3.89.42 \(\int \frac {-1-2 \log (\frac {1}{81+x})}{81+x} \, dx\)

Optimal. Leaf size=18 \[ \left (\frac {1}{2}+\log \left (\frac {x}{81 x+x^2}\right )\right )^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2390, 2301} \begin {gather*} \frac {1}{4} \left (2 \log \left (\frac {1}{x+81}\right )+1\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 2*Log[(81 + x)^(-1)])/(81 + x),x]

[Out]

(1 + 2*Log[(81 + x)^(-1)])^2/4

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 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {-1-2 \log \left (\frac {1}{x}\right )}{x} \, dx,x,81+x\right )\\ &=\frac {1}{4} \left (1+2 \log \left (\frac {1}{81+x}\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.89 \begin {gather*} \frac {1}{4} \left (1+2 \log \left (\frac {1}{81+x}\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 2*Log[(81 + x)^(-1)])/(81 + x),x]

[Out]

(1 + 2*Log[(81 + x)^(-1)])^2/4

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fricas [A]  time = 0.52, size = 15, normalized size = 0.83 \begin {gather*} \log \left (\frac {1}{x + 81}\right )^{2} + \log \left (\frac {1}{x + 81}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(1/(81+x))-1)/(81+x),x, algorithm="fricas")

[Out]

log(1/(x + 81))^2 + log(1/(x + 81))

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giac [A]  time = 0.12, size = 13, normalized size = 0.72 \begin {gather*} \log \left (x + 81\right )^{2} - \log \left (x + 81\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(1/(81+x))-1)/(81+x),x, algorithm="giac")

[Out]

log(x + 81)^2 - log(x + 81)

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maple [A]  time = 0.47, size = 16, normalized size = 0.89

method	result	size
derivativedivides	\(\ln \left (\frac {1}{81+x}\right )^{2}+\ln \left (\frac {1}{81+x}\right )\)	\(16\)
default	\(\ln \left (\frac {1}{81+x}\right )^{2}+\ln \left (\frac {1}{81+x}\right )\)	\(16\)
norman	\(\ln \left (\frac {1}{81+x}\right )^{2}+\ln \left (\frac {1}{81+x}\right )\)	\(16\)
risch	\(-\ln \left (81+x \right )+\ln \left (\frac {1}{81+x}\right )^{2}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(1/(81+x))-1)/(81+x),x,method=_RETURNVERBOSE)

[Out]

ln(1/(81+x))^2+ln(1/(81+x))

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maxima [A]  time = 0.37, size = 12, normalized size = 0.67 \begin {gather*} \frac {1}{4} \, {\left (2 \, \log \left (x + 81\right ) - 1\right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(1/(81+x))-1)/(81+x),x, algorithm="maxima")

[Out]

1/4*(2*log(x + 81) - 1)^2

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mupad [B]  time = 5.11, size = 15, normalized size = 0.83 \begin {gather*} {\ln \left (\frac {1}{x+81}\right )}^2+\ln \left (\frac {1}{x+81}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(1/(x + 81)) + 1)/(x + 81),x)

[Out]

log(1/(x + 81)) + log(1/(x + 81))^2

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sympy [A]  time = 0.09, size = 12, normalized size = 0.67 \begin {gather*} \log {\left (\frac {1}{x + 81} \right )}^{2} - \log {\left (x + 81 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(1/(81+x))-1)/(81+x),x)

[Out]

log(1/(x + 81))**2 - log(x + 81)

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