3.686 \(\int \frac{\sec ^{-1}(x)}{(-1+x^2)^{5/2}} \, dx\)

Optimal. Leaf size=65 \[ \frac{\sqrt{x^2}}{6 \left (1-x^2\right )}+\frac{2 x \sec ^{-1}(x)}{3 \sqrt{x^2-1}}-\frac{x \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac{5}{6} \coth ^{-1}\left (\sqrt{x^2}\right ) \]

[Out]

Sqrt[x^2]/(6*(1 - x^2)) + (5*ArcCoth[Sqrt[x^2]])/6 - (x*ArcSec[x])/(3*(-1 + x^2)^(3/2)) + (2*x*ArcSec[x])/(3*S
qrt[-1 + x^2])

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Rubi [A]  time = 0.0306868, antiderivative size = 67, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {192, 191, 5228, 12, 385, 206} \[ \frac{\sqrt{x^2}}{6 \left (1-x^2\right )}+\frac{2 x \sec ^{-1}(x)}{3 \sqrt{x^2-1}}-\frac{x \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac{5 x \tanh ^{-1}(x)}{6 \sqrt{x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Int[ArcSec[x]/(-1 + x^2)^(5/2),x]

[Out]

Sqrt[x^2]/(6*(1 - x^2)) - (x*ArcSec[x])/(3*(-1 + x^2)^(3/2)) + (2*x*ArcSec[x])/(3*Sqrt[-1 + x^2]) + (5*x*ArcTa
nh[x])/(6*Sqrt[x^2])

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 5228

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyIntegrand[u/(x*Sqrt[c^2*x^2
- 1]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx &=-\frac{x \sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac{2 x \sec ^{-1}(x)}{3 \sqrt{-1+x^2}}-\frac{x \int \frac{-3+2 x^2}{3 \left (1-x^2\right )^2} \, dx}{\sqrt{x^2}}\\ &=-\frac{x \sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac{2 x \sec ^{-1}(x)}{3 \sqrt{-1+x^2}}-\frac{x \int \frac{-3+2 x^2}{\left (1-x^2\right )^2} \, dx}{3 \sqrt{x^2}}\\ &=\frac{\sqrt{x^2}}{6 \left (1-x^2\right )}-\frac{x \sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac{2 x \sec ^{-1}(x)}{3 \sqrt{-1+x^2}}+\frac{(5 x) \int \frac{1}{1-x^2} \, dx}{6 \sqrt{x^2}}\\ &=\frac{\sqrt{x^2}}{6 \left (1-x^2\right )}-\frac{x \sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac{2 x \sec ^{-1}(x)}{3 \sqrt{-1+x^2}}+\frac{5 x \tanh ^{-1}(x)}{6 \sqrt{x^2}}\\ \end{align*}

Mathematica [A]  time = 0.130161, size = 67, normalized size = 1.03 \[ \frac{\sqrt{1-\frac{1}{x^2}} x \left (-5 \left (x^2-1\right ) \log (1-x)+5 \left (x^2-1\right ) \log (x+1)-2 x\right )+4 x \left (2 x^2-3\right ) \sec ^{-1}(x)}{12 \left (x^2-1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[x]/(-1 + x^2)^(5/2),x]

[Out]

(4*x*(-3 + 2*x^2)*ArcSec[x] + Sqrt[1 - x^(-2)]*x*(-2*x - 5*(-1 + x^2)*Log[1 - x] + 5*(-1 + x^2)*Log[1 + x]))/(
12*(-1 + x^2)^(3/2))

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Maple [C]  time = 0.24, size = 128, normalized size = 2. \begin{align*}{\frac{x}{6\,{x}^{4}-12\,{x}^{2}+6}\sqrt{{x}^{2}-1} \left ( 4\,{\rm arcsec} \left (x\right ){x}^{2}-\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-6\,{\rm arcsec} \left (x\right ) \right ) }-{\frac{5\,x}{6}\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}\ln \left ({x}^{-1}+i\sqrt{1-{x}^{-2}}-1 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}}+{\frac{5\,x}{6}\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}\ln \left ({x}^{-1}+i\sqrt{1-{x}^{-2}}+1 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(x)/(x^2-1)^(5/2),x)

[Out]

1/6*(x^2-1)^(1/2)*x/(x^4-2*x^2+1)*(4*arcsec(x)*x^2-((x^2-1)/x^2)^(1/2)*x-6*arcsec(x))-5/6/(x^2-1)^(1/2)*((x^2-
1)/x^2)^(1/2)*x*ln(1/x+I*(1-1/x^2)^(1/2)-1)+5/6/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*ln(1/x+I*(1-1/x^2)^(1/2)+1
)

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Maxima [A]  time = 1.18249, size = 65, normalized size = 1. \begin{align*} \frac{1}{3} \,{\left (\frac{2 \, x}{\sqrt{x^{2} - 1}} - \frac{x}{{\left (x^{2} - 1\right )}^{\frac{3}{2}}}\right )} \operatorname{arcsec}\left (x\right ) - \frac{x}{6 \,{\left (x^{2} - 1\right )}} + \frac{5}{12} \, \log \left (x + 1\right ) - \frac{5}{12} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)/(x^2-1)^(5/2),x, algorithm="maxima")

[Out]

1/3*(2*x/sqrt(x^2 - 1) - x/(x^2 - 1)^(3/2))*arcsec(x) - 1/6*x/(x^2 - 1) + 5/12*log(x + 1) - 5/12*log(x - 1)

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Fricas [A]  time = 2.35359, size = 198, normalized size = 3.05 \begin{align*} -\frac{2 \, x^{3} - 4 \,{\left (2 \, x^{3} - 3 \, x\right )} \sqrt{x^{2} - 1} \operatorname{arcsec}\left (x\right ) - 5 \,{\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right ) + 5 \,{\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x - 1\right ) - 2 \, x}{12 \,{\left (x^{4} - 2 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)/(x^2-1)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(2*x^3 - 4*(2*x^3 - 3*x)*sqrt(x^2 - 1)*arcsec(x) - 5*(x^4 - 2*x^2 + 1)*log(x + 1) + 5*(x^4 - 2*x^2 + 1)*
log(x - 1) - 2*x)/(x^4 - 2*x^2 + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(x)/(x**2-1)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11606, size = 78, normalized size = 1.2 \begin{align*} \frac{{\left (2 \, x^{2} - 3\right )} x \arccos \left (\frac{1}{x}\right )}{3 \,{\left (x^{2} - 1\right )}^{\frac{3}{2}}} + \frac{5 \, \log \left ({\left | x + 1 \right |}\right )}{12 \, \mathrm{sgn}\left (x\right )} - \frac{5 \, \log \left ({\left | x - 1 \right |}\right )}{12 \, \mathrm{sgn}\left (x\right )} - \frac{x}{6 \,{\left (x^{2} - 1\right )} \mathrm{sgn}\left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)/(x^2-1)^(5/2),x, algorithm="giac")

[Out]

1/3*(2*x^2 - 3)*x*arccos(1/x)/(x^2 - 1)^(3/2) + 5/12*log(abs(x + 1))/sgn(x) - 5/12*log(abs(x - 1))/sgn(x) - 1/
6*x/((x^2 - 1)*sgn(x))