RETURN
VERBOSE)

[Out]

-ln(x)+x*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I*ln(x))/(exp(5)*Pi*csgn(
I*x)^2*csgn(I*x^2)-2*exp(5)*Pi*csgn(I*x)*csgn(I*x^2)^2+exp(5)*Pi*csgn(I*x^2)^3+4*I*exp(5)*ln(x)-6*I)/ln(2)/ln(
1-x)

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maxima [A]  time = 0.70, size = 33, normalized size = 0.97 \begin {gather*} \frac {2 \, x \log \relax (x)}{{\left (2 \, e^{5} \log \relax (2) \log \relax (x) - 3 \, \log \relax (2)\right )} \log \left (-x + 1\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*exp(5)^2*log(2)*log(-x+1)^2+(x^2-x)*exp(5)*log(-x+1)-x^2*exp(5))*log(x^2)^2+((6*x-6)*exp(5)
*log(2)*log(-x+1)^2+(-3*x^2+3*x)*log(-x+1)+3*x^2)*log(x^2)+(-9*x+9)*log(2)*log(-x+1)^2+(-6*x^2+6*x)*log(-x+1))
/((x^2-x)*exp(5)^2*log(2)*log(-x+1)^2*log(x^2)^2+(-6*x^2+6*x)*exp(5)*log(2)*log(-x+1)^2*log(x^2)+(9*x^2-9*x)*l
og(2)*log(-x+1)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.01, size = 407, normalized size = 11.97 \begin {gather*} \frac {{\mathrm {e}}^{10}\,\ln \left (16\right )-4\,{\mathrm {e}}^{10}\,\ln \relax (2)+3}{4\,{\mathrm {e}}^{10}\,\ln \relax (2)-4\,x\,{\mathrm {e}}^{10}\,\ln \relax (2)}-\frac {\frac {3\,x^2\,{\mathrm {e}}^{-10}}{2\,\ln \relax (2)\,\left (x-1\right )}+\frac {{\mathrm {e}}^{-10}\,{\ln \left (1-x\right )}^2\,\left (2\,{\mathrm {e}}^5+3\right )\,\left (x-1\right )}{4\,\ln \relax (2)}-\frac {x\,{\mathrm {e}}^{-10}\,\ln \left (1-x\right )\,\left (6\,x-2\,{\mathrm {e}}^5+2\,x\,{\mathrm {e}}^5-9\right )}{4\,\ln \relax (2)\,\left (x-1\right )}}{{\ln \left (1-x\right )}^2}-\frac {\frac {3\,x\,{\mathrm {e}}^{-15}\,\left (3\,x+3\,\ln \left (1-x\right )+2\,{\mathrm {e}}^5\,\ln \left (1-x\right )-3\,x\,\ln \left (1-x\right )-2\,x\,{\mathrm {e}}^5\,\ln \left (1-x\right )\right )}{2\,\ln \relax (2)\,{\ln \left (1-x\right )}^2\,\left (x-1\right )}-\frac {3\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{-10}\,\left (x+\ln \left (1-x\right )-x\,\ln \left (1-x\right )\right )}{2\,\ln \relax (2)\,{\ln \left (1-x\right )}^2\,\left (x-1\right )}}{\ln \left (x^2\right )-3\,{\mathrm {e}}^{-5}}-\frac {\frac {{\mathrm {e}}^{-10}\,\ln \left (1-x\right )\,\left (2\,{\mathrm {e}}^5\,x^2-4\,{\mathrm {e}}^5\,x+2\,{\mathrm {e}}^5-3\right )}{4\,\ln \relax (2)\,\left (x-1\right )}-\frac {{\mathrm {e}}^{-10}\,\left (3\,x-2\,x\,{\mathrm {e}}^5+2\,x^2\,{\mathrm {e}}^5\right )}{4\,\ln \relax (2)\,\left (x-1\right )}+\frac {{\mathrm {e}}^{-10}\,{\ln \left (1-x\right )}^2\,\left (2\,{\mathrm {e}}^5+3\right )\,\left (x-1\right )}{4\,\ln \relax (2)}}{\ln \left (1-x\right )}-\ln \relax (x)-\ln \left (x-1\right )\,\left (\frac {{\mathrm {e}}^{-10}\,\left (\frac {{\mathrm {e}}^5}{2}+\frac {{\mathrm {e}}^{10}\,\ln \left (16\right )}{4}+\frac {3}{4}\right )}{\ln \relax (2)}-1\right )+\frac {x\,{\mathrm {e}}^{-10}\,\left (4\,{\mathrm {e}}^5+3\right )}{4\,\ln \relax (2)}+\frac {x\,{\mathrm {e}}^{-10}\,\ln \left (1-x\right )\,\left (2\,{\mathrm {e}}^5+3\right )}{4\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2)*(log(1 - x)*(3*x - 3*x^2) + 3*x^2 + exp(5)*log(2)*log(1 - x)^2*(6*x - 6)) + log(1 - x)*(6*x - 6
*x^2) - log(x^2)^2*(x^2*exp(5) + exp(5)*log(1 - x)*(x - x^2) + exp(10)*log(2)*log(1 - x)^2*(x - 1)) - log(2)*l
og(1 - x)^2*(9*x - 9))/(log(2)*log(1 - x)^2*(9*x - 9*x^2) - log(x^2)*exp(5)*log(2)*log(1 - x)^2*(6*x - 6*x^2)
+ log(x^2)^2*exp(10)*log(2)*log(1 - x)^2*(x - x^2)),x)

[Out]

(exp(10)*log(16) - 4*exp(10)*log(2) + 3)/(4*exp(10)*log(2) - 4*x*exp(10)*log(2)) - ((3*x^2*exp(-10))/(2*log(2)
*(x - 1)) + (exp(-10)*log(1 - x)^2*(2*exp(5) + 3)*(x - 1))/(4*log(2)) - (x*exp(-10)*log(1 - x)*(6*x - 2*exp(5)
+ 2*x*exp(5) - 9))/(4*log(2)*(x - 1)))/log(1 - x)^2 - ((3*x*exp(-15)*(3*x + 3*log(1 - x) + 2*exp(5)*log(1 - x
) - 3*x*log(1 - x) - 2*x*exp(5)*log(1 - x)))/(2*log(2)*log(1 - x)^2*(x - 1)) - (3*x*log(x^2)*exp(-10)*(x + log
(1 - x) - x*log(1 - x)))/(2*log(2)*log(1 - x)^2*(x - 1)))/(log(x^2) - 3*exp(-5)) - ((exp(-10)*log(1 - x)*(2*ex
p(5) - 4*x*exp(5) + 2*x^2*exp(5) - 3))/(4*log(2)*(x - 1)) - (exp(-10)*(3*x - 2*x*exp(5) + 2*x^2*exp(5)))/(4*lo
g(2)*(x - 1)) + (exp(-10)*log(1 - x)^2*(2*exp(5) + 3)*(x - 1))/(4*log(2)))/log(1 - x) - log(x) - log(x - 1)*((
exp(-10)*(exp(5)/2 + (exp(10)*log(16))/4 + 3/4))/log(2) - 1) + (x*exp(-10)*(4*exp(5) + 3))/(4*log(2)) + (x*exp
(-10)*log(1 - x)*(2*exp(5) + 3))/(4*log(2))

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sympy [A]  time = 0.36, size = 31, normalized size = 0.91 \begin {gather*} \frac {x \log {\left (x^{2} \right )}}{\left (e^{5} \log {\relax (2 )} \log {\left (x^{2} \right )} - 3 \log {\relax (2 )}\right ) \log {\left (1 - x \right )}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*exp(5)**2*ln(2)*ln(-x+1)**2+(x**2-x)*exp(5)*ln(-x+1)-x**2*exp(5))*ln(x**2)**2+((6*x-6)*exp(
5)*ln(2)*ln(-x+1)**2+(-3*x**2+3*x)*ln(-x+1)+3*x**2)*ln(x**2)+(-9*x+9)*ln(2)*ln(-x+1)**2+(-6*x**2+6*x)*ln(-x+1)
)/((x**2-x)*exp(5)**2*ln(2)*ln(-x+1)**2*ln(x**2)**2+(-6*x**2+6*x)*exp(5)*ln(2)*ln(-x+1)**2*ln(x**2)+(9*x**2-9*
x)*ln(2)*ln(-x+1)**2),x)

[Out]

x*log(x**2)/((exp(5)*log(2)*log(x**2) - 3*log(2))*log(1 - x)) - log(x)

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