3.41.78 \(\int \frac {2\ 3^{e^x (12-6 x)}+3^{e^x (12-6 x)} (2+e^x (12 x-12 x^2) \log (3)) \log (x) \log (\log (x))}{\log (x)} \, dx\)

Optimal. Leaf size=18 \[ 2\ 3^{6 e^x (2-x)} x \log (\log (x)) \]

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Rubi [F]  time = 1.18, 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 {2\ 3^{e^x (12-6 x)}+3^{e^x (12-6 x)} \left (2+e^x \left (12 x-12 x^2\right ) \log (3)\right ) \log (x) \log (\log (x))}{\log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*3^(E^x*(12 - 6*x)) + 3^(E^x*(12 - 6*x))*(2 + E^x*(12*x - 12*x^2)*Log[3])*Log[x]*Log[Log[x]])/Log[x],x]

[Out]

2*Defer[Int][1/(3^(6*E^x*(-2 + x))*Log[x]), x] + 2*Defer[Int][Log[Log[x]]/3^(6*E^x*(-2 + x)), x] + 4*Log[3]*De
fer[Int][3^(1 - 6*E^x*(-2 + x))*E^x*x*Log[Log[x]], x] - 4*Log[3]*Defer[Int][3^(1 - 6*E^x*(-2 + x))*E^x*x^2*Log
[Log[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3^{-6 e^x (-2+x)} \left (2-2 \left (-1+6 e^x (-1+x) x \log (3)\right ) \log (x) \log (\log (x))\right )}{\log (x)} \, dx\\ &=\int \left (-4 3^{1-6 e^x (-2+x)} e^x (-1+x) x \log (3) \log (\log (x))+\frac {2\ 3^{-6 e^x (-2+x)} (1+\log (x) \log (\log (x)))}{\log (x)}\right ) \, dx\\ &=2 \int \frac {3^{-6 e^x (-2+x)} (1+\log (x) \log (\log (x)))}{\log (x)} \, dx-(4 \log (3)) \int 3^{1-6 e^x (-2+x)} e^x (-1+x) x \log (\log (x)) \, dx\\ &=2 \int \left (\frac {3^{-6 e^x (-2+x)}}{\log (x)}+3^{-6 e^x (-2+x)} \log (\log (x))\right ) \, dx-(4 \log (3)) \int \left (-3^{1-6 e^x (-2+x)} e^x x \log (\log (x))+3^{1-6 e^x (-2+x)} e^x x^2 \log (\log (x))\right ) \, dx\\ &=2 \int \frac {3^{-6 e^x (-2+x)}}{\log (x)} \, dx+2 \int 3^{-6 e^x (-2+x)} \log (\log (x)) \, dx+(4 \log (3)) \int 3^{1-6 e^x (-2+x)} e^x x \log (\log (x)) \, dx-(4 \log (3)) \int 3^{1-6 e^x (-2+x)} e^x x^2 \log (\log (x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 16, normalized size = 0.89 \begin {gather*} 2\ 3^{-6 e^x (-2+x)} x \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*3^(E^x*(12 - 6*x)) + 3^(E^x*(12 - 6*x))*(2 + E^x*(12*x - 12*x^2)*Log[3])*Log[x]*Log[Log[x]])/Log[
x],x]

[Out]

(2*x*Log[Log[x]])/3^(6*E^x*(-2 + x))

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fricas [A]  time = 0.63, size = 17, normalized size = 0.94 \begin {gather*} \frac {2 \, x \log \left (\log \relax (x)\right )}{3^{6 \, {\left (x - 2\right )} e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^2+12*x)*log(3)*exp(x)+2)*log(x)*exp((12-6*x)*log(3)*exp(x))*log(log(x))+2*exp((12-6*x)*log(
3)*exp(x)))/log(x),x, algorithm="fricas")

[Out]

2*x*log(log(x))/3^(6*(x - 2)*e^x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (\frac {{\left (6 \, {\left (x^{2} - x\right )} e^{x} \log \relax (3) - 1\right )} \log \relax (x) \log \left (\log \relax (x)\right )}{3^{6 \, {\left (x - 2\right )} e^{x}}} - \frac {1}{3^{6 \, {\left (x - 2\right )} e^{x}}}\right )}}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^2+12*x)*log(3)*exp(x)+2)*log(x)*exp((12-6*x)*log(3)*exp(x))*log(log(x))+2*exp((12-6*x)*log(
3)*exp(x)))/log(x),x, algorithm="giac")

[Out]

integrate(-2*((6*(x^2 - x)*e^x*log(3) - 1)*log(x)*log(log(x))/3^(6*(x - 2)*e^x) - 1/3^(6*(x - 2)*e^x))/log(x),
x)

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maple [A]  time = 0.06, size = 15, normalized size = 0.83

method	result	size
risch	\(2 x \left (\frac {1}{729}\right )^{{\mathrm e}^{x} \left (x -2\right )} \ln \left (\ln \relax (x )\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-12*x^2+12*x)*ln(3)*exp(x)+2)*ln(x)*exp((12-6*x)*ln(3)*exp(x))*ln(ln(x))+2*exp((12-6*x)*ln(3)*exp(x)))/
ln(x),x,method=_RETURNVERBOSE)

[Out]

2*x*(1/729)^(exp(x)*(x-2))*ln(ln(x))

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maxima [A]  time = 0.52, size = 21, normalized size = 1.17 \begin {gather*} 2 \, x e^{\left (-6 \, x e^{x} \log \relax (3) + 12 \, e^{x} \log \relax (3)\right )} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^2+12*x)*log(3)*exp(x)+2)*log(x)*exp((12-6*x)*log(3)*exp(x))*log(log(x))+2*exp((12-6*x)*log(
3)*exp(x)))/log(x),x, algorithm="maxima")

[Out]

2*x*e^(-6*x*e^x*log(3) + 12*e^x*log(3))*log(log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {2\,{\mathrm {e}}^{-{\mathrm {e}}^x\,\ln \relax (3)\,\left (6\,x-12\right )}+\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-{\mathrm {e}}^x\,\ln \relax (3)\,\left (6\,x-12\right )}\,\ln \relax (x)\,\left ({\mathrm {e}}^x\,\ln \relax (3)\,\left (12\,x-12\,x^2\right )+2\right )}{\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(-exp(x)*log(3)*(6*x - 12)) + log(log(x))*exp(-exp(x)*log(3)*(6*x - 12))*log(x)*(exp(x)*log(3)*(12*x
- 12*x^2) + 2))/log(x),x)

[Out]

int((2*exp(-exp(x)*log(3)*(6*x - 12)) + log(log(x))*exp(-exp(x)*log(3)*(6*x - 12))*log(x)*(exp(x)*log(3)*(12*x
- 12*x^2) + 2))/log(x), x)

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sympy [A]  time = 74.06, size = 20, normalized size = 1.11 \begin {gather*} 2 x e^{\left (12 - 6 x\right ) e^{x} \log {\relax (3 )}} \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x**2+12*x)*ln(3)*exp(x)+2)*ln(x)*exp((12-6*x)*ln(3)*exp(x))*ln(ln(x))+2*exp((12-6*x)*ln(3)*ex
p(x)))/ln(x),x)

[Out]

2*x*exp((12 - 6*x)*exp(x)*log(3))*log(log(x))

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