3.13.44 \(\int \frac {e^{12 x-3 x \log (\frac {-2+5 x}{x})} (-30+60 x+(6-15 x) \log (\frac {-2+5 x}{x}))}{-2+5 x} \, dx\)

Optimal. Leaf size=23 \[ e^{3 x \left (4-\log \left (5+x-\frac {2+x^2}{x}\right )\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6706} \begin {gather*} e^{12 x} \left (-\frac {2-5 x}{x}\right )^{-3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(12*x - 3*x*Log[(-2 + 5*x)/x])*(-30 + 60*x + (6 - 15*x)*Log[(-2 + 5*x)/x]))/(-2 + 5*x),x]

[Out]

E^(12*x)/(-((2 - 5*x)/x))^(3*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^{12 x} \left (-\frac {2-5 x}{x}\right )^{-3 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.04, size = 17, normalized size = 0.74 \begin {gather*} e^{12 x} \left (5-\frac {2}{x}\right )^{-3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(12*x - 3*x*Log[(-2 + 5*x)/x])*(-30 + 60*x + (6 - 15*x)*Log[(-2 + 5*x)/x]))/(-2 + 5*x),x]

[Out]

E^(12*x)/(5 - 2/x)^(3*x)

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fricas [A]  time = 0.69, size = 18, normalized size = 0.78 \begin {gather*} e^{\left (-3 \, x \log \left (\frac {5 \, x - 2}{x}\right ) + 12 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x+6)*log((5*x-2)/x)+60*x-30)*exp(-3*x*log((5*x-2)/x)+12*x)/(5*x-2),x, algorithm="fricas")

[Out]

e^(-3*x*log((5*x - 2)/x) + 12*x)

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giac [A]  time = 0.63, size = 16, normalized size = 0.70 \begin {gather*} e^{\left (-3 \, x \log \left (-\frac {2}{x} + 5\right ) + 12 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x+6)*log((5*x-2)/x)+60*x-30)*exp(-3*x*log((5*x-2)/x)+12*x)/(5*x-2),x, algorithm="giac")

[Out]

e^(-3*x*log(-2/x + 5) + 12*x)

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maple [A]  time = 0.23, size = 19, normalized size = 0.83




method result size



norman \({\mathrm e}^{-3 x \ln \left (\frac {5 x -2}{x}\right )+12 x}\) \(19\)
risch \(\left (\frac {5 x -2}{x}\right )^{-3 x} {\mathrm e}^{12 x}\) \(19\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x+6)*ln((5*x-2)/x)+60*x-30)*exp(-3*x*ln((5*x-2)/x)+12*x)/(5*x-2),x,method=_RETURNVERBOSE)

[Out]

exp(-3*x*ln((5*x-2)/x)+12*x)

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maxima [A]  time = 0.50, size = 19, normalized size = 0.83 \begin {gather*} e^{\left (-3 \, x \log \left (5 \, x - 2\right ) + 3 \, x \log \relax (x) + 12 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x+6)*log((5*x-2)/x)+60*x-30)*exp(-3*x*log((5*x-2)/x)+12*x)/(5*x-2),x, algorithm="maxima")

[Out]

e^(-3*x*log(5*x - 2) + 3*x*log(x) + 12*x)

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mupad [B]  time = 1.14, size = 18, normalized size = 0.78 \begin {gather*} \frac {{\mathrm {e}}^{12\,x}}{{\left (5-\frac {2}{x}\right )}^{3\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(12*x - 3*x*log((5*x - 2)/x))*(log((5*x - 2)/x)*(15*x - 6) - 60*x + 30))/(5*x - 2),x)

[Out]

exp(12*x)/(5 - 2/x)^(3*x)

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sympy [A]  time = 0.37, size = 15, normalized size = 0.65 \begin {gather*} e^{- 3 x \log {\left (\frac {5 x - 2}{x} \right )} + 12 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x+6)*ln((5*x-2)/x)+60*x-30)*exp(-3*x*ln((5*x-2)/x)+12*x)/(5*x-2),x)

[Out]

exp(-3*x*log((5*x - 2)/x) + 12*x)

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