3.76.90 \(\int \frac {e^{-\frac {x^2}{3}} ((12 e^x-24 x) \log (x)+(-12 e^x+24 x+(48 x+16 x^3+e^x (-12-12 x-8 x^2)) \log (x)) \log (2 x))}{(3 e^{2 x} x^2-12 e^x x^3+12 x^4) \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {4 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \]

________________________________________________________________________________________

Rubi [F]  time = 11.85, 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 {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((12*E^x - 24*x)*Log[x] + (-12*E^x + 24*x + (48*x + 16*x^3 + E^x*(-12 - 12*x - 8*x^2))*Log[x])*Log[2*x])/(
E^(x^2/3)*(3*E^(2*x)*x^2 - 12*E^x*x^3 + 12*x^4)*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{3 \left (e^x-2 x\right )^2 x^2 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (e^x-2 x\right )^2 x^2 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (-\frac {24 e^{-\frac {x^2}{3}} (-1+x) \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)}-\frac {4 e^{-\frac {x^2}{3}} \left (-3 \log (x)+3 \log (2 x)+3 \log (x) \log (2 x)+3 x \log (x) \log (2 x)+2 x^2 \log (x) \log (2 x)\right )}{\left (e^x-2 x\right ) x^2 \log ^2(x)}\right ) \, dx\\ &=-\left (\frac {4}{3} \int \frac {e^{-\frac {x^2}{3}} \left (-3 \log (x)+3 \log (2 x)+3 \log (x) \log (2 x)+3 x \log (x) \log (2 x)+2 x^2 \log (x) \log (2 x)\right )}{\left (e^x-2 x\right ) x^2 \log ^2(x)} \, dx\right )-8 \int \frac {e^{-\frac {x^2}{3}} (-1+x) \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)} \, dx\\ &=-\left (\frac {4}{3} \int \frac {e^{-\frac {x^2}{3}} \left (3 \log (2 x)+\log (x) \left (-3+\left (3+3 x+2 x^2\right ) \log (2 x)\right )\right )}{\left (e^x-2 x\right ) x^2 \log ^2(x)} \, dx\right )-8 \int \left (\frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 \log (x)}-\frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)}\right ) \, dx\\ &=-\left (\frac {4}{3} \int \left (-\frac {3 e^{-\frac {x^2}{3}}}{\left (e^x-2 x\right ) x^2 \log (x)}+\frac {3 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log ^2(x)}+\frac {2 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) \log (x)}+\frac {3 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log (x)}+\frac {3 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)}\right ) \, dx\right )-8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 \log (x)} \, dx+8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)} \, dx\\ &=-\left (\frac {8}{3} \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) \log (x)} \, dx\right )+4 \int \frac {e^{-\frac {x^2}{3}}}{\left (e^x-2 x\right ) x^2 \log (x)} \, dx-4 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log ^2(x)} \, dx-4 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log (x)} \, dx-4 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \, dx-8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 \log (x)} \, dx+8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 31, normalized size = 1.00 \begin {gather*} \frac {4 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((12*E^x - 24*x)*Log[x] + (-12*E^x + 24*x + (48*x + 16*x^3 + E^x*(-12 - 12*x - 8*x^2))*Log[x])*Log[2
*x])/(E^(x^2/3)*(3*E^(2*x)*x^2 - 12*E^x*x^3 + 12*x^4)*Log[x]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 2.55, size = 38, normalized size = 1.23 \begin {gather*} -\frac {4 \, {\left (e^{\left (-\frac {1}{3} \, x^{2}\right )} \log \relax (2) + e^{\left (-\frac {1}{3} \, x^{2}\right )} \log \relax (x)\right )}}{{\left (2 \, x^{2} - x e^{x}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*log(x)-12*exp(x)+24*x)*log(2*x)+(12*exp(x)-24*x)*log(x))/(3*
exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x^2)/log(x)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.18, size = 46, normalized size = 1.48 \begin {gather*} -\frac {4 \, {\left (e^{\left (-\frac {1}{3} \, x^{2} + x\right )} \log \relax (2) + e^{\left (-\frac {1}{3} \, x^{2} + x\right )} \log \relax (x)\right )}}{2 \, x^{2} e^{x} \log \relax (x) - x e^{\left (2 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*log(x)-12*exp(x)+24*x)*log(2*x)+(12*exp(x)-24*x)*log(x))/(3*
exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x^2)/log(x)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (\left (-8 x^{2}-12 x -12\right ) {\mathrm e}^{x}+16 x^{3}+48 x \right ) \ln \relax (x )-12 \,{\mathrm e}^{x}+24 x \right ) \ln \left (2 x \right )+\left (12 \,{\mathrm e}^{x}-24 x \right ) \ln \relax (x )\right ) {\mathrm e}^{-\frac {x^{2}}{3}}}{\left (3 \,{\mathrm e}^{2 x} x^{2}-12 \,{\mathrm e}^{x} x^{3}+12 x^{4}\right ) \ln \relax (x )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*ln(x)-12*exp(x)+24*x)*ln(2*x)+(12*exp(x)-24*x)*ln(x))/(3*exp(x)^2*
x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x^2)/ln(x)^2,x)

[Out]

int(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*ln(x)-12*exp(x)+24*x)*ln(2*x)+(12*exp(x)-24*x)*ln(x))/(3*exp(x)^2*
x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x^2)/ln(x)^2,x)

________________________________________________________________________________________

maxima [A]  time = 0.66, size = 30, normalized size = 0.97 \begin {gather*} -\frac {4 \, {\left (\log \relax (2) + \log \relax (x)\right )} e^{\left (-\frac {1}{3} \, x^{2}\right )}}{2 \, x^{2} \log \relax (x) - x e^{x} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*log(x)-12*exp(x)+24*x)*log(2*x)+(12*exp(x)-24*x)*log(x))/(3*
exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x^2)/log(x)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {x^2}{3}}\,\left (\ln \left (2\,x\right )\,\left (24\,x-12\,{\mathrm {e}}^x+\ln \relax (x)\,\left (48\,x-{\mathrm {e}}^x\,\left (8\,x^2+12\,x+12\right )+16\,x^3\right )\right )-\ln \relax (x)\,\left (24\,x-12\,{\mathrm {e}}^x\right )\right )}{{\ln \relax (x)}^2\,\left (3\,x^2\,{\mathrm {e}}^{2\,x}-12\,x^3\,{\mathrm {e}}^x+12\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x^2/3)*(log(2*x)*(24*x - 12*exp(x) + log(x)*(48*x - exp(x)*(12*x + 8*x^2 + 12) + 16*x^3)) - log(x)*(
24*x - 12*exp(x))))/(log(x)^2*(3*x^2*exp(2*x) - 12*x^3*exp(x) + 12*x^4)),x)

[Out]

int((exp(-x^2/3)*(log(2*x)*(24*x - 12*exp(x) + log(x)*(48*x - exp(x)*(12*x + 8*x^2 + 12) + 16*x^3)) - log(x)*(
24*x - 12*exp(x))))/(log(x)^2*(3*x^2*exp(2*x) - 12*x^3*exp(x) + 12*x^4)), x)

________________________________________________________________________________________

sympy [A]  time = 0.44, size = 34, normalized size = 1.10 \begin {gather*} \frac {\left (- 4 \log {\relax (x )} - 4 \log {\relax (2 )}\right ) e^{- \frac {x^{2}}{3}}}{2 x^{2} \log {\relax (x )} - x e^{x} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-8*x**2-12*x-12)*exp(x)+16*x**3+48*x)*ln(x)-12*exp(x)+24*x)*ln(2*x)+(12*exp(x)-24*x)*ln(x))/(3*e
xp(x)**2*x**2-12*exp(x)*x**3+12*x**4)/exp(1/3*x**2)/ln(x)**2,x)

[Out]

(-4*log(x) - 4*log(2))*exp(-x**2/3)/(2*x**2*log(x) - x*exp(x)*log(x))

________________________________________________________________________________________