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

Optimal. Leaf size=51 \[ \frac {\sqrt {x^2}}{6 \left (1-x^2\right )}-\frac {1}{6} \coth ^{-1}\left (\sqrt {x^2}\right )-\frac {x^3 \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {264, 5238, 12, 288, 207} \[ \frac {\sqrt {x^2}}{6 \left (1-x^2\right )}-\frac {x^3 \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}-\frac {x \tanh ^{-1}(x)}{6 \sqrt {x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[x^2]/(6*(1 - x^2)) - (x^3*ArcSec[x])/(3*(-1 + x^2)^(3/2)) - (x*ArcTanh[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 207

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

Rule 264

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 5238

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 61, normalized size = 1.20 \[ \frac {\sqrt {1-\frac {1}{x^2}} x \left (\left (x^2-1\right ) \log (1-x)-\left (x^2-1\right ) \log (x+1)-2 x\right )-4 x^3 \sec ^{-1}(x)}{12 \left (x^2-1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.19, size = 68, normalized size = 1.33 \[ -\frac {4 \, \sqrt {x^{2} - 1} x^{3} \operatorname {arcsec}\relax (x) + 2 \, x^{3} + {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right ) - {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x - 1\right ) - 2 \, x}{12 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.13, size = 53, normalized size = 1.04 \[ -\frac {x^{3} \arccos \left (\frac {1}{x}\right )}{3 \, {\left (x^{2} - 1\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | x + 1 \right |}\right )}{12 \, \mathrm {sgn}\relax (x)} + \frac {\log \left ({\left | x - 1 \right |}\right )}{12 \, \mathrm {sgn}\relax (x)} - \frac {x}{6 \, {\left (x^{2} - 1\right )} \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.62, size = 121, normalized size = 2.37

method	result	size
default	\(-\frac {\sqrt {x^{2}-1}\, x^{2} \left (2 x \,\mathrm {arcsec}\relax (x )+\sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{6 \left (x^{4}-2 x^{2}+1\right )}+\frac {\sqrt {\frac {x^{2}-1}{x^{2}}}\, x \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}-1\right )}{6 \sqrt {x^{2}-1}}-\frac {\sqrt {\frac {x^{2}-1}{x^{2}}}\, x \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}+1\right )}{6 \sqrt {x^{2}-1}}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(x^2-1)^(1/2)*x^2/(x^4-2*x^2+1)*(2*x*arcsec(x)+((x^2-1)/x^2)^(1/2))+1/6/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)
*x*ln(1/x+I*(1-1/x^2)^(1/2)-1)-1/6/(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 = 0.64, size = 46, normalized size = 0.90 \[ -\frac {1}{3} \, {\left (\frac {x}{\sqrt {x^{2} - 1}} + \frac {x}{{\left (x^{2} - 1\right )}^{\frac {3}{2}}}\right )} \operatorname {arcsec}\relax (x) - \frac {x}{6 \, {\left (x^{2} - 1\right )}} - \frac {1}{12} \, \log \left (x + 1\right ) + \frac {1}{12} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*acos(1/x))/(x^2 - 1)^(5/2),x)

[Out]

int((x^2*acos(1/x))/(x^2 - 1)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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