3.7.86 \(\int \frac {\sec ^{-1}(x)}{(-1+x^2)^{5/2}} \, dx\) [686]

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

[Out]

5/6*arccoth((x^2)^(1/2))-1/3*x*arcsec(x)/(x^2-1)^(3/2)+1/6*(x^2)^(1/2)/(-x^2+1)+2/3*x*arcsec(x)/(x^2-1)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {198, 197, 5336, 12, 393, 212} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[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 12

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

Rule 197

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 198

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 212

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

Rule 393

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 5336

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])

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*}

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Mathematica [A]
time = 0.08, size = 67, normalized size = 1.03 \begin {gather*} \frac {4 x \left (-3+2 x^2\right ) \sec ^{-1}(x)+\sqrt {1-\frac {1}{x^2}} x \left (-2 x-5 \left (-1+x^2\right ) \log (1-x)+5 \left (-1+x^2\right ) \log (1+x)\right )}{12 \left (-1+x^2\right )^{3/2}} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.42, size = 128, normalized size = 1.97

method	result	size
default	\(\frac {\sqrt {x^{2}-1}\, x \left (4 \,\mathrm {arcsec}\left (x \right ) x^{2}-\sqrt {\frac {x^{2}-1}{x^{2}}}\, x -6 \,\mathrm {arcsec}\left (x \right )\right )}{6 x^{4}-12 x^{2}+6}-\frac {5 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}-1\right )}{6 \sqrt {x^{2}-1}}+\frac {5 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}+1\right )}{6 \sqrt {x^{2}-1}}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(x^2-1)^(1/2)*x*(4*arcsec(x)*x^2-((x^2-1)/x^2)^(1/2)*x-6*arcsec(x))/(x^4-2*x^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)+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 = 2.52, size = 48, normalized size = 0.74 \begin {gather*} \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 {gather*}

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 = 1.01, size = 75, normalized size = 1.15 \begin {gather*} -\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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [A]
time = 0.62, size = 58, normalized size = 0.89 \begin {gather*} \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 {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {acos}\left (\frac {1}{x}\right )}{{\left (x^2-1\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/x)/(x^2 - 1)^(5/2),x)

[Out]

int(acos(1/x)/(x^2 - 1)^(5/2), x)

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