3.19 \(\int \frac {\text {csch}^{-1}(\sqrt {x})}{x^2} \, dx\)

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

[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 = 0.02, 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 = {6346, 12, 51, 63, 204} \[ \frac {\sqrt {-x-1}}{2 \sqrt {-x} \sqrt {x}}-\frac {\sqrt {x} \tan ^{-1}\left (\sqrt {-x-1}\right )}{2 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x} \]

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 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 204

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

Rule 6346

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} \operatorname {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 \[ \frac {\sqrt {\frac {x+1}{x}}}{2 \sqrt {x}}-\frac {1}{2} \sinh ^{-1}\left (\frac {1}{\sqrt {x}}\right )-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x} \]

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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fricas [A]  time = 0.63, size = 44, normalized size = 0.70 \[ -\frac {{\left (x + 2\right )} \log \left (\frac {x \sqrt {\frac {x + 1}{x}} + \sqrt {x}}{x}\right ) - \sqrt {x} \sqrt {\frac {x + 1}{x}}}{2 \, x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcsch}\left (\sqrt {x}\right )}{x^{2}}\,{d x} \]

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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maple [A]  time = 0.06, size = 45, normalized size = 0.71 \[ -\frac {\mathrm {arccsch}\left (\sqrt {x}\right )}{x}+\frac {\sqrt {1+x}\, \left (-\arctanh \left (\frac {1}{\sqrt {1+x}}\right ) x +\sqrt {1+x}\right )}{2 \sqrt {\frac {1+x}{x}}\, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.32, size = 65, normalized size = 1.03 \[ \frac {\sqrt {x} \sqrt {\frac {1}{x} + 1}}{2 \, {\left (x {\left (\frac {1}{x} + 1\right )} - 1\right )}} - \frac {\operatorname {arcsch}\left (\sqrt {x}\right )}{x} - \frac {1}{4} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} - 1\right ) \]

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

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

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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