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

Optimal. Leaf size=56 \[ i \text {Li}_2\left (-e^{2 i \sec ^{-1}\left (\sqrt {x}\right )}\right )+i \sec ^{-1}\left (\sqrt {x}\right )^2-2 \sec ^{-1}\left (\sqrt {x}\right ) \log \left (1+e^{2 i \sec ^{-1}\left (\sqrt {x}\right )}\right ) \]

[Out]

I*arcsec(x^(1/2))^2-2*arcsec(x^(1/2))*ln(1+(1/x^(1/2)+I*(1-1/x)^(1/2))^2)+I*polylog(2,-(1/x^(1/2)+I*(1-1/x)^(1
/2))^2)

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Rubi [A]  time = 0.08, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5218, 4626, 3719, 2190, 2279, 2391} \[ i \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (\sqrt {x}\right )}\right )+i \sec ^{-1}\left (\sqrt {x}\right )^2-2 \sec ^{-1}\left (\sqrt {x}\right ) \log \left (1+e^{2 i \sec ^{-1}\left (\sqrt {x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

I*ArcSec[Sqrt[x]]^2 - 2*ArcSec[Sqrt[x]]*Log[1 + E^((2*I)*ArcSec[Sqrt[x]])] + I*PolyLog[2, -E^((2*I)*ArcSec[Sqr
t[x]])]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5218

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCos[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x} \, dx &=2 \operatorname {Subst}\left (\int \frac {\sec ^{-1}(x)}{x} \, dx,x,\sqrt {x}\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {\cos ^{-1}(x)}{x} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=2 \operatorname {Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right )\\ &=i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2-4 i \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right )\\ &=i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2-2 \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )+2 \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right )\\ &=i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2-2 \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )-i \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )\\ &=i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )^2-2 \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )+i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac {1}{\sqrt {x}}\right )}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.96 \[ i \left (\text {Li}_2\left (-e^{2 i \sec ^{-1}\left (\sqrt {x}\right )}\right )+\sec ^{-1}\left (\sqrt {x}\right ) \left (\sec ^{-1}\left (\sqrt {x}\right )+2 i \log \left (1+e^{2 i \sec ^{-1}\left (\sqrt {x}\right )}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

I*(ArcSec[Sqrt[x]]*(ArcSec[Sqrt[x]] + (2*I)*Log[1 + E^((2*I)*ArcSec[Sqrt[x]])]) + PolyLog[2, -E^((2*I)*ArcSec[
Sqrt[x]])])

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsec(sqrt(x))/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(sqrt(x))/x, x)

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maple [A]  time = 0.12, size = 63, normalized size = 1.12 \[ i \mathrm {arcsec}\left (\sqrt {x}\right )^{2}-2 \,\mathrm {arcsec}\left (\sqrt {x}\right ) \ln \left (1+\left (\frac {1}{\sqrt {x}}+i \sqrt {1-\frac {1}{x}}\right )^{2}\right )+i \polylog \left (2, -\left (\frac {1}{\sqrt {x}}+i \sqrt {1-\frac {1}{x}}\right )^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*arcsec(x^(1/2))^2-2*arcsec(x^(1/2))*ln(1+(1/x^(1/2)+I*(1-1/x)^(1/2))^2)+I*polylog(2,-(1/x^(1/2)+I*(1-1/x)^(1
/2))^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(sqrt(x))/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acos(1/x^(1/2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asec}{\left (\sqrt {x} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(sqrt(x))/x, x)

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