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

Optimal. Leaf size=25 \[ \sqrt {x^2-1} \sec ^{-1}(x)-\frac {x \log (x)}{\sqrt {x^2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5236, 29} \[ \sqrt {x^2-1} \sec ^{-1}(x)-\frac {x \log (x)}{\sqrt {x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcSec[x])/Sqrt[-1 + x^2],x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 5236

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p +
1)*(a + b*ArcSec[c*x]))/(2*e*(p + 1)), x] - Dist[(b*c*x)/(2*e*(p + 1)*Sqrt[c^2*x^2]), Int[(d + e*x^2)^(p + 1)/
(x*Sqrt[c^2*x^2 - 1]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 35, normalized size = 1.40 \[ \frac {\left (x^2-1\right ) \sec ^{-1}(x)-\sqrt {1-\frac {1}{x^2}} x \log (x)}{\sqrt {x^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcSec[x])/Sqrt[-1 + x^2],x]

[Out]

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

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fricas [A]  time = 0.44, size = 15, normalized size = 0.60 \[ \sqrt {x^{2} - 1} \operatorname {arcsec}\relax (x) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 - 1)*arcsec(x) - log(x)

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giac [A]  time = 1.09, size = 22, normalized size = 0.88 \[ \sqrt {x^{2} - 1} \arccos \left (\frac {1}{x}\right ) - \frac {\log \left ({\left | x \right |}\right )}{\mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 - 1)*arccos(1/x) - log(abs(x))/sgn(x)

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maple [C]  time = 0.51, size = 97, normalized size = 3.88 \[ -\frac {2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \,\mathrm {arcsec}\relax (x )}{\sqrt {x^{2}-1}}+\frac {\sqrt {\frac {x^{2}-1}{x^{2}}}\, x \ln \left (\left (\frac {1}{x}+i \sqrt {-\frac {1}{x^{2}}+1}\right )^{2}+1\right )}{\sqrt {x^{2}-1}}+\frac {\left (x^{2}+i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -1\right ) \mathrm {arcsec}\relax (x )}{\sqrt {x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 15, normalized size = 0.60 \[ \sqrt {x^{2} - 1} \operatorname {arcsec}\relax (x) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 - 1)*arcsec(x) - log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {x\,\mathrm {acos}\left (\frac {1}{x}\right )}{\sqrt {x^2-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*acos(1/x))/(x^2 - 1)^(1/2),x)

[Out]

int((x*acos(1/x))/(x^2 - 1)^(1/2), x)

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sympy [A]  time = 19.71, size = 22, normalized size = 0.88 \[ \sqrt {x^{2} - 1} \operatorname {asec}{\relax (x )} - \begin {cases} - \log {\left (\frac {1}{x} \right )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x**2 - 1)*asec(x) - Piecewise((-log(1/x), (x > -1) & (x < 1)))

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