3.5 \(\int \sec ^{-1}(\sqrt {x}) \, dx\)

Optimal. Leaf size=18 \[ x \sec ^{-1}\left (\sqrt {x}\right )-\sqrt {x-1} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5268, 12, 32} \[ x \sec ^{-1}\left (\sqrt {x}\right )-\sqrt {x-1} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[Sqrt[x]],x]

[Out]

-Sqrt[-1 + x] + x*ArcSec[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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 5268

Int[ArcSec[u_], x_Symbol] :> Simp[x*ArcSec[u], x] - Dist[u/Sqrt[u^2], Int[SimplifyIntegrand[(x*D[u, x])/(u*Sqr
t[u^2 - 1]), x], x], x] /; InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ x \sec ^{-1}\left (\sqrt {x}\right )-\sqrt {x-1} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[Sqrt[x]],x]

[Out]

-Sqrt[-1 + x] + x*ArcSec[Sqrt[x]]

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fricas [A]  time = 0.93, size = 14, normalized size = 0.78 \[ x \operatorname {arcsec}\left (\sqrt {x}\right ) - \sqrt {x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arcsec(sqrt(x)) - sqrt(x - 1)

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giac [B]  time = 0.13, size = 41, normalized size = 2.28 \[ x \arccos \left (\frac {1}{\sqrt {x}}\right ) - \frac {1}{2} \, \sqrt {x} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )} + \frac {1}{2 \, \sqrt {x} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 25, normalized size = 1.39 \[ x \,\mathrm {arcsec}\left (\sqrt {x}\right )-\frac {x -1}{\sqrt {\frac {x -1}{x}}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 21, normalized size = 1.17 \[ x \operatorname {arcsec}\left (\sqrt {x}\right ) - \sqrt {x} \sqrt {-\frac {1}{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.15, size = 21, normalized size = 1.17 \[ x\,\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right )-\sqrt {x}\,\sqrt {1-\frac {1}{x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 4.42, size = 29, normalized size = 1.61 \[ x \operatorname {asec}{\left (\sqrt {x} \right )} - \frac {\begin {cases} 2 \sqrt {x - 1} & \text {for}\: \left |{x}\right | > 1 \\2 i \sqrt {1 - x} & \text {otherwise} \end {cases}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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