3.82.98 \(\int \frac {-120+120 \log (\frac {2}{x})+e^{e^x-x} (-12 x+12 x^2-12 e^x x^2) \log (\frac {2}{x})}{x \log (\frac {2}{x})} \, dx\)

Optimal. Leaf size=31 \[ 12 \left (1-e^{e^x-x} x+5 \log \left (x^2 \log ^2\left (\frac {2}{x}\right )\right )\right ) \]

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Rubi [F]  time = 0.95, 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 {-120+120 \log \left (\frac {2}{x}\right )+e^{e^x-x} \left (-12 x+12 x^2-12 e^x x^2\right ) \log \left (\frac {2}{x}\right )}{x \log \left (\frac {2}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-120 + 120*Log[2/x] + E^(E^x - x)*(-12*x + 12*x^2 - 12*E^x*x^2)*Log[2/x])/(x*Log[2/x]),x]

[Out]

12*E^(E^x - x) - 12*ExpIntegralEi[E^x] + 120*Log[x] + 120*Log[Log[2/x]] - 12*Defer[Int][E^E^x*x, x] + 12*Defer
[Int][E^(E^x - x)*x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (12 e^{e^x-x} (-1+x)-\frac {12 \left (10-10 \log \left (\frac {2}{x}\right )+e^{e^x} x^2 \log \left (\frac {2}{x}\right )\right )}{x \log \left (\frac {2}{x}\right )}\right ) \, dx\\ &=12 \int e^{e^x-x} (-1+x) \, dx-12 \int \frac {10-10 \log \left (\frac {2}{x}\right )+e^{e^x} x^2 \log \left (\frac {2}{x}\right )}{x \log \left (\frac {2}{x}\right )} \, dx\\ &=12 \int \left (-e^{e^x-x}+e^{e^x-x} x\right ) \, dx-12 \int \frac {-10+e^{e^x} x^2+\frac {10}{\log \left (\frac {2}{x}\right )}}{x} \, dx\\ &=-\left (12 \int e^{e^x-x} \, dx\right )+12 \int e^{e^x-x} x \, dx-12 \int \left (e^{e^x} x-\frac {10 \left (-1+\log \left (\frac {2}{x}\right )\right )}{x \log \left (\frac {2}{x}\right )}\right ) \, dx\\ &=-\left (12 \int e^{e^x} x \, dx\right )+12 \int e^{e^x-x} x \, dx-12 \operatorname {Subst}\left (\int \frac {e^x}{x^2} \, dx,x,e^x\right )+120 \int \frac {-1+\log \left (\frac {2}{x}\right )}{x \log \left (\frac {2}{x}\right )} \, dx\\ &=12 e^{e^x-x}-12 \int e^{e^x} x \, dx+12 \int e^{e^x-x} x \, dx-12 \operatorname {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right )-120 \operatorname {Subst}\left (\int \frac {-1+x}{x} \, dx,x,\log \left (\frac {2}{x}\right )\right )\\ &=12 e^{e^x-x}-12 \text {Ei}\left (e^x\right )-12 \int e^{e^x} x \, dx+12 \int e^{e^x-x} x \, dx-120 \operatorname {Subst}\left (\int \left (1-\frac {1}{x}\right ) \, dx,x,\log \left (\frac {2}{x}\right )\right )\\ &=12 e^{e^x-x}-12 \text {Ei}\left (e^x\right )+120 \log (x)+120 \log \left (\log \left (\frac {2}{x}\right )\right )-12 \int e^{e^x} x \, dx+12 \int e^{e^x-x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 28, normalized size = 0.90 \begin {gather*} 12 \left (-e^{e^x-x} x+10 \log (x)+10 \log \left (\log \left (\frac {2}{x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-120 + 120*Log[2/x] + E^(E^x - x)*(-12*x + 12*x^2 - 12*E^x*x^2)*Log[2/x])/(x*Log[2/x]),x]

[Out]

12*(-(E^(E^x - x)*x) + 10*Log[x] + 10*Log[Log[2/x]])

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fricas [A]  time = 0.72, size = 28, normalized size = 0.90 \begin {gather*} -12 \, x e^{\left (-x + e^{x}\right )} - 120 \, \log \left (\frac {2}{x}\right ) + 120 \, \log \left (\log \left (\frac {2}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(x)*x^2+12*x^2-12*x)*log(2/x)*exp(exp(x)-x)+120*log(2/x)-120)/x/log(2/x),x, algorithm="fric
as")

[Out]

-12*x*e^(-x + e^x) - 120*log(2/x) + 120*log(log(2/x))

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giac [A]  time = 0.14, size = 24, normalized size = 0.77 \begin {gather*} -12 \, x e^{\left (-x + e^{x}\right )} + 120 \, \log \relax (x) + 120 \, \log \left (\log \left (\frac {2}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(x)*x^2+12*x^2-12*x)*log(2/x)*exp(exp(x)-x)+120*log(2/x)-120)/x/log(2/x),x, algorithm="giac
")

[Out]

-12*x*e^(-x + e^x) + 120*log(x) + 120*log(log(2/x))

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maple [A]  time = 0.10, size = 26, normalized size = 0.84

method	result	size
risch	\(120 \ln \relax (x )+120 \ln \left (\ln \relax (x )-\ln \relax (2)\right )-12 \,{\mathrm e}^{{\mathrm e}^{x}-x} x\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*exp(x)*x^2+12*x^2-12*x)*ln(2/x)*exp(exp(x)-x)+120*ln(2/x)-120)/x/ln(2/x),x,method=_RETURNVERBOSE)

[Out]

120*ln(x)+120*ln(ln(x)-ln(2))-12*exp(exp(x)-x)*x

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maxima [A]  time = 0.40, size = 24, normalized size = 0.77 \begin {gather*} -12 \, x e^{\left (-x + e^{x}\right )} + 120 \, \log \relax (x) + 120 \, \log \left (\log \left (\frac {2}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(x)*x^2+12*x^2-12*x)*log(2/x)*exp(exp(x)-x)+120*log(2/x)-120)/x/log(2/x),x, algorithm="maxi
ma")

[Out]

-12*x*e^(-x + e^x) + 120*log(x) + 120*log(log(2/x))

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mupad [B]  time = 7.32, size = 24, normalized size = 0.77 \begin {gather*} 120\,\ln \left (\ln \left (\frac {2}{x}\right )\right )+120\,\ln \relax (x)-12\,x\,{\mathrm {e}}^{{\mathrm {e}}^x-x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x) - x)*log(2/x)*(12*x + 12*x^2*exp(x) - 12*x^2) - 120*log(2/x) + 120)/(x*log(2/x)),x)

[Out]

120*log(log(2/x)) + 120*log(x) - 12*x*exp(exp(x) - x)

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sympy [A]  time = 0.40, size = 22, normalized size = 0.71 \begin {gather*} - 12 x e^{- x + e^{x}} + 120 \log {\relax (x )} + 120 \log {\left (\log {\left (\frac {2}{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(x)*x**2+12*x**2-12*x)*ln(2/x)*exp(exp(x)-x)+120*ln(2/x)-120)/x/ln(2/x),x)

[Out]

-12*x*exp(-x + exp(x)) + 120*log(x) + 120*log(log(2/x))

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