3.3.38 \(\int \tanh ^{-1}(\frac {1}{x}) \, dx\) [238]

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

[Out]

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

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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 = {6021, 269, 266} \begin {gather*} \frac {1}{2} \log \left (1-x^2\right )+x \tanh ^{-1}\left (\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcTanh[x^(-1)] + Log[1 - x^2]/2

Rule 266

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 269

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 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

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*}

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

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcTanh[x^(-1)] + Log[-1 + x^2]/2

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Maple [A]
time = 0.14, size = 30, normalized size = 1.58

method	result	size
derivativedivides	\(x \arctanh \left (\frac {1}{x}\right )+\frac {\ln \left (\frac {1}{x}+1\right )}{2}+\frac {\ln \left (\frac {1}{x}-1\right )}{2}-\ln \left (\frac {1}{x}\right )\)	\(30\)
default	\(x \arctanh \left (\frac {1}{x}\right )+\frac {\ln \left (\frac {1}{x}+1\right )}{2}+\frac {\ln \left (\frac {1}{x}-1\right )}{2}-\ln \left (\frac {1}{x}\right )\)	\(30\)
meijerg	\(-\frac {\ln \left (1-\sqrt {\frac {1}{x^{2}}}\right )-\ln \left (1+\sqrt {\frac {1}{x^{2}}}\right )}{2 \sqrt {\frac {1}{x^{2}}}}+\frac {\ln \left (-\frac {1}{x^{2}}+1\right )}{2}+\ln \left (x \right )-\frac {i \pi }{2}\)	\(46\)
risch	\(-\frac {x \ln \left (x -1\right )}{2}+\frac {\ln \left (1+x \right ) x}{2}+\frac {i \pi \,\mathrm {csgn}\left (i \left (1+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (1+x \right )}{x}\right )^{2} x}{4}-\frac {i \pi \,\mathrm {csgn}\left (i \left (1+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (1+x \right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) x}{4}-\frac {i \pi \,\mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} x}{4}+\frac {i \pi \,\mathrm {csgn}\left (i \left (x -1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) x}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i \left (1+x \right )}{x}\right )^{3} x}{4}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (1+x \right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) x}{4}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{3} x}{4}-\frac {i \pi \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) x}{4}+\frac {\ln \left (x^{2}-1\right )}{2}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(1/x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 15, normalized size = 0.79 \begin {gather*} x \operatorname {artanh}\left (\frac {1}{x}\right ) + \frac {1}{2} \, \log \left (x^{2} - 1\right ) \end {gather*}

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)

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Fricas [A]
time = 0.35, size = 22, normalized size = 1.16 \begin {gather*} \frac {1}{2} \, x \log \left (\frac {x + 1}{x - 1}\right ) + \frac {1}{2} \, \log \left (x^{2} - 1\right ) \end {gather*}

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)

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Sympy [A]
time = 0.08, size = 15, normalized size = 0.79 \begin {gather*} x \operatorname {atanh}{\left (\frac {1}{x} \right )} + \log {\left (x + 1 \right )} - \operatorname {atanh}{\left (\frac {1}{x} \right )} \end {gather*}

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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (17) = 34\).
time = 0.41, size = 101, normalized size = 5.32 \begin {gather*} \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 ) \end {gather*}

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))

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

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)

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