3.238 \(\int \tanh ^{-1}(\frac {1}{x}) \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{2} \log \left (1-x^2\right )+x \tanh ^{-1}\left (\frac {1}{x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6091, 263, 260} \[ \frac {1}{2} \log \left (1-x^2\right )+x \tanh ^{-1}\left (\frac {1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[x^(-1)],x]

[Out]

x*ArcTanh[x^(-1)] + Log[1 - x^2]/2

Rule 260

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

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 6091

Int[ArcTanh[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTanh[c*x^n], x] - Dist[c*n, Int[x^n/(1 - c^2*x^(2*n)), x]
, x] /; FreeQ[{c, n}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 17, normalized size = 0.89 \[ \frac {1}{2} \log \left (x^2-1\right )+x \tanh ^{-1}\left (\frac {1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[x^(-1)],x]

[Out]

x*ArcTanh[x^(-1)] + Log[-1 + x^2]/2

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 22, normalized size = 1.16 \[ \frac {1}{2} \, x \log \left (\frac {x + 1}{x - 1}\right ) + \frac {1}{2} \, \log \left (x^{2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(1/x),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.16, size = 101, normalized size = 5.32 \[ \frac {\log \left (-\frac {\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} + 1}{\frac {\frac {x + 1}{x - 1} - 1}{\frac {x + 1}{x - 1} + 1} - 1}\right )}{\frac {x + 1}{x - 1} - 1} + \log \left (\frac {{\left | x + 1 \right |}}{{\left | x - 1 \right |}}\right ) - \log \left ({\left | \frac {x + 1}{x - 1} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(1/x),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 30, normalized size = 1.58 \[ x \arctanh \left (\frac {1}{x}\right )-\ln \left (\frac {1}{x}\right )+\frac {\ln \left (\frac {1}{x}-1\right )}{2}+\frac {\ln \left (\frac {1}{x}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(1/x),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 15, normalized size = 0.79 \[ x \operatorname {artanh}\left (\frac {1}{x}\right ) + \frac {1}{2} \, \log \left (x^{2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(1/x),x, algorithm="maxima")

[Out]

x*arctanh(1/x) + 1/2*log(x^2 - 1)

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 15, normalized size = 0.79 \[ \frac {\ln \left (x^2-1\right )}{2}+x\,\mathrm {atanh}\left (\frac {1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(1/x),x)

[Out]

log(x^2 - 1)/2 + x*atanh(1/x)

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 15, normalized size = 0.79 \[ x \operatorname {atanh}{\left (\frac {1}{x} \right )} + \log {\left (x + 1 \right )} - \operatorname {atanh}{\left (\frac {1}{x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(1/x),x)

[Out]

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

________________________________________________________________________________________