3.1.19 \(\int \frac {\text {csch}^{-1}(\sqrt {x})}{x^2} \, dx\) [19]

Optimal. Leaf size=63 \[ \frac {\sqrt {-1-x}}{2 \sqrt {-x} \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {x} \text {ArcTan}\left (\sqrt {-1-x}\right )}{2 \sqrt {-x}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6481, 12, 44, 65, 210} \begin {gather*} -\frac {\sqrt {x} \text {ArcTan}\left (\sqrt {-x-1}\right )}{2 \sqrt {-x}}+\frac {\sqrt {-x-1}}{2 \sqrt {-x} \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCsch[Sqrt[x]]/x^2,x]

[Out]

Sqrt[-1 - x]/(2*Sqrt[-x]*Sqrt[x]) - ArcCsch[Sqrt[x]]/x - (Sqrt[x]*ArcTan[Sqrt[-1 - x]])/(2*Sqrt[-x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 6481

Int[((a_.) + ArcCsch[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((a + b*ArcCsc
h[u])/(d*(m + 1))), x] - Dist[b*(u/(d*(m + 1)*Sqrt[-u^2])), Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/(
u*Sqrt[-1 - u^2])), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !F
unctionOfQ[(c + d*x)^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^2} \, dx &=-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x}+\frac {\sqrt {x} \int \frac {1}{2 \sqrt {-1-x} x^2} \, dx}{\sqrt {-x}}\\ &=-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x}+\frac {\sqrt {x} \int \frac {1}{\sqrt {-1-x} x^2} \, dx}{2 \sqrt {-x}}\\ &=\frac {\sqrt {-1-x}}{2 \sqrt {-x} \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {x} \int \frac {1}{\sqrt {-1-x} x} \, dx}{4 \sqrt {-x}}\\ &=\frac {\sqrt {-1-x}}{2 \sqrt {-x} \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x}+\frac {\sqrt {x} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {-1-x}\right )}{2 \sqrt {-x}}\\ &=\frac {\sqrt {-1-x}}{2 \sqrt {-x} \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {x} \tan ^{-1}\left (\sqrt {-1-x}\right )}{2 \sqrt {-x}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 42, normalized size = 0.67 \begin {gather*} \frac {\sqrt {\frac {1+x}{x}}}{2 \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x}-\frac {1}{2} \sinh ^{-1}\left (\frac {1}{\sqrt {x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCsch[Sqrt[x]]/x^2,x]

[Out]

Sqrt[(1 + x)/x]/(2*Sqrt[x]) - ArcCsch[Sqrt[x]]/x - ArcSinh[1/Sqrt[x]]/2

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Maple [A]
time = 0.16, size = 46, normalized size = 0.73

method	result	size
derivativedivides	\(-\frac {\mathrm {arccsch}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {1+x}\, \left (\arctanh \left (\frac {1}{\sqrt {1+x}}\right ) x -\sqrt {1+x}\right )}{2 \sqrt {\frac {1+x}{x}}\, x^{\frac {3}{2}}}\)	\(46\)
default	\(-\frac {\mathrm {arccsch}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {1+x}\, \left (\arctanh \left (\frac {1}{\sqrt {1+x}}\right ) x -\sqrt {1+x}\right )}{2 \sqrt {\frac {1+x}{x}}\, x^{\frac {3}{2}}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccsch(x^(1/2))/x^2,x,method=_RETURNVERBOSE)

[Out]

-arccsch(x^(1/2))/x-1/2*(1+x)^(1/2)*(arctanh(1/(1+x)^(1/2))*x-(1+x)^(1/2))/((1+x)/x)^(1/2)/x^(3/2)

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Maxima [A]
time = 0.26, size = 65, normalized size = 1.03 \begin {gather*} \frac {\sqrt {x} \sqrt {\frac {1}{x} + 1}}{2 \, {\left (x {\left (\frac {1}{x} + 1\right )} - 1\right )}} - \frac {\operatorname {arcsch}\left (\sqrt {x}\right )}{x} - \frac {1}{4} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(x^(1/2))/x^2,x, algorithm="maxima")

[Out]

1/2*sqrt(x)*sqrt(1/x + 1)/(x*(1/x + 1) - 1) - arccsch(sqrt(x))/x - 1/4*log(sqrt(x)*sqrt(1/x + 1) + 1) + 1/4*lo
g(sqrt(x)*sqrt(1/x + 1) - 1)

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Fricas [A]
time = 0.36, size = 44, normalized size = 0.70 \begin {gather*} -\frac {{\left (x + 2\right )} \log \left (\frac {x \sqrt {\frac {x + 1}{x}} + \sqrt {x}}{x}\right ) - \sqrt {x} \sqrt {\frac {x + 1}{x}}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(x^(1/2))/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acsch}{\left (\sqrt {x} \right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(x**(1/2))/x**2,x)

[Out]

Integral(acsch(sqrt(x))/x**2, x)

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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(x^(1/2))/x^2,x, algorithm="giac")

[Out]

integrate(arccsch(sqrt(x))/x^2, x)

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Mupad [B]
time = 2.22, size = 33, normalized size = 0.52 \begin {gather*} \frac {\sqrt {\frac {1}{x}+1}}{2\,\sqrt {x}}-\frac {2\,\mathrm {asinh}\left (\frac {1}{\sqrt {x}}\right )\,\left (\frac {1}{2\,\sqrt {x}}+\frac {\sqrt {x}}{4}\right )}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(1/x^(1/2))/x^2,x)

[Out]

(1/x + 1)^(1/2)/(2*x^(1/2)) - (2*asinh(1/x^(1/2))*(1/(2*x^(1/2)) + x^(1/2)/4))/x^(1/2)

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