3.60.11 \(\int \frac {e^{3-x \log (\frac {1}{3} (e^5 x-3 \log (x)))} (-12+4 e^5 x+(4 e^5 x-12 \log (x)) \log (\frac {1}{3} (e^5 x-3 \log (x))))}{-e^5 x+3 \log (x)} \, dx\)

Optimal. Leaf size=25 \[ 4 \left (-4+e^{3-x \log \left (\frac {e^5 x}{3}-\log (x)\right )}\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 1, number of rules used = 1, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6706} \begin {gather*} 4 e^3 3^x \left (e^5 x-3 \log (x)\right )^{-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 - x*Log[(E^5*x - 3*Log[x])/3])*(-12 + 4*E^5*x + (4*E^5*x - 12*Log[x])*Log[(E^5*x - 3*Log[x])/3]))/(-
(E^5*x) + 3*Log[x]),x]

[Out]

(4*3^x*E^3)/(E^5*x - 3*Log[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} &=4\ 3^x e^3 \left (e^5 x-3 \log (x)\right )^{-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.73, size = 22, normalized size = 0.88 \begin {gather*} 4 e^3 \left (\frac {e^5 x}{3}-\log (x)\right )^{-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 - x*Log[(E^5*x - 3*Log[x])/3])*(-12 + 4*E^5*x + (4*E^5*x - 12*Log[x])*Log[(E^5*x - 3*Log[x])/3
]))/(-(E^5*x) + 3*Log[x]),x]

[Out]

(4*E^3)/((E^5*x)/3 - Log[x])^x

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fricas [A]  time = 0.69, size = 19, normalized size = 0.76 \begin {gather*} 4 \, e^{\left (-x \log \left (\frac {1}{3} \, x e^{5} - \log \relax (x)\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(x)+4*x*exp(5))*log(-log(x)+1/3*x*exp(5))+4*x*exp(5)-12)*exp(-x*log(-log(x)+1/3*x*exp(5))+3
)/(3*log(x)-x*exp(5)),x, algorithm="fricas")

[Out]

4*e^(-x*log(1/3*x*e^5 - log(x)) + 3)

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giac [A]  time = 0.24, size = 19, normalized size = 0.76 \begin {gather*} 4 \, e^{\left (-x \log \left (\frac {1}{3} \, x e^{5} - \log \relax (x)\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(x)+4*x*exp(5))*log(-log(x)+1/3*x*exp(5))+4*x*exp(5)-12)*exp(-x*log(-log(x)+1/3*x*exp(5))+3
)/(3*log(x)-x*exp(5)),x, algorithm="giac")

[Out]

4*e^(-x*log(1/3*x*e^5 - log(x)) + 3)

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maple [A]  time = 0.10, size = 19, normalized size = 0.76

method	result	size
risch	\(4 \left (-\ln \relax (x )+\frac {x \,{\mathrm e}^{5}}{3}\right )^{-x} {\mathrm e}^{3}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*ln(x)+4*x*exp(5))*ln(-ln(x)+1/3*x*exp(5))+4*x*exp(5)-12)*exp(-x*ln(-ln(x)+1/3*x*exp(5))+3)/(3*ln(x)-
x*exp(5)),x,method=_RETURNVERBOSE)

[Out]

4*(-ln(x)+1/3*x*exp(5))^(-x)*exp(3)

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maxima [A]  time = 0.51, size = 22, normalized size = 0.88 \begin {gather*} 4 \, e^{\left (x \log \relax (3) - x \log \left (x e^{5} - 3 \, \log \relax (x)\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(x)+4*x*exp(5))*log(-log(x)+1/3*x*exp(5))+4*x*exp(5)-12)*exp(-x*log(-log(x)+1/3*x*exp(5))+3
)/(3*log(x)-x*exp(5)),x, algorithm="maxima")

[Out]

4*e^(x*log(3) - x*log(x*e^5 - 3*log(x)) + 3)

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mupad [B]  time = 4.48, size = 18, normalized size = 0.72 \begin {gather*} \frac {4\,{\mathrm {e}}^3}{{\left (\frac {x\,{\mathrm {e}}^5}{3}-\ln \relax (x)\right )}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3 - x*log((x*exp(5))/3 - log(x)))*(log((x*exp(5))/3 - log(x))*(12*log(x) - 4*x*exp(5)) - 4*x*exp(5)
+ 12))/(3*log(x) - x*exp(5)),x)

[Out]

(4*exp(3))/((x*exp(5))/3 - log(x))^x

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sympy [A]  time = 0.79, size = 17, normalized size = 0.68 \begin {gather*} 4 e^{- x \log {\left (\frac {x e^{5}}{3} - \log {\relax (x )} \right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*ln(x)+4*x*exp(5))*ln(-ln(x)+1/3*x*exp(5))+4*x*exp(5)-12)*exp(-x*ln(-ln(x)+1/3*x*exp(5))+3)/(3*
ln(x)-x*exp(5)),x)

[Out]

4*exp(-x*log(x*exp(5)/3 - log(x)) + 3)

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