3.13 \(\int \frac {\sec ^{-1}(\frac {a}{x})}{x} \, dx\)

Optimal. Leaf size=59 \[ -\frac {1}{2} i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+\cos ^{-1}\left (\frac {x}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right ) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5264, 4626, 3719, 2190, 2279, 2391} \[ -\frac {1}{2} i \text {PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right )-\frac {1}{2} i \cos ^{-1}\left (\frac {x}{a}\right )^2+\cos ^{-1}\left (\frac {x}{a}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {x}{a}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a/x]/x,x]

[Out]

(-I/2)*ArcCos[x/a]^2 + ArcCos[x/a]*Log[1 + E^((2*I)*ArcCos[x/a])] - (I/2)*PolyLog[2, -E^((2*I)*ArcCos[x/a])]

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 5264

Int[ArcSec[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcCos[a/c + (b*x^n)/c]^m, x] /;
FreeQ[{a, b, c, n, m}, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 59, normalized size = 1.00 \[ -\frac {1}{2} i \text {Li}_2\left (-e^{2 i \sec ^{-1}\left (\frac {a}{x}\right )}\right )-\frac {1}{2} i \sec ^{-1}\left (\frac {a}{x}\right )^2+\sec ^{-1}\left (\frac {a}{x}\right ) \log \left (1+e^{2 i \sec ^{-1}\left (\frac {a}{x}\right )}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSec[a/x]/x,x]

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a/x)/x,x, algorithm="fricas")

[Out]

integral(arcsec(a/x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a/x)/x,x, algorithm="giac")

[Out]

integrate(arcsec(a/x)/x, x)

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maple [A]  time = 0.12, size = 76, normalized size = 1.29 \[ -\frac {i \mathrm {arcsec}\left (\frac {a}{x}\right )^{2}}{2}+\mathrm {arcsec}\left (\frac {a}{x}\right ) \ln \left (1+\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )^{2}\right )-\frac {i \polylog \left (2, -\left (\frac {x}{a}+i \sqrt {1-\frac {x^{2}}{a^{2}}}\right )^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(a/x)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a/x)/x,x, algorithm="maxima")

[Out]

integrate(arcsec(a/x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(x/a)/x,x)

[Out]

int(acos(x/a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(a/x)/x,x)

[Out]

Integral(asec(a/x)/x, x)

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