3.23.41 \(\int \frac {e^{\frac {e^{2+4 x}+(-x^2+x \log ^2(x)) \log (\log (5))}{\log (\log (5))}} (4 e^{2+4 x} x+(1-2 x^2+2 x \log (x)+x \log ^2(x)) \log (\log (5)))}{\log (\log (5))} \, dx\)

Optimal. Leaf size=28 \[ e^{x \left (-x+\log ^2(x)\right )+\frac {e^{2+4 x}}{\log (\log (5))}} x \]

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Rubi [F]  time = 1.76, 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 {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5]])*(4*E^(2 + 4*x)*x + (1 - 2*x^2 + 2*x*Log[x
] + x*Log[x]^2)*Log[Log[5]]))/Log[Log[5]],x]

[Out]

Defer[Int][E^((E^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5]]), x] + (4*Defer[Int][E^(2 + 4*x + (E
^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5]])*x, x])/Log[Log[5]] - 2*Defer[Int][E^((E^(2 + 4*x) +
(-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5]])*x^2, x] + 2*Defer[Int][E^((E^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Lo
g[Log[5]])/Log[Log[5]])*x*Log[x], x] + Defer[Int][E^((E^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5
]])*x*Log[x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right ) \, dx}{\log (\log (5))}\\ &=\frac {\int \left (4 \exp \left (2+4 x+\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x-\exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) \left (-1+2 x^2-2 x \log (x)-x \log ^2(x)\right ) \log (\log (5))\right ) \, dx}{\log (\log (5))}\\ &=\frac {4 \int \exp \left (2+4 x+\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x \, dx}{\log (\log (5))}-\int \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) \left (-1+2 x^2-2 x \log (x)-x \log ^2(x)\right ) \, dx\\ &=\frac {4 \int \exp \left (2+4 x+\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x \, dx}{\log (\log (5))}-\int \left (-\exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right )+2 \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x^2-2 \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x \log (x)-\exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x \log ^2(x)\right ) \, dx\\ &=-\left (2 \int \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x^2 \, dx\right )+2 \int \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x \log (x) \, dx+\frac {4 \int \exp \left (2+4 x+\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x \, dx}{\log (\log (5))}+\int \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) \, dx+\int \exp \left (\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}\right ) x \log ^2(x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.91, size = 29, normalized size = 1.04 \begin {gather*} e^{-x^2+x \log ^2(x)+\frac {e^{2+4 x}}{\log (\log (5))}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5]])*(4*E^(2 + 4*x)*x + (1 - 2*x^2 + 2*x
*Log[x] + x*Log[x]^2)*Log[Log[5]]))/Log[Log[5]],x]

[Out]

E^(-x^2 + x*Log[x]^2 + E^(2 + 4*x)/Log[Log[5]])*x

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fricas [A]  time = 0.79, size = 32, normalized size = 1.14 \begin {gather*} x e^{\left (\frac {{\left (x \log \relax (x)^{2} - x^{2}\right )} \log \left (\log \relax (5)\right ) + e^{\left (4 \, x + 2\right )}}{\log \left (\log \relax (5)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+2*x*log(x)-2*x^2+1)*log(log(5))+4*x*exp(2)*exp(4*x))*exp(((x*log(x)^2-x^2)*log(log(5))+
exp(2)*exp(4*x))/log(log(5)))/log(log(5)),x, algorithm="fricas")

[Out]

x*e^(((x*log(x)^2 - x^2)*log(log(5)) + e^(4*x + 2))/log(log(5)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+2*x*log(x)-2*x^2+1)*log(log(5))+4*x*exp(2)*exp(4*x))*exp(((x*log(x)^2-x^2)*log(log(5))+
exp(2)*exp(4*x))/log(log(5)))/log(log(5)),x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.05, size = 34, normalized size = 1.21

method	result	size
risch	\(x \,{\mathrm e}^{\frac {\ln \relax (x )^{2} \ln \left (\ln \relax (5)\right ) x -\ln \left (\ln \relax (5)\right ) x^{2}+{\mathrm e}^{4 x +2}}{\ln \left (\ln \relax (5)\right )}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)^2+2*x*ln(x)-2*x^2+1)*ln(ln(5))+4*x*exp(2)*exp(4*x))*exp(((x*ln(x)^2-x^2)*ln(ln(5))+exp(2)*exp(4*
x))/ln(ln(5)))/ln(ln(5)),x,method=_RETURNVERBOSE)

[Out]

x*exp((ln(x)^2*ln(ln(5))*x-ln(ln(5))*x^2+exp(4*x+2))/ln(ln(5)))

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maxima [A]  time = 0.62, size = 27, normalized size = 0.96 \begin {gather*} x e^{\left (x \log \relax (x)^{2} - x^{2} + \frac {e^{\left (4 \, x + 2\right )}}{\log \left (\log \relax (5)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+2*x*log(x)-2*x^2+1)*log(log(5))+4*x*exp(2)*exp(4*x))*exp(((x*log(x)^2-x^2)*log(log(5))+
exp(2)*exp(4*x))/log(log(5)))/log(log(5)),x, algorithm="maxima")

[Out]

x*e^(x*log(x)^2 - x^2 + e^(4*x + 2)/log(log(5)))

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mupad [B]  time = 1.56, size = 28, normalized size = 1.00 \begin {gather*} x\,{\mathrm {e}}^{x\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^2}{\ln \left (\ln \relax (5)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(4*x)*exp(2) + log(log(5))*(x*log(x)^2 - x^2))/log(log(5)))*(log(log(5))*(x*log(x)^2 + 2*x*log(x)
- 2*x^2 + 1) + 4*x*exp(4*x)*exp(2)))/log(log(5)),x)

[Out]

x*exp(x*log(x)^2)*exp(-x^2)*exp((exp(4*x)*exp(2))/log(log(5)))

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sympy [A]  time = 0.52, size = 31, normalized size = 1.11 \begin {gather*} x e^{\frac {\left (- x^{2} + x \log {\relax (x )}^{2}\right ) \log {\left (\log {\relax (5 )} \right )} + e^{2} e^{4 x}}{\log {\left (\log {\relax (5 )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)**2+2*x*ln(x)-2*x**2+1)*ln(ln(5))+4*x*exp(2)*exp(4*x))*exp(((x*ln(x)**2-x**2)*ln(ln(5))+exp
(2)*exp(4*x))/ln(ln(5)))/ln(ln(5)),x)

[Out]

x*exp(((-x**2 + x*log(x)**2)*log(log(5)) + exp(2)*exp(4*x))/log(log(5)))

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