3.5.94 \(\int \frac {e^{\frac {\log (-\frac {6}{e x-x^2})-x \log (\log (1-x))}{\log (\log (1-x))}} ((-e x+x^2) \log (-\frac {6}{e x-x^2})+(e (1-x)-2 x+2 x^2) \log (1-x) \log (\log (1-x))+(x-2 x^2+x^3+e (-1+2 x-x^2)) \log (1-x) \log ^2(\log (1-x)))}{(e (-1+x)+x-x^2) \log (1-x) \log ^2(\log (1-x))} \, dx\)

Optimal. Leaf size=31 \[ e^{-x+\frac {\log \left (\frac {6}{x (-e+x)}\right )}{\log (\log (1-x))}} x \]

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Rubi [F]  time = 8.30, 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 {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right ) \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 0.99, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 40, normalized size = 1.29 \begin {gather*} x e^{\left (-\frac {x \log \left (\log \left (-x + 1\right )\right ) - \log \left (\frac {6}{x^{2} - x e}\right )}{\log \left (\log \left (-x + 1\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*log(-x+1)*log(log(-x+1))^2+((-x+1)*exp(1)+2*x^2-2*x)*log(-x+1)*lo
g(log(-x+1))+(-x*exp(1)+x^2)*log(-6/(x*exp(1)-x^2)))*exp((-x*log(log(-x+1))+log(-6/(x*exp(1)-x^2)))/log(log(-x
+1)))/((x-1)*exp(1)-x^2+x)/log(-x+1)/log(log(-x+1))^2,x, algorithm="fricas")

[Out]

x*e^(-(x*log(log(-x + 1)) - log(6/(x^2 - x*e)))/log(log(-x + 1)))

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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((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*log(-x+1)*log(log(-x+1))^2+((-x+1)*exp(1)+2*x^2-2*x)*log(-x+1)*lo
g(log(-x+1))+(-x*exp(1)+x^2)*log(-6/(x*exp(1)-x^2)))*exp((-x*log(log(-x+1))+log(-6/(x*exp(1)-x^2)))/log(log(-x
+1)))/((x-1)*exp(1)-x^2+x)/log(-x+1)/log(log(-x+1))^2,x, algorithm="giac")

[Out]

undef

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maple [C]  time = 0.55, size = 192, normalized size = 6.19

method	result	size
risch	\(x \,{\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+i \pi \mathrm {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}-x}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}-x}\right )-2 i \mathrm {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{2} \pi +2 i \pi -2 x \ln \left (\ln \left (1-x \right )\right )-2 \ln \relax (x )+2 \ln \relax (2)+2 \ln \relax (3)-2 \ln \left ({\mathrm e}-x \right )}{2 \ln \left (\ln \left (1-x \right )\right )}}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*ln(1-x)*ln(ln(1-x))^2+((1-x)*exp(1)+2*x^2-2*x)*ln(1-x)*ln(ln(1-x))+(-x*
exp(1)+x^2)*ln(-6/(x*exp(1)-x^2)))*exp((-x*ln(ln(1-x))+ln(-6/(x*exp(1)-x^2)))/ln(ln(1-x)))/((x-1)*exp(1)-x^2+x
)/ln(1-x)/ln(ln(1-x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left ({\left (x^{3} - 2 \, x^{2} - {\left (x^{2} - 2 \, x + 1\right )} e + x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right )^{2} + {\left (2 \, x^{2} - {\left (x - 1\right )} e - 2 \, x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right ) + {\left (x^{2} - x e\right )} \log \left (\frac {6}{x^{2} - x e}\right )\right )} e^{\left (-\frac {x \log \left (\log \left (-x + 1\right )\right ) - \log \left (\frac {6}{x^{2} - x e}\right )}{\log \left (\log \left (-x + 1\right )\right )}\right )}}{{\left (x^{2} - {\left (x - 1\right )} e - x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*log(-x+1)*log(log(-x+1))^2+((-x+1)*exp(1)+2*x^2-2*x)*log(-x+1)*lo
g(log(-x+1))+(-x*exp(1)+x^2)*log(-6/(x*exp(1)-x^2)))*exp((-x*log(log(-x+1))+log(-6/(x*exp(1)-x^2)))/log(log(-x
+1)))/((x-1)*exp(1)-x^2+x)/log(-x+1)/log(log(-x+1))^2,x, algorithm="maxima")

[Out]

-integrate(((x^3 - 2*x^2 - (x^2 - 2*x + 1)*e + x)*log(-x + 1)*log(log(-x + 1))^2 + (2*x^2 - (x - 1)*e - 2*x)*l
og(-x + 1)*log(log(-x + 1)) + (x^2 - x*e)*log(6/(x^2 - x*e)))*e^(-(x*log(log(-x + 1)) - log(6/(x^2 - x*e)))/lo
g(log(-x + 1)))/((x^2 - (x - 1)*e - x)*log(-x + 1)*log(log(-x + 1))^2), x)

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mupad [B]  time = 1.30, size = 30, normalized size = 0.97 \begin {gather*} x\,{\mathrm {e}}^{-x}\,{\left (-\frac {6}{x\,\mathrm {e}-x^2}\right )}^{\frac {1}{\ln \left (\ln \left (1-x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((log(-6/(x*exp(1) - x^2)) - x*log(log(1 - x)))/log(log(1 - x)))*(log(-6/(x*exp(1) - x^2))*(x*exp(1)
- x^2) + log(log(1 - x))*log(1 - x)*(2*x + exp(1)*(x - 1) - 2*x^2) - log(log(1 - x))^2*log(1 - x)*(x - exp(1)*
(x^2 - 2*x + 1) - 2*x^2 + x^3)))/(log(log(1 - x))^2*log(1 - x)*(x + exp(1)*(x - 1) - x^2)),x)

[Out]

x*exp(-x)*(-6/(x*exp(1) - x^2))^(1/log(log(1 - x)))

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sympy [A]  time = 113.59, size = 31, normalized size = 1.00 \begin {gather*} x e^{\frac {- x \log {\left (\log {\left (1 - x \right )} \right )} + \log {\left (- \frac {6}{- x^{2} + e x} \right )}}{\log {\left (\log {\left (1 - x \right )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2+2*x-1)*exp(1)+x**3-2*x**2+x)*ln(-x+1)*ln(ln(-x+1))**2+((-x+1)*exp(1)+2*x**2-2*x)*ln(-x+1)*l
n(ln(-x+1))+(-x*exp(1)+x**2)*ln(-6/(x*exp(1)-x**2)))*exp((-x*ln(ln(-x+1))+ln(-6/(x*exp(1)-x**2)))/ln(ln(-x+1))
)/((x-1)*exp(1)-x**2+x)/ln(-x+1)/ln(ln(-x+1))**2,x)

[Out]

x*exp((-x*log(log(1 - x)) + log(-6/(-x**2 + E*x)))/log(log(1 - x)))

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