3.34.94 \(\int \frac {e^{-1-x+\log ^2(\frac {-60+20 x}{x})} (3 x-x^2+6 \log (\frac {-60+20 x}{x}))}{-3 x+x^2} \, dx\)

Optimal. Leaf size=18 \[ e^{-1-x+\log ^2\left (\frac {20 (-3+x)}{x}\right )} \]

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Rubi [A]  time = 0.56, antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {1593, 6706} \begin {gather*} e^{-x+\log ^2\left (-\frac {20 (3-x)}{x}\right )-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 - x + Log[(-60 + 20*x)/x]^2)*(3*x - x^2 + 6*Log[(-60 + 20*x)/x]))/(-3*x + x^2),x]

[Out]

E^(-1 - x + Log[(-20*(3 - x))/x]^2)

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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} &=\int \frac {e^{-1-x+\log ^2\left (\frac {-60+20 x}{x}\right )} \left (3 x-x^2+6 \log \left (\frac {-60+20 x}{x}\right )\right )}{(-3+x) x} \, dx\\ &=e^{-1-x+\log ^2\left (-\frac {20 (3-x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.04, size = 18, normalized size = 1.00 \begin {gather*} e^{-1-x+\log ^2\left (\frac {20 (-3+x)}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 - x + Log[(-60 + 20*x)/x]^2)*(3*x - x^2 + 6*Log[(-60 + 20*x)/x]))/(-3*x + x^2),x]

[Out]

E^(-1 - x + Log[(20*(-3 + x))/x]^2)

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fricas [A]  time = 0.88, size = 17, normalized size = 0.94 \begin {gather*} e^{\left (\log \left (\frac {20 \, {\left (x - 3\right )}}{x}\right )^{2} - x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log((20*x-60)/x)-x^2+3*x)*exp(log((20*x-60)/x)^2-x-1)/(x^2-3*x),x, algorithm="fricas")

[Out]

e^(log(20*(x - 3)/x)^2 - x - 1)

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giac [A]  time = 0.29, size = 16, normalized size = 0.89 \begin {gather*} e^{\left (\log \left (-\frac {60}{x} + 20\right )^{2} - x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log((20*x-60)/x)-x^2+3*x)*exp(log((20*x-60)/x)^2-x-1)/(x^2-3*x),x, algorithm="giac")

[Out]

e^(log(-60/x + 20)^2 - x - 1)

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maple [A]  time = 0.47, size = 19, normalized size = 1.06

method	result	size
norman	\({\mathrm e}^{\ln \left (\frac {20 x -60}{x}\right )^{2}-x -1}\)	\(19\)
risch	\({\mathrm e}^{\ln \left (\frac {20 x -60}{x}\right )^{2}-x -1}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*ln((20*x-60)/x)-x^2+3*x)*exp(ln((20*x-60)/x)^2-x-1)/(x^2-3*x),x,method=_RETURNVERBOSE)

[Out]

exp(ln((20*x-60)/x)^2-x-1)

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maxima [B]  time = 0.99, size = 69, normalized size = 3.83 \begin {gather*} 2^{4 \, \log \relax (5)} e^{\left (\log \relax (5)^{2} + 4 \, \log \relax (2)^{2} + 2 \, \log \relax (5) \log \left (x - 3\right ) + 4 \, \log \relax (2) \log \left (x - 3\right ) + \log \left (x - 3\right )^{2} - 2 \, \log \relax (5) \log \relax (x) - 4 \, \log \relax (2) \log \relax (x) - 2 \, \log \left (x - 3\right ) \log \relax (x) + \log \relax (x)^{2} - x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log((20*x-60)/x)-x^2+3*x)*exp(log((20*x-60)/x)^2-x-1)/(x^2-3*x),x, algorithm="maxima")

[Out]

2^(4*log(5))*e^(log(5)^2 + 4*log(2)^2 + 2*log(5)*log(x - 3) + 4*log(2)*log(x - 3) + log(x - 3)^2 - 2*log(5)*lo
g(x) - 4*log(2)*log(x) - 2*log(x - 3)*log(x) + log(x)^2 - x - 1)

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mupad [B]  time = 2.21, size = 20, normalized size = 1.11 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{{\ln \left (\frac {20\,x-60}{x}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log((20*x - 60)/x)^2 - x - 1)*(3*x + 6*log((20*x - 60)/x) - x^2))/(3*x - x^2),x)

[Out]

exp(-x)*exp(-1)*exp(log((20*x - 60)/x)^2)

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sympy [A]  time = 0.38, size = 14, normalized size = 0.78 \begin {gather*} e^{- x + \log {\left (\frac {20 x - 60}{x} \right )}^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*ln((20*x-60)/x)-x**2+3*x)*exp(ln((20*x-60)/x)**2-x-1)/(x**2-3*x),x)

[Out]

exp(-x + log((20*x - 60)/x)**2 - 1)

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