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]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, 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 = {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 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])

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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 1.59, size = 19, normalized size = 0.50 \[ \frac {{\left (x - 2\right )} \operatorname {arcsec}\left (\sqrt {x}\right ) + \sqrt {x - 1}}{2 \, x} \]

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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giac [A]  time = 0.13, size = 30, normalized size = 0.79 \[ \frac {\sqrt {-\frac {1}{x} + 1}}{2 \, \sqrt {x}} - \frac {\arccos \left (\frac {1}{\sqrt {x}}\right )}{x} + \frac {1}{2} \, \arccos \left (\frac {1}{\sqrt {x}}\right ) \]

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))

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maple [A]  time = 0.05, size = 46, normalized size = 1.21 \[ -\frac {\mathrm {arcsec}\left (\sqrt {x}\right )}{x}-\frac {\sqrt {x -1}\, \left (\arctan \left (\frac {1}{\sqrt {x -1}}\right ) x -\sqrt {x -1}\right )}{2 \sqrt {\frac {x -1}{x}}\, x^{\frac {3}{2}}} \]

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 = 0.42, size = 51, normalized size = 1.34 \[ -\frac {\sqrt {x} \sqrt {-\frac {1}{x} + 1}}{2 \, {\left (x {\left (\frac {1}{x} - 1\right )} - 1\right )}} - \frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{x} + \frac {1}{2} \, \arctan \left (\sqrt {x} \sqrt {-\frac {1}{x} + 1}\right ) \]

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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mupad [B]  time = 0.68, size = 28, normalized size = 0.74 \[ \frac {\sqrt {1-\frac {1}{x}}}{2\,\sqrt {x}}-\frac {\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right )\,\left (\frac {2}{x}-1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - 1/x)^(1/2)/(2*x^(1/2)) - (acos(1/x^(1/2))*(2/x - 1))/2

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sympy [C]  time = 20.05, size = 76, normalized size = 2.00 \[ \frac {\begin {cases} i \operatorname {acosh}{\left (\frac {1}{\sqrt {x}} \right )} + \frac {i \sqrt {-1 + \frac {1}{x}}}{\sqrt {x}} & \text {for}\: \frac {1}{\left |{x}\right |} > 1 \\- \operatorname {asin}{\left (\frac {1}{\sqrt {x}} \right )} + \frac {1}{\sqrt {x} \sqrt {1 - \frac {1}{x}}} - \frac {1}{x^{\frac {3}{2}} \sqrt {1 - \frac {1}{x}}} & \text {otherwise} \end {cases}}{2} - \frac {\operatorname {asec}{\left (\sqrt {x} \right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*acosh(1/sqrt(x)) + I*sqrt(-1 + 1/x)/sqrt(x), 1/Abs(x) > 1), (-asin(1/sqrt(x)) + 1/(sqrt(x)*sqrt(1
- 1/x)) - 1/(x**(3/2)*sqrt(1 - 1/x)), True))/2 - asec(sqrt(x))/x

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