\(\int \frac {x^2 \sec ^{-1}(x)}{(-1+x^2)^{5/2}} \, dx\) [687]

Optimal result

Integrand size = 15, antiderivative size = 51 \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {\sqrt {x^2}}{6 \left (1-x^2\right )}-\frac {1}{6} \coth ^{-1}\left (\sqrt {x^2}\right )-\frac {x^3 \sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {270, 5346, 12, 294, 213} \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {x \text {arctanh}(x)}{6 \sqrt {x^2}}+\frac {\sqrt {x^2}}{6 \left (1-x^2\right )}-\frac {x^3 \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}} \]

[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 213

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

Rule 270

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 294

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 5346

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*} \text {integral}& = -\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 \text {arctanh}(x)}{6 \sqrt {x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20 \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {-4 x^3 \sec ^{-1}(x)+\sqrt {1-\frac {1}{x^2}} x \left (-2 x+\left (-1+x^2\right ) \log (1-x)-\left (-1+x^2\right ) \log (1+x)\right )}{12 \left (-1+x^2\right )^{3/2}} \]

[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))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.49 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.63

method	result	size
default	\(-\frac {\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \left (2 \,\operatorname {arcsec}\left (x \right ) x^{4} \sqrt {\frac {x^{2}-1}{x^{2}}}-\ln \left (\frac {1}{\sqrt {1-\frac {1}{x^{2}}}}-\frac {1}{x \sqrt {1-\frac {1}{x^{2}}}}\right ) x^{4}+2 \ln \left (\frac {1}{\sqrt {1-\frac {1}{x^{2}}}}-\frac {1}{x \sqrt {1-\frac {1}{x^{2}}}}\right ) x^{2}+x^{3}-\ln \left (\frac {1}{\sqrt {1-\frac {1}{x^{2}}}}-\frac {1}{x \sqrt {1-\frac {1}{x^{2}}}}\right )-x \right )}{6 \left (x^{2}-1\right )^{2}}\)	\(134\)

[In]

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

[Out]

-1/6*csgn(x*(1-1/x^2)^(1/2))*(2*arcsec(x)*x^4*((x^2-1)/x^2)^(1/2)-ln(1/(1-1/x^2)^(1/2)-1/x/(1-1/x^2)^(1/2))*x^
4+2*ln(1/(1-1/x^2)^(1/2)-1/x/(1-1/x^2)^(1/2))*x^2+x^3-ln(1/(1-1/x^2)^(1/2)-1/x/(1-1/x^2)^(1/2))-x)/(x^2-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {4 \, \sqrt {x^{2} - 1} x^{3} \operatorname {arcsec}\left (x\right ) + 2 \, x^{3} + {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right ) - {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x - 1\right ) - 2 \, x}{12 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, {\left (\frac {x}{\sqrt {x^{2} - 1}} + \frac {x}{{\left (x^{2} - 1\right )}^{\frac {3}{2}}}\right )} \operatorname {arcsec}\left (x\right ) - \frac {x}{6 \, {\left (x^{2} - 1\right )}} - \frac {1}{12} \, \log \left (x + 1\right ) + \frac {1}{12} \, \log \left (x - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04 \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {x^{3} \arccos \left (\frac {1}{x}\right )}{3 \, {\left (x^{2} - 1\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | x + 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} + \frac {\log \left ({\left | x - 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} - \frac {x}{6 \, {\left (x^{2} - 1\right )} \mathrm {sgn}\left (x\right )} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int \frac {x^2\,\mathrm {acos}\left (\frac {1}{x}\right )}{{\left (x^2-1\right )}^{5/2}} \,d x \]

[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)