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

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

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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} $$\begin{array}{c}{\log }^2(-\frac{1}{x+5})\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-(2\int \frac{\log (-\frac{1}{5+x})}{5+x}dx)\\ & =-\left(2\mathrm{Subst}(\int \frac{\log (-\frac{1}{x})}{x}dx,x,5+x)\right)\\ & ={\log }^2(-\frac{1}{5+x})\end{array}\end{array}$$

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

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 $$\begin{array}{c}\log {(-\frac{1}{x+5})}^2\end{array}$$

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 $$\begin{array}{c}\log {(-\frac{1}{x+5})}^2\end{array}$$

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	$\ln {(-\frac{1}{5+x})}^2$	$11$
default	$\ln {(-\frac{1}{5+x})}^2$	$11$
norman	$\ln {(-\frac{1}{5+x})}^2$	$11$
risch	$\ln {(-\frac{1}{5+x})}^2$	$11$

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 $$\begin{array}{c}-\log {(x+5)}^2-2\log (x+5)\log (-\frac{1}{x+5})\end{array}$$

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 $$\begin{array}{c}{\ln (-\frac{1}{x+5})}^2\end{array}$$

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 $$\begin{array}{c}\log {(-\frac{1}{x+5})}^2\end{array}$$

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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