\(\int \frac {e^{8+2 x^2-e^{2 x^2} x^2} (-2 x-4 x^3) \log (x)+\log ^{x^2}(x) (x+2 x \log (x) \log (\log (x)))}{\log (x)} \, dx\) [8252]

Optimal result

Integrand size = 56, antiderivative size = 24 \[ \int \frac {e^{8+2 x^2-e^{2 x^2} x^2} \left (-2 x-4 x^3\right ) \log (x)+\log ^{x^2}(x) (x+2 x \log (x) \log (\log (x)))}{\log (x)} \, dx=25+e^{8-e^{2 x^2} x^2}+\log ^{x^2}(x) \]

[Out]

exp(-x^2*exp(x^2)^2+8)+exp(x^2*ln(ln(x)))+25

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{8+2 x^2-e^{2 x^2} x^2} \left (-2 x-4 x^3\right ) \log (x)+\log ^{x^2}(x) (x+2 x \log (x) \log (\log (x)))}{\log (x)} \, dx=e^{8-e^{2 x^2} x^2}+\log ^{x^2}(x) \]

[In]

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

[Out]

E^(8 - E^(2*x^2)*x^2) + Log[x]^x^2

Maple [A] (verified)

Time = 13.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

method	result	size
risch	\(\ln \left (x \right )^{x^{2}}+{\mathrm e}^{-{\mathrm e}^{2 x^{2}} x^{2}+8}\)	\(22\)
default	\({\mathrm e}^{x^{2} \ln \left (\ln \left (x \right )\right )}+{\mathrm e}^{-{\mathrm e}^{2 x^{2}} x^{2}+8}\)	\(24\)
parallelrisch	\({\mathrm e}^{x^{2} \ln \left (\ln \left (x \right )\right )}+{\mathrm e}^{-{\mathrm e}^{2 x^{2}} x^{2}+8}\)	\(24\)
parts	\({\mathrm e}^{x^{2} \ln \left (\ln \left (x \right )\right )}+{\mathrm e}^{-{\mathrm e}^{2 x^{2}} x^{2}+8}\)	\(24\)

[In]

int(((2*x*ln(x)*ln(ln(x))+x)*exp(x^2*ln(ln(x)))+(-4*x^3-2*x)*exp(x^2)^2*ln(x)*exp(-x^2*exp(x^2)^2+8))/ln(x),x,
method=_RETURNVERBOSE)

[Out]

ln(x)^(x^2)+exp(-exp(2*x^2)*x^2+8)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{8+2 x^2-e^{2 x^2} x^2} \left (-2 x-4 x^3\right ) \log (x)+\log ^{x^2}(x) (x+2 x \log (x) \log (\log (x)))}{\log (x)} \, dx={\left (\log \left (x\right )^{\left (x^{2}\right )} e^{\left (2 \, x^{2}\right )} + e^{\left (-x^{2} e^{\left (2 \, x^{2}\right )} + 2 \, x^{2} + 8\right )}\right )} e^{\left (-2 \, x^{2}\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{8+2 x^2-e^{2 x^2} x^2} \left (-2 x-4 x^3\right ) \log (x)+\log ^{x^2}(x) (x+2 x \log (x) \log (\log (x)))}{\log (x)} \, dx=e^{x^{2} \log {\left (\log {\left (x \right )} \right )}} + e^{- x^{2} e^{2 x^{2}} + 8} \]

[In]

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

[Out]

exp(x**2*log(log(x))) + exp(-x**2*exp(2*x**2) + 8)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^{8+2 x^2-e^{2 x^2} x^2} \left (-2 x-4 x^3\right ) \log (x)+\log ^{x^2}(x) (x+2 x \log (x) \log (\log (x)))}{\log (x)} \, dx={\left (e^{8} + e^{\left (x^{2} e^{\left (2 \, x^{2}\right )} + x^{2} \log \left (\log \left (x\right )\right )\right )}\right )} e^{\left (-x^{2} e^{\left (2 \, x^{2}\right )}\right )} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(-(2*(2*x^3 + x)*e^(-x^2*e^(2*x^2) + 2*x^2 + 8)*log(x) - (2*x*log(x)*log(log(x)) + x)*log(x)^(x^2))/l
og(x), x)

Mupad [B] (verification not implemented)

Time = 13.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{8+2 x^2-e^{2 x^2} x^2} \left (-2 x-4 x^3\right ) \log (x)+\log ^{x^2}(x) (x+2 x \log (x) \log (\log (x)))}{\log (x)} \, dx={\ln \left (x\right )}^{x^2}+{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^8 \]

[In]

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

[Out]

log(x)^(x^2) + exp(-x^2*exp(2*x^2))*exp(8)