3.69.35 \(\int -\frac {2 \log (-\frac {1}{5+x})}{5+x} \, dx\)

Optimal. Leaf size=10 \[ \log ^2\left (-\frac {1}{5+x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2390, 2301} \begin {gather*} \log ^2\left (-\frac {1}{x+5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*Log[-(5 + x)^(-1)])/(5 + x),x]

[Out]

Log[-(5 + x)^(-1)]^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 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} &=-\left (2 \int \frac {\log \left (-\frac {1}{5+x}\right )}{5+x} \, dx\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {\log \left (-\frac {1}{x}\right )}{x} \, dx,x,5+x\right )\right )\\ &=\log ^2\left (-\frac {1}{5+x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \begin {gather*} \log ^2\left (-\frac {1}{5+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*Log[-(5 + x)^(-1)])/(5 + x),x]

[Out]

Log[-(5 + x)^(-1)]^2

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fricas [A]  time = 0.57, size = 10, normalized size = 1.00 \begin {gather*} \log \left (-\frac {1}{x + 5}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-1/(x + 5))^2

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giac [A]  time = 0.15, size = 10, normalized size = 1.00 \begin {gather*} \log \left (-\frac {1}{x + 5}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-1/(x + 5))^2

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maple [A]  time = 0.07, size = 11, normalized size = 1.10

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(-1/(5+x))^2

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maxima [B]  time = 0.37, size = 23, normalized size = 2.30 \begin {gather*} -\log \left (x + 5\right )^{2} - 2 \, \log \left (x + 5\right ) \log \left (-\frac {1}{x + 5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x + 5)^2 - 2*log(x + 5)*log(-1/(x + 5))

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mupad [B]  time = 4.55, size = 10, normalized size = 1.00 \begin {gather*} {\ln \left (-\frac {1}{x+5}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(-1/(x + 5)))/(x + 5),x)

[Out]

log(-1/(x + 5))^2

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sympy [A]  time = 0.13, size = 8, normalized size = 0.80 \begin {gather*} \log {\left (- \frac {1}{x + 5} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-1/(x + 5))**2

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