3.7.69 \(\int \frac {2 \log (\frac {1}{3} (3 x-e^5 x))}{x} \, dx\)

Optimal. Leaf size=13 \[ \log ^2\left (x-\frac {e^5 x}{3}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {12, 2421, 2301} \begin {gather*} \log ^2\left (\frac {1}{3} \left (3-e^5\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*Log[(3*x - E^5*x)/3])/x,x]

[Out]

Log[((3 - E^5)*x)/3]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2421

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int \frac {\log \left (\frac {1}{3} \left (3 x-e^5 x\right )\right )}{x} \, dx\\ &=2 \int \frac {\log \left (\frac {1}{3} \left (3-e^5\right ) x\right )}{x} \, dx\\ &=\log ^2\left (\frac {1}{3} \left (3-e^5\right ) x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 25, normalized size = 1.92 \begin {gather*} 2 \left (-\frac {1}{2} \log ^2(x)+\log (x) \log \left (x-\frac {e^5 x}{3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*Log[(3*x - E^5*x)/3])/x,x]

[Out]

2*(-1/2*Log[x]^2 + Log[x]*Log[x - (E^5*x)/3])

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fricas [A]  time = 0.63, size = 10, normalized size = 0.77 \begin {gather*} \log \left (-\frac {1}{3} \, x e^{5} + x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(-1/3*x*exp(5)+x)/x,x, algorithm="fricas")

[Out]

log(-1/3*x*e^5 + x)^2

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giac [A]  time = 0.31, size = 10, normalized size = 0.77 \begin {gather*} \log \left (-\frac {1}{3} \, x e^{5} + x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(-1/3*x*exp(5)+x)/x,x, algorithm="giac")

[Out]

log(-1/3*x*e^5 + x)^2

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maple [A]  time = 0.05, size = 11, normalized size = 0.85

method	result	size
norman	\(\ln \left (-\frac {x \,{\mathrm e}^{5}}{3}+x \right )^{2}\)	\(11\)
risch	\(\ln \left (-\frac {x \,{\mathrm e}^{5}}{3}+x \right )^{2}\)	\(11\)
derivativedivides	\(\ln \left (\left (-\frac {{\mathrm e}^{5}}{3}+1\right ) x \right )^{2}\)	\(12\)
default	\(\ln \left (\left (-\frac {{\mathrm e}^{5}}{3}+1\right ) x \right )^{2}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*ln(-1/3*x*exp(5)+x)/x,x,method=_RETURNVERBOSE)

[Out]

ln(-1/3*x*exp(5)+x)^2

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maxima [B]  time = 0.65, size = 21, normalized size = 1.62 \begin {gather*} -2 \, {\left (\log \relax (3) - \log \left (-e^{5} + 3\right )\right )} \log \relax (x) + \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(-1/3*x*exp(5)+x)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.02, size = 10, normalized size = 0.77 \begin {gather*} {\ln \left (x-\frac {x\,{\mathrm {e}}^5}{3}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(x - (x*exp(5))/3))/x,x)

[Out]

log(x - (x*exp(5))/3)^2

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sympy [A]  time = 0.10, size = 10, normalized size = 0.77 \begin {gather*} \log {\left (- \frac {x e^{5}}{3} + x \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*ln(-1/3*x*exp(5)+x)/x,x)

[Out]

log(-x*exp(5)/3 + x)**2

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