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

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

[Out]

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

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Rubi [A]  time = 0.063826, 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 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 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]))

Rule 12

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

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 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])

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*}

Mathematica [A]  time = 0.116783, size = 61, normalized size = 1.2 \[ \frac{\sqrt{1-\frac{1}{x^2}} x \left (\left (x^2-1\right ) \log (1-x)-\left (x^2-1\right ) \log (x+1)-2 x\right )-4 x^3 \sec ^{-1}(x)}{12 \left (x^2-1\right )^{3/2}} \]

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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Maple [C]  time = 0.248, size = 121, normalized size = 2.4 \begin{align*} -{\frac{{x}^{2}}{6\,{x}^{4}-12\,{x}^{2}+6}\sqrt{{x}^{2}-1} \left ( 2\,{\rm arcsec} \left (x\right )x+\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}} \right ) }+{\frac{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{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(x^2*arcsec(x)/(x^2-1)^(5/2),x)

[Out]

-1/6*(x^2-1)^(1/2)*x^2/(x^4-2*x^2+1)*(2*arcsec(x)*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 = 1.15586, size = 62, normalized size = 1.22 \begin{align*} -\frac{1}{3} \,{\left (\frac{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{1}{12} \, \log \left (x + 1\right ) + \frac{1}{12} \, \log \left (x - 1\right ) \end{align*}

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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Fricas [A]  time = 2.64376, size = 180, normalized size = 3.53 \begin{align*} -\frac{4 \, \sqrt{x^{2} - 1} x^{3} \operatorname{arcsec}\left (x\right ) + 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 )}} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.09206, size = 72, normalized size = 1.41 \begin{align*} -\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}\left (x\right )} + \frac{\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(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))