### 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]

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

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Rubi [A]  time = 0.0238083, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, 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 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]

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])

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*}

Mathematica [A]  time = 0.0230363, 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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Maple [A]  time = 0.112, size = 46, normalized size = 0.7 \begin{align*} -{\frac{1}{x}{\rm arccsch} \left (\sqrt{x}\right )}-{\frac{1}{2}\sqrt{1+x} \left ({\it Artanh} \left ({\frac{1}{\sqrt{1+x}}} \right ) x-\sqrt{1+x} \right ){\frac{1}{\sqrt{{\frac{1+x}{x}}}}}{x}^{-{\frac{3}{2}}}} \end{align*}

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 = 0.997511, size = 88, normalized size = 1.4 \begin{align*} \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{align*}

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 = 2.69195, size = 109, normalized size = 1.73 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acsch}{\left (\sqrt{x} \right )}}{x^{2}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsch}\left (\sqrt{x}\right )}{x^{2}}\,{d x} \end{align*}

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)