\(\int -\frac {2 \log (-\frac {1}{5+x})}{5+x} \, dx\) [6835]

Optimal result

Integrand size = 15, antiderivative size = 10 \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx=\log ^2\left (-\frac {1}{5+x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2437, 2338} \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx=\log ^2\left (-\frac {1}{x+5}\right ) \]

[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 2338

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 2437

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{align*} \text {integral}& = -\left (2 \int \frac {\log \left (-\frac {1}{5+x}\right )}{5+x} \, dx\right ) \\ & = -\left (2 \text {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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx=\log ^2\left (-\frac {1}{5+x}\right ) \]

[In]

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

[Out]

Log[-(5 + x)^(-1)]^2

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 11, normalized size of antiderivative = 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	\(\ln \left (-\frac {1}{5+x}\right )^{2}\)	\(11\)
parts	\(-2 \ln \left (-\frac {1}{5+x}\right ) \ln \left (5+x \right )-\ln \left (5+x \right )^{2}\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx=\log \left (-\frac {1}{x + 5}\right )^{2} \]

[In]

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

[Out]

log(-1/(x + 5))^2

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx=\log {\left (- \frac {1}{x + 5} \right )}^{2} \]

[In]

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

[Out]

log(-1/(x + 5))**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 23 vs. \(2 (10) = 20\).

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 2.30 \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx=-\log \left (x + 5\right )^{2} - 2 \, \log \left (x + 5\right ) \log \left (-\frac {1}{x + 5}\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx=\log \left (-\frac {1}{x + 5}\right )^{2} \]

[In]

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

[Out]

log(-1/(x + 5))^2

Mupad [B] (verification not implemented)

Time = 12.82 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int -\frac {2 \log \left (-\frac {1}{5+x}\right )}{5+x} \, dx={\ln \left (-\frac {1}{x+5}\right )}^2 \]

[In]

int(-(2*log(-1/(x + 5)))/(x + 5),x)

[Out]

log(-1/(x + 5))^2