3.45.47 \(\int \frac {e^{-2 x} \log (x) (-128 \log (x)+(64+(-64-64 x) \log (x)) \log (\frac {x^2}{4}))}{3 x^3 \log ^3(\frac {x^2}{4})} \, dx\)

Optimal. Leaf size=26 \[ \frac {32 e^{-2 x} \log ^2(x)}{3 x^2 \log ^2\left (\frac {x^2}{4}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 3.45, 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^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{3 x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Log[x]*(-128*Log[x] + (64 + (-64 - 64*x)*Log[x])*Log[x^2/4]))/(3*E^(2*x)*x^3*Log[x^2/4]^3),x]

[Out]

(-128*Defer[Int][Log[x]^2/(E^(2*x)*x^3*Log[x^2/4]^3), x])/3 + (64*Defer[Int][Log[x]/(E^(2*x)*x^3*Log[x^2/4]^2)
, x])/3 - (64*Defer[Int][Log[x]^2/(E^(2*x)*x^3*Log[x^2/4]^2), x])/3 - (64*Defer[Int][Log[x]^2/(E^(2*x)*x^2*Log
[x^2/4]^2), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^{-2 x} \log (x) \left (-128 \log (x)+(64+(-64-64 x) \log (x)) \log \left (\frac {x^2}{4}\right )\right )}{x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx\\ &=\frac {1}{3} \int \left (-\frac {128 e^{-2 x} \log ^2(x)}{x^3 \log ^3\left (\frac {x^2}{4}\right )}-\frac {64 e^{-2 x} \log (x) (-1+\log (x)+x \log (x))}{x^3 \log ^2\left (\frac {x^2}{4}\right )}\right ) \, dx\\ &=-\left (\frac {64}{3} \int \frac {e^{-2 x} \log (x) (-1+\log (x)+x \log (x))}{x^3 \log ^2\left (\frac {x^2}{4}\right )} \, dx\right )-\frac {128}{3} \int \frac {e^{-2 x} \log ^2(x)}{x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx\\ &=-\left (\frac {64}{3} \int \left (-\frac {e^{-2 x} \log (x)}{x^3 \log ^2\left (\frac {x^2}{4}\right )}+\frac {e^{-2 x} \log ^2(x)}{x^3 \log ^2\left (\frac {x^2}{4}\right )}+\frac {e^{-2 x} \log ^2(x)}{x^2 \log ^2\left (\frac {x^2}{4}\right )}\right ) \, dx\right )-\frac {128}{3} \int \frac {e^{-2 x} \log ^2(x)}{x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx\\ &=\frac {64}{3} \int \frac {e^{-2 x} \log (x)}{x^3 \log ^2\left (\frac {x^2}{4}\right )} \, dx-\frac {64}{3} \int \frac {e^{-2 x} \log ^2(x)}{x^3 \log ^2\left (\frac {x^2}{4}\right )} \, dx-\frac {64}{3} \int \frac {e^{-2 x} \log ^2(x)}{x^2 \log ^2\left (\frac {x^2}{4}\right )} \, dx-\frac {128}{3} \int \frac {e^{-2 x} \log ^2(x)}{x^3 \log ^3\left (\frac {x^2}{4}\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.55, size = 26, normalized size = 1.00 \begin {gather*} \frac {32 e^{-2 x} \log ^2(x)}{3 x^2 \log ^2\left (\frac {x^2}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[x]*(-128*Log[x] + (64 + (-64 - 64*x)*Log[x])*Log[x^2/4]))/(3*E^(2*x)*x^3*Log[x^2/4]^3),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 26, normalized size = 1.00 \begin {gather*} \frac {32}{3} \, e^{\left (-2 \, x + 2 \, \log \left (-\frac {\log \relax (x)}{2 \, {\left (x \log \relax (2) - x \log \relax (x)\right )}}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((-64*x-64)*log(x)+64)*log(1/4*x^2)-128*log(x))*exp(log(log(x)/x/log(1/4*x^2))-x)^2/x/log(x)/lo
g(1/4*x^2),x, algorithm="fricas")

[Out]

32/3*e^(-2*x + 2*log(-1/2*log(x)/(x*log(2) - x*log(x))))

________________________________________________________________________________________

giac [A]  time = 0.68, size = 24, normalized size = 0.92 \begin {gather*} \frac {32}{3} \, e^{\left (-2 \, x + 2 \, \log \left (\frac {\log \relax (x)}{x \log \left (\frac {1}{4} \, x^{2}\right )}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((-64*x-64)*log(x)+64)*log(1/4*x^2)-128*log(x))*exp(log(log(x)/x/log(1/4*x^2))-x)^2/x/log(x)/lo
g(1/4*x^2),x, algorithm="giac")

[Out]

32/3*e^(-2*x + 2*log(log(x)/(x*log(1/4*x^2))))

________________________________________________________________________________________

maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (-64 x -64\right ) \ln \relax (x )+64\right ) \ln \left (\frac {x^{2}}{4}\right )-128 \ln \relax (x )\right ) \ln \relax (x ) {\mathrm e}^{-2 x}}{3 \ln \left (\frac {x^{2}}{4}\right )^{3} x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(((-64*x-64)*ln(x)+64)*ln(1/4*x^2)-128*ln(x))*exp(ln(ln(x)/x/ln(1/4*x^2))-x)^2/x/ln(x)/ln(1/4*x^2),x)

[Out]

int(1/3*(((-64*x-64)*ln(x)+64)*ln(1/4*x^2)-128*ln(x))*exp(ln(ln(x)/x/ln(1/4*x^2))-x)^2/x/ln(x)/ln(1/4*x^2),x)

________________________________________________________________________________________

maxima [A]  time = 0.52, size = 38, normalized size = 1.46 \begin {gather*} \frac {8 \, e^{\left (-2 \, x\right )} \log \relax (x)^{2}}{3 \, {\left (x^{2} \log \relax (2)^{2} - 2 \, x^{2} \log \relax (2) \log \relax (x) + x^{2} \log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((-64*x-64)*log(x)+64)*log(1/4*x^2)-128*log(x))*exp(log(log(x)/x/log(1/4*x^2))-x)^2/x/log(x)/lo
g(1/4*x^2),x, algorithm="maxima")

[Out]

8/3*e^(-2*x)*log(x)^2/(x^2*log(2)^2 - 2*x^2*log(2)*log(x) + x^2*log(x)^2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{2\,\ln \left (\frac {\ln \relax (x)}{x\,\ln \left (\frac {x^2}{4}\right )}\right )-2\,x}\,\left (128\,\ln \relax (x)+\ln \left (\frac {x^2}{4}\right )\,\left (\ln \relax (x)\,\left (64\,x+64\right )-64\right )\right )}{3\,x\,\ln \left (\frac {x^2}{4}\right )\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*log(log(x)/(x*log(x^2/4))) - 2*x)*(128*log(x) + log(x^2/4)*(log(x)*(64*x + 64) - 64)))/(3*x*log(x^
2/4)*log(x)),x)

[Out]

int(-(exp(2*log(log(x)/(x*log(x^2/4))) - 2*x)*(128*log(x) + log(x^2/4)*(log(x)*(64*x + 64) - 64)))/(3*x*log(x^
2/4)*log(x)), x)

________________________________________________________________________________________

sympy [A]  time = 0.40, size = 42, normalized size = 1.62 \begin {gather*} \frac {8 e^{- 2 x} \log {\relax (x )}^{2}}{3 x^{2} \log {\relax (x )}^{2} - 6 x^{2} \log {\relax (2 )} \log {\relax (x )} + 3 x^{2} \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((-64*x-64)*ln(x)+64)*ln(1/4*x**2)-128*ln(x))*exp(ln(ln(x)/x/ln(1/4*x**2))-x)**2/x/ln(x)/ln(1/4
*x**2),x)

[Out]

8*exp(-2*x)*log(x)**2/(3*x**2*log(x)**2 - 6*x**2*log(2)*log(x) + 3*x**2*log(2)**2)

________________________________________________________________________________________