3.56.100 \(\int \frac {e^{20 x+x \log (\frac {1-x \log (\frac {e^{-x} \log (2)}{x})}{x})} (-19-x-x^2+20 x \log (\frac {e^{-x} \log (2)}{x})+(-1+x \log (\frac {e^{-x} \log (2)}{x})) \log (\frac {1-x \log (\frac {e^{-x} \log (2)}{x})}{x}))}{-1+x \log (\frac {e^{-x} \log (2)}{x})} \, dx\)

Optimal. Leaf size=25 \[ e^{x \left (20+\log \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )\right )} \]

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Rubi [A]  time = 1.63, antiderivative size = 29, normalized size of antiderivative = 1.16, number of steps used = 1, number of rules used = 1, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6706} \begin {gather*} e^{20 x} \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(20*x + x*Log[(1 - x*Log[Log[2]/(E^x*x)])/x])*(-19 - x - x^2 + 20*x*Log[Log[2]/(E^x*x)] + (-1 + x*Log[L
og[2]/(E^x*x)])*Log[(1 - x*Log[Log[2]/(E^x*x)])/x]))/(-1 + x*Log[Log[2]/(E^x*x)]),x]

[Out]

E^(20*x)*((1 - x*Log[Log[2]/(E^x*x)])/x)^x

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{20 x} \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )^x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 26, normalized size = 1.04 \begin {gather*} e^{20 x} \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )^x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(20*x + x*Log[(1 - x*Log[Log[2]/(E^x*x)])/x])*(-19 - x - x^2 + 20*x*Log[Log[2]/(E^x*x)] + (-1 + x
*Log[Log[2]/(E^x*x)])*Log[(1 - x*Log[Log[2]/(E^x*x)])/x]))/(-1 + x*Log[Log[2]/(E^x*x)]),x]

[Out]

E^(20*x)*(x^(-1) - Log[Log[2]/(E^x*x)])^x

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fricas [A]  time = 0.54, size = 28, normalized size = 1.12 \begin {gather*} e^{\left (x \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \relax (2)}{x}\right ) - 1}{x}\right ) + 20 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x*log(log(2)/exp(x)/x)-x^2-x-19)*e
xp(x*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm="fricas")

[Out]

e^(x*log(-(x*log(e^(-x)*log(2)/x) - 1)/x) + 20*x)

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giac [A]  time = 0.64, size = 25, normalized size = 1.00 \begin {gather*} e^{\left (x \log \left (\frac {1}{x} - \log \left (\frac {e^{\left (-x\right )} \log \relax (2)}{x}\right )\right ) + 20 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x*log(log(2)/exp(x)/x)-x^2-x-19)*e
xp(x*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm="giac")

[Out]

e^(x*log(1/x - log(e^(-x)*log(2)/x)) + 20*x)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (x \ln \left (\frac {\ln \relax (2) {\mathrm e}^{-x}}{x}\right )-1\right ) \ln \left (\frac {-x \ln \left (\frac {\ln \relax (2) {\mathrm e}^{-x}}{x}\right )+1}{x}\right )+20 x \ln \left (\frac {\ln \relax (2) {\mathrm e}^{-x}}{x}\right )-x^{2}-x -19\right ) {\mathrm e}^{x \ln \left (\frac {-x \ln \left (\frac {\ln \relax (2) {\mathrm e}^{-x}}{x}\right )+1}{x}\right )+20 x}}{x \ln \left (\frac {\ln \relax (2) {\mathrm e}^{-x}}{x}\right )-1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(ln(2)/exp(x)/x)-1)*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x*ln(ln(2)/exp(x)/x)-x^2-x-19)*exp(x*ln((-x*l
n(ln(2)/exp(x)/x)+1)/x)+20*x)/(x*ln(ln(2)/exp(x)/x)-1),x)

[Out]

int(((x*ln(ln(2)/exp(x)/x)-1)*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x*ln(ln(2)/exp(x)/x)-x^2-x-19)*exp(x*ln((-x*l
n(ln(2)/exp(x)/x)+1)/x)+20*x)/(x*ln(ln(2)/exp(x)/x)-1),x)

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maxima [A]  time = 0.63, size = 28, normalized size = 1.12 \begin {gather*} e^{\left (x \log \left (x^{2} + x \log \relax (x) - x \log \left (\log \relax (2)\right ) + 1\right ) - x \log \relax (x) + 20 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x*log(log(2)/exp(x)/x)-x^2-x-19)*e
xp(x*log((-x*log(log(2)/exp(x)/x)+1)/x)+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm="maxima")

[Out]

e^(x*log(x^2 + x*log(x) - x*log(log(2)) + 1) - x*log(x) + 20*x)

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mupad [B]  time = 4.90, size = 21, normalized size = 0.84 \begin {gather*} {\mathrm {e}}^{20\,x}\,{\left (x-\ln \left (\frac {\ln \relax (2)}{x}\right )+\frac {1}{x}\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(20*x + x*log(-(x*log((exp(-x)*log(2))/x) - 1)/x))*(x - log(-(x*log((exp(-x)*log(2))/x) - 1)/x)*(x*lo
g((exp(-x)*log(2))/x) - 1) - 20*x*log((exp(-x)*log(2))/x) + x^2 + 19))/(x*log((exp(-x)*log(2))/x) - 1),x)

[Out]

exp(20*x)*(x - log(log(2)/x) + 1/x)^x

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(ln(2)/exp(x)/x)-1)*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x*ln(ln(2)/exp(x)/x)-x**2-x-19)*exp(x*l
n((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x)/(x*ln(ln(2)/exp(x)/x)-1),x)

[Out]

Timed out

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