3.76.54 \(\int \frac {e^{\frac {16 \log (\frac {1}{x^2})+(-x+x^2) \log (x)}{4 \log (\frac {1}{x^2})}} ((-1+x) \log (\frac {1}{x^2})+(-2+2 x+(-1+2 x) \log (\frac {1}{x^2})) \log (x))}{4 \log ^2(\frac {1}{x^2})} \, dx\)

Optimal. Leaf size=25 \[ 36+e^{4+\frac {\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \]

________________________________________________________________________________________

Rubi [F]  time = 2.82, 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 {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}\right ) \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((16*Log[x^(-2)] + (-x + x^2)*Log[x])/(4*Log[x^(-2)]))*((-1 + x)*Log[x^(-2)] + (-2 + 2*x + (-1 + 2*x)*L
og[x^(-2)])*Log[x]))/(4*Log[x^(-2)]^2),x]

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.29, size = 20, normalized size = 0.80 \begin {gather*} e^4 x^{\frac {(-1+x) x}{4 \log \left (\frac {1}{x^2}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((16*Log[x^(-2)] + (-x + x^2)*Log[x])/(4*Log[x^(-2)]))*((-1 + x)*Log[x^(-2)] + (-2 + 2*x + (-1 +
2*x)*Log[x^(-2)])*Log[x]))/(4*Log[x^(-2)]^2),x]

[Out]

E^4*x^(((-1 + x)*x)/(4*Log[x^(-2)]))

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 11, normalized size = 0.44 \begin {gather*} e^{\left (-\frac {1}{8} \, x^{2} + \frac {1}{8} \, x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-1/8*x^2 + 1/8*x + 4)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 27, normalized size = 1.08 \begin {gather*} e^{\left (-\frac {x^{2} \log \relax (x)}{4 \, \log \left (x^{2}\right )} + \frac {x \log \relax (x)}{4 \, \log \left (x^{2}\right )} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.29, size = 77, normalized size = 3.08

method	result	size
default	\(-\frac {-4 \left (\ln \left (\frac {1}{x^{2}}\right )+2 \ln \relax (x )\right ) {\mathrm e}^{\frac {\ln \relax (x ) \left (x^{2}-x \right )+16 \ln \left (\frac {1}{x^{2}}\right )}{4 \ln \left (\frac {1}{x^{2}}\right )}}+8 \ln \relax (x ) {\mathrm e}^{\frac {\ln \relax (x ) \left (x^{2}-x \right )+16 \ln \left (\frac {1}{x^{2}}\right )}{4 \ln \left (\frac {1}{x^{2}}\right )}}}{4 \ln \left (\frac {1}{x^{2}}\right )}\)	\(77\)
risch	\({\mathrm e}^{-\frac {8 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-16 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+8 i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+x^{2} \ln \relax (x )-x \ln \relax (x )-32 \ln \relax (x )}{2 \left (-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 \ln \relax (x )\right )}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(((2*x-1)*ln(1/x^2)+2*x-2)*ln(x)+(x-1)*ln(1/x^2))*exp(1/4*(ln(x)*(x^2-x)+16*ln(1/x^2))/ln(1/x^2))/ln(1
/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*(-4*(ln(1/x^2)+2*ln(x))*exp(1/4*(ln(x)*(x^2-x)+16*ln(1/x^2))/ln(1/x^2))+8*ln(x)*exp(1/4*(ln(x)*(x^2-x)+16
*ln(1/x^2))/ln(1/x^2)))/ln(1/x^2)

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 11, normalized size = 0.44 \begin {gather*} e^{\left (-\frac {1}{8} \, x^{2} + \frac {1}{8} \, x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-1/8*x^2 + 1/8*x + 4)

________________________________________________________________________________________

mupad [B]  time = 4.86, size = 29, normalized size = 1.16 \begin {gather*} {\mathrm {e}}^{\frac {x^2\,\ln \relax (x)}{4\,\ln \left (\frac {1}{x^2}\right )}}\,{\mathrm {e}}^4\,{\mathrm {e}}^{-\frac {x\,\ln \relax (x)}{4\,\ln \left (\frac {1}{x^2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((4*log(1/x^2) - (log(x)*(x - x^2))/4)/log(1/x^2))*(log(x)*(2*x + log(1/x^2)*(2*x - 1) - 2) + log(1/x^
2)*(x - 1)))/(4*log(1/x^2)^2),x)

[Out]

exp((x^2*log(x))/(4*log(1/x^2)))*exp(4)*exp(-(x*log(x))/(4*log(1/x^2)))

________________________________________________________________________________________

sympy [A]  time = 0.32, size = 22, normalized size = 0.88 \begin {gather*} e^{- \frac {\frac {\left (x^{2} - x\right ) \log {\relax (x )}}{4} - 8 \log {\relax (x )}}{2 \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((2*x-1)*ln(1/x**2)+2*x-2)*ln(x)+(x-1)*ln(1/x**2))*exp(1/4*(ln(x)*(x**2-x)+16*ln(1/x**2))/ln(1/
x**2))/ln(1/x**2)**2,x)

[Out]

exp(-((x**2 - x)*log(x)/4 - 8*log(x))/(2*log(x)))

________________________________________________________________________________________