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

Optimal. Leaf size=24 \[ 25+e^{8-e^{2 x^2} x^2}+\log ^{x^2}(x) \]

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Rubi [F]  time = 0.79, 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^{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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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-\operatorname {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-\operatorname {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 \operatorname {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-\operatorname {Subst}\left (\int e^{8-\left (-2+e^{2 x}\right ) x} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 23, normalized size = 0.96 \begin {gather*} e^{8-e^{2 x^2} x^2}+\log ^{x^2}(x) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.57, size = 40, normalized size = 1.67 \begin {gather*} {\left (\log \relax (x)^{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \relax (x) - {\left (2 \, x \log \relax (x) \log \left (\log \relax (x)\right ) + x\right )} \log \relax (x)^{\left (x^{2}\right )}}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.08, size = 22, normalized size = 0.92

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.42, size = 35, normalized size = 1.46 \begin {gather*} {\left (e^{8} + e^{\left (x^{2} e^{\left (2 \, x^{2}\right )} + x^{2} \log \left (\log \relax (x)\right )\right )}\right )} e^{\left (-x^{2} e^{\left (2 \, x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 4.93, size = 22, normalized size = 0.92 \begin {gather*} {\ln \relax (x)}^{x^2}+{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^8 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 2.55, size = 22, normalized size = 0.92 \begin {gather*} e^{x^{2} \log {\left (\log {\relax (x )} \right )}} + e^{- x^{2} e^{2 x^{2}} + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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