3.3.93 \(\int \frac {-5+5 x+(-5-10 x) \log (x)+5 \log (e^{-1+x} x)}{6 x-12 x^2+6 x^3+(-12 x+12 x^2) \log (e^{-1+x} x)+6 x \log ^2(e^{-1+x} x)} \, dx\)

Optimal. Leaf size=21 \[ -5+\frac {5 \log (x)}{6 \left (-1+x+\log \left (e^{-1+x} x\right )\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 0.73, 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 {-5+5 x+(-5-10 x) \log (x)+5 \log \left (e^{-1+x} x\right )}{6 x-12 x^2+6 x^3+\left (-12 x+12 x^2\right ) \log \left (e^{-1+x} x\right )+6 x \log ^2\left (e^{-1+x} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5 + 5*x + (-5 - 10*x)*Log[x] + 5*Log[E^(-1 + x)*x])/(6*x - 12*x^2 + 6*x^3 + (-12*x + 12*x^2)*Log[E^(-1 +
x)*x] + 6*x*Log[E^(-1 + x)*x]^2),x]

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.29, size = 32, normalized size = 1.52 \begin {gather*} -\frac {5 \left (-1+x-\log (x)+\log \left (e^{-1+x} x\right )\right )}{6 \left (-1+x+\log \left (e^{-1+x} x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 5*x + (-5 - 10*x)*Log[x] + 5*Log[E^(-1 + x)*x])/(6*x - 12*x^2 + 6*x^3 + (-12*x + 12*x^2)*Log[E
^(-1 + x)*x] + 6*x*Log[E^(-1 + x)*x]^2),x]

[Out]

(-5*(-1 + x - Log[x] + Log[E^(-1 + x)*x]))/(6*(-1 + x + Log[E^(-1 + x)*x]))

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 14, normalized size = 0.67 \begin {gather*} -\frac {5 \, {\left (x - 1\right )}}{3 \, {\left (2 \, x + \log \relax (x) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(x*exp(x-1))+(-10*x-5)*log(x)+5*x-5)/(6*x*log(x*exp(x-1))^2+(12*x^2-12*x)*log(x*exp(x-1))+6*x^
3-12*x^2+6*x),x, algorithm="fricas")

[Out]

-5/3*(x - 1)/(2*x + log(x) - 2)

________________________________________________________________________________________

giac [A]  time = 0.64, size = 14, normalized size = 0.67 \begin {gather*} -\frac {5 \, {\left (x - 1\right )}}{3 \, {\left (2 \, x + \log \relax (x) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(x*exp(x-1))+(-10*x-5)*log(x)+5*x-5)/(6*x*log(x*exp(x-1))^2+(12*x^2-12*x)*log(x*exp(x-1))+6*x^
3-12*x^2+6*x),x, algorithm="giac")

[Out]

-5/3*(x - 1)/(2*x + log(x) - 2)

________________________________________________________________________________________

maple [C]  time = 0.10, size = 107, normalized size = 5.10

method	result	size
risch	\(\frac {5 \ln \relax (x )}{3 \left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )^{2}-i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )^{3}+2 x +2 \ln \relax (x )+2 \ln \left ({\mathrm e}^{x -1}\right )-2\right )}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*ln(x*exp(x-1))+(-10*x-5)*ln(x)+5*x-5)/(6*x*ln(x*exp(x-1))^2+(12*x^2-12*x)*ln(x*exp(x-1))+6*x^3-12*x^2+6
*x),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 14, normalized size = 0.67 \begin {gather*} -\frac {5 \, {\left (x - 1\right )}}{3 \, {\left (2 \, x + \log \relax (x) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(x*exp(x-1))+(-10*x-5)*log(x)+5*x-5)/(6*x*log(x*exp(x-1))^2+(12*x^2-12*x)*log(x*exp(x-1))+6*x^
3-12*x^2+6*x),x, algorithm="maxima")

[Out]

-5/3*(x - 1)/(2*x + log(x) - 2)

________________________________________________________________________________________

mupad [B]  time = 0.68, size = 15, normalized size = 0.71 \begin {gather*} \frac {5\,\ln \relax (x)}{6\,\left (2\,x+\ln \relax (x)-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 5*log(x*exp(x - 1)) - log(x)*(10*x + 5) - 5)/(6*x - log(x*exp(x - 1))*(12*x - 12*x^2) + 6*x*log(x*e
xp(x - 1))^2 - 12*x^2 + 6*x^3),x)

[Out]

(5*log(x))/(6*(2*x + log(x) - 2))

________________________________________________________________________________________

sympy [A]  time = 0.36, size = 14, normalized size = 0.67 \begin {gather*} \frac {5 - 5 x}{6 x + 3 \log {\relax (x )} - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*ln(x*exp(x-1))+(-10*x-5)*ln(x)+5*x-5)/(6*x*ln(x*exp(x-1))**2+(12*x**2-12*x)*ln(x*exp(x-1))+6*x**3
-12*x**2+6*x),x)

[Out]

(5 - 5*x)/(6*x + 3*log(x) - 6)

________________________________________________________________________________________