3.26.20 \(\int \frac {(-2-x) \log (5-\log (4))}{x} \, dx\)

Optimal. Leaf size=19 \[ \left (-4-x+\log \left (\frac {4}{x^2}\right )\right ) \log (5-\log (4)) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 43} \begin {gather*} x (-\log (5-\log (4)))-2 \log (5-\log (4)) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-2 - x)*Log[5 - Log[4]])/x,x]

[Out]

-(x*Log[5 - Log[4]]) - 2*Log[x]*Log[5 - Log[4]]

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (5-\log (4)) \int \frac {-2-x}{x} \, dx\\ &=\log (5-\log (4)) \int \left (-1-\frac {2}{x}\right ) \, dx\\ &=-x \log (5-\log (4))-2 \log (x) \log (5-\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.79 \begin {gather*} -((x+2 \log (x)) \log (5-\log (4))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2 - x)*Log[5 - Log[4]])/x,x]

[Out]

-((x + 2*Log[x])*Log[5 - Log[4]])

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fricas [A]  time = 0.67, size = 22, normalized size = 1.16 \begin {gather*} -x \log \left (-2 \, \log \relax (2) + 5\right ) - 2 \, \log \relax (x) \log \left (-2 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(-2*log(2) + 5) - 2*log(x)*log(-2*log(2) + 5)

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giac [A]  time = 0.12, size = 16, normalized size = 0.84 \begin {gather*} -{\left (x + 2 \, \log \left ({\left | x \right |}\right )\right )} \log \left (-2 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x + 2*log(abs(x)))*log(-2*log(2) + 5)

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maple [A]  time = 0.02, size = 17, normalized size = 0.89

method	result	size
default	\(\ln \left (-2 \ln \relax (2)+5\right ) \left (-x -2 \ln \relax (x )\right )\)	\(17\)
norman	\(-\ln \left (-2 \ln \relax (2)+5\right ) x -2 \ln \left (-2 \ln \relax (2)+5\right ) \ln \relax (x )\)	\(23\)
risch	\(-\ln \left (-2 \ln \relax (2)+5\right ) x -2 \ln \left (-2 \ln \relax (2)+5\right ) \ln \relax (x )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(-2*ln(2)+5)*(-x-2*ln(x))

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maxima [A]  time = 0.39, size = 15, normalized size = 0.79 \begin {gather*} -{\left (x + 2 \, \log \relax (x)\right )} \log \left (-2 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x + 2*log(x))*log(-2*log(2) + 5)

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mupad [B]  time = 0.05, size = 15, normalized size = 0.79 \begin {gather*} -\ln \left (5-\ln \relax (4)\right )\,\left (x+2\,\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5 - 2*log(2))*(x + 2))/x,x)

[Out]

-log(5 - log(4))*(x + 2*log(x))

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sympy [A]  time = 0.11, size = 24, normalized size = 1.26 \begin {gather*} - x \log {\left (5 - 2 \log {\relax (2 )} \right )} - 2 \log {\relax (x )} \log {\left (5 - 2 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(5 - 2*log(2)) - 2*log(x)*log(5 - 2*log(2))

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