\(\int \frac {x^3 \sec ^{-1}(x)}{(-1+x^2)^{5/2}} \, dx\) [688]

Optimal result

Integrand size = 15, antiderivative size = 82 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {x}{6 \sqrt {x^2} \left (1-x^2\right )}-\frac {\sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}-\frac {\sec ^{-1}(x)}{\sqrt {-1+x^2}}-\frac {2 x \log (x)}{3 \sqrt {x^2}}+\frac {x \log \left (-1+x^2\right )}{3 \sqrt {x^2}} \]

[Out]

-1/3*arcsec(x)/(x^2-1)^(3/2)+1/6*x/(-x^2+1)/(x^2)^(1/2)-2/3*x*ln(x)/(x^2)^(1/2)+1/3*x*ln(x^2-1)/(x^2)^(1/2)-ar
csec(x)/(x^2-1)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 45, 5346, 12, 457, 78} \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {x}{6 \sqrt {x^2} \left (1-x^2\right )}-\frac {2 x \log (x)}{3 \sqrt {x^2}}+\frac {x \log \left (1-x^2\right )}{3 \sqrt {x^2}}-\frac {\sec ^{-1}(x)}{\sqrt {x^2-1}}-\frac {\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}} \]

[In]

Int[(x^3*ArcSec[x])/(-1 + x^2)^(5/2),x]

[Out]

x/(6*Sqrt[x^2]*(1 - x^2)) - ArcSec[x]/(3*(-1 + x^2)^(3/2)) - ArcSec[x]/Sqrt[-1 + x^2] - (2*x*Log[x])/(3*Sqrt[x
^2]) + (x*Log[1 - x^2])/(3*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 272

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {-2 \left (-2+3 x^2\right ) \sec ^{-1}(x)+\sqrt {1-\frac {1}{x^2}} x \left (-1+2 \left (-1+x^2\right ) \log (-1+x)-4 \left (-1+x^2\right ) \log (x)-2 \log (1+x)+2 x^2 \log (1+x)\right )}{6 \left (-1+x^2\right )^{3/2}} \]

[In]

Integrate[(x^3*ArcSec[x])/(-1 + x^2)^(5/2),x]

[Out]

(-2*(-2 + 3*x^2)*ArcSec[x] + Sqrt[1 - x^(-2)]*x*(-1 + 2*(-1 + x^2)*Log[-1 + x] - 4*(-1 + x^2)*Log[x] - 2*Log[1
+ x] + 2*x^2*Log[1 + x]))/(6*(-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 = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.23

method	result	size
default	\(-\frac {\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \left (6 x^{3} \operatorname {arcsec}\left (x \right ) \sqrt {\frac {x^{2}-1}{x^{2}}}-2 \ln \left (1-\frac {1}{x^{2}}\right ) x^{4}+x^{4}-4 \,\operatorname {arcsec}\left (x \right ) x \sqrt {\frac {x^{2}-1}{x^{2}}}+4 \ln \left (1-\frac {1}{x^{2}}\right ) x^{2}-x^{2}-2 \ln \left (1-\frac {1}{x^{2}}\right )\right )}{6 \left (x^{2}-1\right )^{2}}\)	\(101\)

[In]

int(x^3*arcsec(x)/(x^2-1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, x^{2} - 2\right )} \sqrt {x^{2} - 1} \operatorname {arcsec}\left (x\right ) + x^{2} - 2 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x^{2} - 1\right ) + 4 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x\right ) - 1}{6 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

[In]

integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(2*(3*x^2 - 2)*sqrt(x^2 - 1)*arcsec(x) + x^2 - 2*(x^4 - 2*x^2 + 1)*log(x^2 - 1) + 4*(x^4 - 2*x^2 + 1)*log
(x) - 1)/(x^4 - 2*x^2 + 1)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(x**3*asec(x)/(x**2-1)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int { \frac {x^{3} \operatorname {arcsec}\left (x\right )}{{\left (x^{2} - 1\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3*arcsec(x)/(x^2 - 1)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {{\left (3 \, x^{2} - 2\right )} \arccos \left (\frac {1}{x}\right )}{3 \, {\left (x^{2} - 1\right )}^{\frac {3}{2}}} - \frac {\log \left (x^{2}\right )}{3 \, \mathrm {sgn}\left (x\right )} + \frac {\log \left ({\left | x^{2} - 1 \right |}\right )}{3 \, \mathrm {sgn}\left (x\right )} - \frac {2 \, x^{2} - 1}{6 \, {\left (x^{2} - 1\right )} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\mathrm {acos}\left (\frac {1}{x}\right )}{{\left (x^2-1\right )}^{5/2}} \,d x \]

[In]

int((x^3*acos(1/x))/(x^2 - 1)^(5/2),x)

[Out]

int((x^3*acos(1/x))/(x^2 - 1)^(5/2), x)