\(\int \frac {e^{2 x^2-4 x \log (\frac {1}{x^2})+2 \log ^2(\frac {1}{x^2})} (-4+(-8+32 x+16 x^2+(-32-16 x) \log (\frac {1}{x^2})) \log (x))}{x^3 \log ^2(x)} \, dx\) [8055]

Optimal result

Integrand size = 56, antiderivative size = 23 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^2 \log (x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx \]

[In]

Int[(E^(2*x^2 - 4*x*Log[x^(-2)] + 2*Log[x^(-2)]^2)*(-4 + (-8 + 32*x + 16*x^2 + (-32 - 16*x)*Log[x^(-2)])*Log[x
]))/(x^3*Log[x]^2),x]

[Out]

-4*Defer[Int][E^(2*(x - Log[x^(-2)])^2)/(x^3*Log[x]^2), x] - 8*Defer[Int][E^(2*(x - Log[x^(-2)])^2)/(x^3*Log[x
]), x] + 32*Defer[Int][E^(2*(x - Log[x^(-2)])^2)/(x^2*Log[x]), x] + 16*Defer[Int][E^(2*(x - Log[x^(-2)])^2)/(x
*Log[x]), x] - 32*Defer[Int][(E^(2*(x - Log[x^(-2)])^2)*Log[x^(-2)])/(x^3*Log[x]), x] - 16*Defer[Int][(E^(2*(x
- Log[x^(-2)])^2)*Log[x^(-2)])/(x^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx \\ & = \int \left (-\frac {4 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log ^2(x)}+\frac {8 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \left (-1+4 x+2 x^2-4 \log \left (\frac {1}{x^2}\right )-2 x \log \left (\frac {1}{x^2}\right )\right )}{x^3 \log (x)}\right ) \, dx \\ & = -\left (4 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log ^2(x)} \, dx\right )+8 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \left (-1+4 x+2 x^2-4 \log \left (\frac {1}{x^2}\right )-2 x \log \left (\frac {1}{x^2}\right )\right )}{x^3 \log (x)} \, dx \\ & = -\left (4 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log ^2(x)} \, dx\right )+8 \int \left (-\frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log (x)}+\frac {4 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^2 \log (x)}+\frac {2 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x \log (x)}-\frac {4 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \log \left (\frac {1}{x^2}\right )}{x^3 \log (x)}-\frac {2 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \log \left (\frac {1}{x^2}\right )}{x^2 \log (x)}\right ) \, dx \\ & = -\left (4 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log ^2(x)} \, dx\right )-8 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log (x)} \, dx+16 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x \log (x)} \, dx-16 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \log \left (\frac {1}{x^2}\right )}{x^2 \log (x)} \, dx+32 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^2 \log (x)} \, dx-32 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \log \left (\frac {1}{x^2}\right )}{x^3 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 e^{2 \left (x^2+\log ^2\left (\frac {1}{x^2}\right )\right )} \left (\frac {1}{x^2}\right )^{1-4 x}}{\log (x)} \]

[In]

Integrate[(E^(2*x^2 - 4*x*Log[x^(-2)] + 2*Log[x^(-2)]^2)*(-4 + (-8 + 32*x + 16*x^2 + (-32 - 16*x)*Log[x^(-2)])
*Log[x]))/(x^3*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30

method	result	size
parallelrisch	\(\frac {4 \,{\mathrm e}^{2 \ln \left (\frac {1}{x^{2}}\right )^{2}-4 x \ln \left (\frac {1}{x^{2}}\right )+2 x^{2}}}{x^{2} \ln \left (x \right )}\)	\(30\)
risch	\(\frac {4 \,{\mathrm e}^{\frac {{\left (-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 \ln \left (x \right )+2 x \right )}^{2}}{2}}}{x^{2} \ln \left (x \right )}\)	\(72\)

[In]

int((((-16*x-32)*ln(1/x^2)+16*x^2+32*x-8)*ln(x)-4)*exp(ln(1/x^2)^2-2*x*ln(1/x^2)+x^2)^2/x^3/ln(x)^2,x,method=_
RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=-\frac {8 \, e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {1}{x^{2}}\right ) + 2 \, \log \left (\frac {1}{x^{2}}\right )^{2}\right )}}{x^{2} \log \left (\frac {1}{x^{2}}\right )} \]

[In]

integrate((((-16*x-32)*log(1/x^2)+16*x^2+32*x-8)*log(x)-4)*exp(log(1/x^2)^2-2*x*log(1/x^2)+x^2)^2/x^3/log(x)^2
,x, algorithm="fricas")

[Out]

-8*e^(2*x^2 - 4*x*log(x^(-2)) + 2*log(x^(-2))^2)/(x^2*log(x^(-2)))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 e^{2 x^{2} + 8 x \log {\left (x \right )} + 8 \log {\left (x \right )}^{2}}}{x^{2} \log {\left (x \right )}} \]

[In]

integrate((((-16*x-32)*ln(1/x**2)+16*x**2+32*x-8)*ln(x)-4)*exp(ln(1/x**2)**2-2*x*ln(1/x**2)+x**2)**2/x**3/ln(x
)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 \, e^{\left (2 \, x^{2} + 8 \, x \log \left (x\right ) + 8 \, \log \left (x\right )^{2}\right )}}{x^{2} \log \left (x\right )} \]

[In]

integrate((((-16*x-32)*log(1/x^2)+16*x^2+32*x-8)*log(x)-4)*exp(log(1/x^2)^2-2*x*log(1/x^2)+x^2)^2/x^3/log(x)^2
,x, algorithm="maxima")

[Out]

4*e^(2*x^2 + 8*x*log(x) + 8*log(x)^2)/(x^2*log(x))

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 \, e^{\left (2 \, x^{2} + 8 \, x \log \left (x\right ) + 8 \, \log \left (x\right )^{2}\right )}}{x^{2} \log \left (x\right )} \]

[In]

integrate((((-16*x-32)*log(1/x^2)+16*x^2+32*x-8)*log(x)-4)*exp(log(1/x^2)^2-2*x*log(1/x^2)+x^2)^2/x^3/log(x)^2
,x, algorithm="giac")

[Out]

4*e^(2*x^2 + 8*x*log(x) + 8*log(x)^2)/(x^2*log(x))

Mupad [B] (verification not implemented)

Time = 13.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x^2}\right )}^2}\,{\mathrm {e}}^{2\,x^2}\,{\left (x^8\right )}^x}{x^2\,\ln \left (x\right )} \]

[In]

int((exp(2*log(1/x^2)^2 - 4*x*log(1/x^2) + 2*x^2)*(log(x)*(32*x + 16*x^2 - log(1/x^2)*(16*x + 32) - 8) - 4))/(
x^3*log(x)^2),x)

[Out]

(4*exp(2*log(1/x^2)^2)*exp(2*x^2)*(x^8)^x)/(x^2*log(x))