3.7 \(\int \frac{\sec ^{-1}(\sqrt{x})}{x^2} \, dx\)

Optimal. Leaf size=38 \[ \frac{\sqrt{x-1}}{2 x}+\frac{1}{2} \tan ^{-1}\left (\sqrt{x-1}\right )-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0156021, antiderivative size = 38, 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 = {5270, 12, 51, 63, 203} \[ \frac{\sqrt{x-1}}{2 x}+\frac{1}{2} \tan ^{-1}\left (\sqrt{x-1}\right )-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5270

Int[((a_.) + ArcSec[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSec[
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[u^2 - 1]), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !Funct
ionOfQ[(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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0203111, size = 32, normalized size = 0.84 \[ \frac{\sqrt{x-1}-x \sin ^{-1}\left (\frac{1}{\sqrt{x}}\right )-2 \sec ^{-1}\left (\sqrt{x}\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-1 + x] - 2*ArcSec[Sqrt[x]] - x*ArcSin[1/Sqrt[x]])/(2*x)

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Maple [A]  time = 0.11, size = 45, normalized size = 1.2 \begin{align*} -{\frac{1}{x}{\rm arcsec} \left (\sqrt{x}\right )}+{\frac{1}{2}\sqrt{x-1} \left ( -\arctan \left ({\frac{1}{\sqrt{x-1}}} \right ) x+\sqrt{x-1} \right ){\frac{1}{\sqrt{{\frac{x-1}{x}}}}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsec(x^(1/2))/x+1/2*(x-1)^(1/2)*(-arctan(1/(x-1)^(1/2))*x+(x-1)^(1/2))/((x-1)/x)^(1/2)/x^(3/2)

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Maxima [A]  time = 1.46345, size = 69, normalized size = 1.82 \begin{align*} -\frac{\sqrt{x} \sqrt{-\frac{1}{x} + 1}}{2 \,{\left (x{\left (\frac{1}{x} - 1\right )} - 1\right )}} - \frac{\operatorname{arcsec}\left (\sqrt{x}\right )}{x} + \frac{1}{2} \, \arctan \left (\sqrt{x} \sqrt{-\frac{1}{x} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(x)*sqrt(-1/x + 1)/(x*(1/x - 1) - 1) - arcsec(sqrt(x))/x + 1/2*arctan(sqrt(x)*sqrt(-1/x + 1))

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Fricas [A]  time = 2.19028, size = 63, normalized size = 1.66 \begin{align*} \frac{{\left (x - 2\right )} \operatorname{arcsec}\left (\sqrt{x}\right ) + \sqrt{x - 1}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(sqrt(x))/x**2, x)

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Giac [A]  time = 1.10828, size = 41, normalized size = 1.08 \begin{align*} \frac{\sqrt{-\frac{1}{x} + 1}}{2 \, \sqrt{x}} - \frac{\arccos \left (\frac{1}{\sqrt{x}}\right )}{x} + \frac{1}{2} \, \arccos \left (\frac{1}{\sqrt{x}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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