\(\int \frac {e^{\frac {16 \log (\frac {1}{x^2})+(-x+x^2) \log (x)}{4 \log (\frac {1}{x^2})}} ((-1+x) \log (\frac {1}{x^2})+(-2+2 x+(-1+2 x) \log (\frac {1}{x^2})) \log (x))}{4 \log ^2(\frac {1}{x^2})} \, dx\) [7554]

Optimal result

Integrand size = 66, antiderivative size = 25 \[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx=36+e^{4+\frac {\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx=\int \frac {\exp \left (\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}\right ) \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx=e^4 x^{\frac {(-1+x) x}{4 \log \left (\frac {1}{x^2}\right )}} \]

[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)]))

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\ln \left (x \right ) \left (x^{2}-x \right )+16 \ln \left (\frac {1}{x^{2}}\right )}{4 \ln \left (\frac {1}{x^{2}}\right )}}\)	\(27\)
default	\(-\frac {-4 \left (\ln \left (\frac {1}{x^{2}}\right )+2 \ln \left (x \right )\right ) {\mathrm e}^{\frac {\ln \left (x \right ) \left (x^{2}-x \right )+16 \ln \left (\frac {1}{x^{2}}\right )}{4 \ln \left (\frac {1}{x^{2}}\right )}}+8 \ln \left (x \right ) {\mathrm e}^{\frac {\ln \left (x \right ) \left (x^{2}-x \right )+16 \ln \left (\frac {1}{x^{2}}\right )}{4 \ln \left (\frac {1}{x^{2}}\right )}}}{4 \ln \left (\frac {1}{x^{2}}\right )}\)	\(77\)
risch	\({\mathrm e}^{-\frac {8 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-16 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+8 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+x^{2} \ln \left (x \right )-x \ln \left (x \right )-32 \ln \left (x \right )}{2 \left (-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 \ln \left (x \right )\right )}}\)	\(125\)

[In]

int(1/4*(((-1+2*x)*ln(1/x^2)+2*x-2)*ln(x)+(-1+x)*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]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.44 \[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx=e^{\left (-\frac {1}{8} \, x^{2} + \frac {1}{8} \, x + 4\right )} \]

[In]

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

[Out]

e^(-1/8*x^2 + 1/8*x + 4)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx=e^{- \frac {\frac {\left (x^{2} - x\right ) \log {\left (x \right )}}{4} - 8 \log {\left (x \right )}}{2 \log {\left (x \right )}}} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.44 \[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx=e^{\left (-\frac {1}{8} \, x^{2} + \frac {1}{8} \, x + 4\right )} \]

[In]

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

[Out]

e^(-1/8*x^2 + 1/8*x + 4)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx=e^{\left (-\frac {x^{2} \log \left (x\right )}{4 \, \log \left (x^{2}\right )} + \frac {x \log \left (x\right )}{4 \, \log \left (x^{2}\right )} + 4\right )} \]

[In]

integrate(1/4*(((-1+2*x)*log(1/x^2)+2*x-2)*log(x)+(-1+x)*log(1/x^2))*exp(1/4*(log(x)*(x^2-x)+16*log(1/x^2))/lo
g(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)

Mupad [B] (verification not implemented)

Time = 13.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {16 \log \left (\frac {1}{x^2}\right )+\left (-x+x^2\right ) \log (x)}{4 \log \left (\frac {1}{x^2}\right )}} \left ((-1+x) \log \left (\frac {1}{x^2}\right )+\left (-2+2 x+(-1+2 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{4 \log ^2\left (\frac {1}{x^2}\right )} \, dx={\mathrm {e}}^{\frac {x^2\,\ln \left (x\right )}{4\,\ln \left (\frac {1}{x^2}\right )}}\,{\mathrm {e}}^4\,{\mathrm {e}}^{-\frac {x\,\ln \left (x\right )}{4\,\ln \left (\frac {1}{x^2}\right )}} \]

[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)))