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

Optimal. Leaf size=98 \[ \frac {1-x}{2 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{3/2}}+\frac {\sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{2 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 98, 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 = {6345, 12, 51, 63, 206} \[ \frac {1-x}{2 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{3/2}}+\frac {\sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{2 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - x)/(2*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*x^(3/2)) - ArcSech[Sqrt[x]]/x + (Sqrt[1 - x]*ArcTanh[Sqrt[
1 - x]])/(2*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*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 206

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

Rule 6345

Int[((a_.) + ArcSech[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSec
h[u]))/(d*(m + 1)), x] + Dist[(b*Sqrt[1 - u^2])/(d*(m + 1)*u*Sqrt[-1 + 1/u]*Sqrt[1 + 1/u]), Int[SimplifyIntegr
and[((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] && In
verseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^2} \, dx &=-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {1-x} \int \frac {1}{2 \sqrt {1-x} x^2} \, dx}{\sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {1-x} \int \frac {1}{\sqrt {1-x} x^2} \, dx}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {1-x} \int \frac {1}{\sqrt {1-x} x} \, dx}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x}+\frac {\sqrt {1-x} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x}\right )}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x}+\frac {\sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 111, normalized size = 1.13 \[ \frac {\sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}} \left (\sqrt {x}+1\right )+x \log \left (\sqrt {x} \sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}}+\sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}}+1\right )-\frac {1}{2} x \log (x)-2 \text {sech}^{-1}\left (\sqrt {x}\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(1 + Sqrt[x]) - 2*ArcSech[Sqrt[x]] + x*Log[1 + Sqrt[(1 - Sqrt[x])/(1 + Sqrt
[x])] + Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*Sqrt[x]] - (x*Log[x])/2)/(2*x)

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fricas [A]  time = 0.61, size = 45, normalized size = 0.46 \[ \frac {{\left (x - 2\right )} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right ) + \sqrt {x} \sqrt {-\frac {x - 1}{x}}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(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 \[ \int \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(x^(1/2))/x^2,x, algorithm="giac")

[Out]

integrate(arcsech(sqrt(x))/x^2, x)

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maple [A]  time = 0.07, size = 64, normalized size = 0.65 \[ -\frac {\mathrm {arcsech}\left (\sqrt {x}\right )}{x}+\frac {\sqrt {-\frac {-1+\sqrt {x}}{\sqrt {x}}}\, \sqrt {\frac {1+\sqrt {x}}{\sqrt {x}}}\, \left (\arctanh \left (\frac {1}{\sqrt {1-x}}\right ) x +\sqrt {1-x}\right )}{2 \sqrt {x}\, \sqrt {1-x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 65, normalized size = 0.66 \[ -\frac {\sqrt {x} \sqrt {\frac {1}{x} - 1}}{2 \, {\left (x {\left (\frac {1}{x} - 1\right )} - 1\right )}} - \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x} + \frac {1}{4} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} + 1\right ) - \frac {1}{4} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(x)*sqrt(1/x - 1)/(x*(1/x - 1) - 1) - arcsech(sqrt(x))/x + 1/4*log(sqrt(x)*sqrt(1/x - 1) + 1) - 1/4*l
og(sqrt(x)*sqrt(1/x - 1) - 1)

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mupad [B]  time = 2.15, size = 40, normalized size = 0.41 \[ \frac {\sqrt {\frac {1}{\sqrt {x}}-1}\,\sqrt {\frac {1}{\sqrt {x}}+1}}{2\,\sqrt {x}}-\frac {2\,\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right )\,\left (\frac {1}{2\,\sqrt {x}}-\frac {\sqrt {x}}{4}\right )}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1/x^(1/2) - 1)^(1/2)*(1/x^(1/2) + 1)^(1/2))/(2*x^(1/2)) - (2*acosh(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 \[ \int \frac {\operatorname {asech}{\left (\sqrt {x} \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(sqrt(x))/x**2, x)

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