3.1.22 \(\int \text {csch}^{-1}(\frac {1}{x}) \, dx\) [22]

Optimal. Leaf size=16 \[ -\sqrt {1+x^2}+x \sinh ^{-1}(x) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 16, 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 = {6463, 5772, 267} \begin {gather*} x \sinh ^{-1}(x)-\sqrt {x^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + x^2] + x*ArcSinh[x]

Rule 267

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

Rule 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[x*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 6463

Int[ArcCsch[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcSinh[a/c + b*(x^n/c)]^m, x] /
; FreeQ[{a, b, c, n, m}, x]

Rubi steps

\begin {align*} \int \text {csch}^{-1}\left (\frac {1}{x}\right ) \, dx &=\int \sinh ^{-1}(x) \, dx\\ &=x \sinh ^{-1}(x)-\int \frac {x}{\sqrt {1+x^2}} \, dx\\ &=-\sqrt {1+x^2}+x \sinh ^{-1}(x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 18, normalized size = 1.12 \begin {gather*} -\sqrt {1+x^2}+x \text {csch}^{-1}\left (\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + x^2] + x*ArcCsch[x^(-1)]

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Maple [A]
time = 0.15, size = 29, normalized size = 1.81

method	result	size
derivativedivides	\(x \,\mathrm {arccsch}\left (\frac {1}{x}\right )-\frac {x^{2} \left (\frac {1}{x^{2}}+1\right )}{\sqrt {\left (\frac {1}{x^{2}}+1\right ) x^{2}}}\)	\(29\)
default	\(x \,\mathrm {arccsch}\left (\frac {1}{x}\right )-\frac {x^{2} \left (\frac {1}{x^{2}}+1\right )}{\sqrt {\left (\frac {1}{x^{2}}+1\right ) x^{2}}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 16, normalized size = 1.00 \begin {gather*} x \operatorname {arcsch}\left (\frac {1}{x}\right ) - \sqrt {x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccsch(1/x) - sqrt(x^2 + 1)

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Fricas [A]
time = 0.36, size = 22, normalized size = 1.38 \begin {gather*} x \log \left (x + \sqrt {x^{2} + 1}\right ) - \sqrt {x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.04, size = 14, normalized size = 0.88 \begin {gather*} x \operatorname {acsch}{\left (\frac {1}{x} \right )} - \sqrt {x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(1/x),x)

[Out]

x*acsch(1/x) - sqrt(x**2 + 1)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsch(1/x), x)

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Mupad [B]
time = 0.07, size = 14, normalized size = 0.88 \begin {gather*} x\,\mathrm {asinh}\left (x\right )-\sqrt {x^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(x),x)

[Out]

x*asinh(x) - (x^2 + 1)^(1/2)

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