### 3.275 $$\int \log (\frac{-1+x}{1+x}) \, dx$$

Optimal. Leaf size=27 $-(1-x) \log \left (-\frac{1-x}{x+1}\right )-2 \log (x+1)$

[Out]

-((1 - x)*Log[-((1 - x)/(1 + x))]) - 2*Log[1 + x]

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Rubi [A]  time = 0.0049812, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2486, 31} $-(1-x) \log \left (-\frac{1-x}{x+1}\right )-2 \log (x+1)$

Antiderivative was successfully veriﬁed.

[In]

Int[Log[(-1 + x)/(1 + x)],x]

[Out]

-((1 - x)*Log[-((1 - x)/(1 + x))]) - 2*Log[1 + x]

Rule 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0027194, size = 21, normalized size = 0.78 $(x-1) \log \left (\frac{x-1}{x+1}\right )-2 \log (x+1)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Log[(-1 + x)/(1 + x)],x]

[Out]

(-1 + x)*Log[(-1 + x)/(1 + x)] - 2*Log[1 + x]

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Maple [A]  time = 0.007, size = 35, normalized size = 1.3 \begin{align*} 2\,\ln \left ( -2\, \left ( 1+x \right ) ^{-1} \right ) +\ln \left ( 1-2\, \left ( 1+x \right ) ^{-1} \right ) \left ( 1-2\, \left ( 1+x \right ) ^{-1} \right ) \left ( 1+x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(ln((-1+x)/(1+x)),x)

[Out]

2*ln(-2/(1+x))+ln(1-2/(1+x))*(1-2/(1+x))*(1+x)

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Maxima [A]  time = 1.00724, size = 34, normalized size = 1.26 \begin{align*} x \log \left (\frac{x - 1}{x + 1}\right ) - \log \left (x + 1\right ) - \log \left (x - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log((x - 1)/(x + 1)) - log(x + 1) - log(x - 1)

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Fricas [A]  time = 2.10258, size = 53, normalized size = 1.96 \begin{align*} x \log \left (\frac{x - 1}{x + 1}\right ) - \log \left (x^{2} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log((x - 1)/(x + 1)) - log(x^2 - 1)

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Sympy [A]  time = 0.119236, size = 15, normalized size = 0.56 \begin{align*} x \log{\left (\frac{x - 1}{x + 1} \right )} - \log{\left (x^{2} - 1 \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln((-1+x)/(1+x)),x)

[Out]

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

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Giac [A]  time = 1.32349, size = 30, normalized size = 1.11 \begin{align*} x \log \left (\frac{x - 1}{x + 1}\right ) - \log \left ({\left | x^{2} - 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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