\(\int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx\) [694]

Optimal result

Integrand size = 17, antiderivative size = 110 \[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=\frac {2 \left (1-21 x^2\right )}{27 \left (x^2\right )^{3/2}}-\frac {4 \sqrt {-1+x^2} \sec ^{-1}(x)}{3 x}-\frac {2 \left (-1+x^2\right )^{3/2} \sec ^{-1}(x)}{9 x^3}+\frac {2 \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (-1+x^2\right ) \sec ^{-1}(x)^2}{3 \left (x^2\right )^{3/2}}+\frac {\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)^3}{3 x^3} \]

[Out]

-2/9*(x^2-1)^(3/2)*arcsec(x)/x^3+1/3*(x^2-1)^(3/2)*arcsec(x)^3/x^3+2/27*(-21*x^2+1)/x^2/(x^2)^(1/2)+2/3*arcsec
(x)^2/(x^2)^(1/2)+1/3*(x^2-1)*arcsec(x)^2/x^2/(x^2)^(1/2)-4/3*arcsec(x)*(x^2-1)^(1/2)/x

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5350, 4768, 4744, 4716, 8} \[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=-\frac {14}{9 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)^3}{3 x}+\frac {\left (1-\frac {1}{x^2}\right ) \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {2 \sec ^{-1}(x)^2}{3 \sqrt {x^2}}-\frac {2 \left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)}{9 x}-\frac {4 \sqrt {1-\frac {1}{x^2}} \sqrt {x^2} \sec ^{-1}(x)}{3 x}+\frac {2 \sqrt {x^2}}{27 x^4} \]

[In]

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

[Out]

-14/(9*Sqrt[x^2]) + (2*Sqrt[x^2])/(27*x^4) - (4*Sqrt[1 - x^(-2)]*Sqrt[x^2]*ArcSec[x])/(3*x) - (2*(1 - x^(-2))^
(3/2)*Sqrt[x^2]*ArcSec[x])/(9*x) + (2*ArcSec[x]^2)/(3*Sqrt[x^2]) + ((1 - x^(-2))*ArcSec[x]^2)/(3*Sqrt[x^2]) +
((1 - x^(-2))^(3/2)*Sqrt[x^2]*ArcSec[x]^3)/(3*x)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4716

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcCos[c*x])^n, x] + Dist[b*c*n, Int[
x*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4744

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[x*(d + e*x^2)^p*((
a + b*ArcCos[c*x])^n/(2*p + 1)), x] + (Dist[2*d*(p/(2*p + 1)), Int[(d + e*x^2)^(p - 1)*(a + b*ArcCos[c*x])^n,
x], x] + Dist[b*c*(n/(2*p + 1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcC
os[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0]

Rule 4768

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(
p + 1)*((a + b*ArcCos[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p
], Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*
d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 5350

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist[-Sqrt[x
^2]/x, Subst[Int[(e + d*x^2)^p*((a + b*ArcCos[x/c])^n/x^(m + 2*(p + 1))), x], x, 1/x], x] /; FreeQ[{a, b, c, d
, e, n}, x] && IGtQ[n, 0] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ[p + 1/2] && GtQ[e, 0] && LtQ[d, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {x^2} \text {Subst}\left (\int x \sqrt {1-x^2} \arccos (x)^3 \, dx,x,\frac {1}{x}\right )}{x} \\ & = \frac {\left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)^3}{3 x}+\frac {\sqrt {x^2} \text {Subst}\left (\int \left (1-x^2\right ) \arccos (x)^2 \, dx,x,\frac {1}{x}\right )}{x} \\ & = \frac {\left (1-\frac {1}{x^2}\right ) \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)^3}{3 x}+\frac {\left (2 \sqrt {x^2}\right ) \text {Subst}\left (\int x \sqrt {1-x^2} \arccos (x) \, dx,x,\frac {1}{x}\right )}{3 x}+\frac {\left (2 \sqrt {x^2}\right ) \text {Subst}\left (\int \arccos (x)^2 \, dx,x,\frac {1}{x}\right )}{3 x} \\ & = -\frac {2 \left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)}{9 x}+\frac {2 \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right ) \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)^3}{3 x}-\frac {\left (2 \sqrt {x^2}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\frac {1}{x}\right )}{9 x}+\frac {\left (4 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {x \arccos (x)}{\sqrt {1-x^2}} \, dx,x,\frac {1}{x}\right )}{3 x} \\ & = -\frac {2}{9 \sqrt {x^2}}+\frac {2 \sqrt {x^2}}{27 x^4}-\frac {4 \sqrt {1-\frac {1}{x^2}} \sqrt {x^2} \sec ^{-1}(x)}{3 x}-\frac {2 \left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)}{9 x}+\frac {2 \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right ) \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)^3}{3 x}-\frac {\left (4 \sqrt {x^2}\right ) \text {Subst}\left (\int 1 \, dx,x,\frac {1}{x}\right )}{3 x} \\ & = -\frac {14}{9 \sqrt {x^2}}+\frac {2 \sqrt {x^2}}{27 x^4}-\frac {4 \sqrt {1-\frac {1}{x^2}} \sqrt {x^2} \sec ^{-1}(x)}{3 x}-\frac {2 \left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)}{9 x}+\frac {2 \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right ) \sec ^{-1}(x)^2}{3 \sqrt {x^2}}+\frac {\left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \sec ^{-1}(x)^3}{3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=\frac {2 \sqrt {1-\frac {1}{x^2}} x \left (1-21 x^2\right )-6 \left (1-8 x^2+7 x^4\right ) \sec ^{-1}(x)+9 \sqrt {1-\frac {1}{x^2}} x \left (-1+3 x^2\right ) \sec ^{-1}(x)^2+9 \left (-1+x^2\right )^2 \sec ^{-1}(x)^3}{27 x^3 \sqrt {-1+x^2}} \]

[In]

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

[Out]

(2*Sqrt[1 - x^(-2)]*x*(1 - 21*x^2) - 6*(1 - 8*x^2 + 7*x^4)*ArcSec[x] + 9*Sqrt[1 - x^(-2)]*x*(-1 + 3*x^2)*ArcSe
c[x]^2 + 9*(-1 + x^2)^2*ArcSec[x]^3)/(27*x^3*Sqrt[-1 + x^2])

Maple [C] (warning: unable to verify)

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

Time = 0.54 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.34

method	result	size
default	\(\frac {\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \sqrt {\frac {x^{2}-1}{x^{2}}}\, \left (9 \operatorname {arcsec}\left (x \right )^{3} x^{4}+27 \operatorname {arcsec}\left (x \right )^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-18 x^{2} \operatorname {arcsec}\left (x \right )^{3}-42 \,\operatorname {arcsec}\left (x \right ) x^{4}-9 \operatorname {arcsec}\left (x \right )^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -42 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}+9 \operatorname {arcsec}\left (x \right )^{3}+48 \,\operatorname {arcsec}\left (x \right ) x^{2}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -6 \,\operatorname {arcsec}\left (x \right )\right )}{27 \left (x^{2}-1\right ) x^{2}}\)	\(147\)

[In]

int(arcsec(x)^3*(x^2-1)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/27*csgn(x*(1-1/x^2)^(1/2))*((x^2-1)/x^2)^(1/2)/(x^2-1)/x^2*(9*arcsec(x)^3*x^4+27*arcsec(x)^2*((x^2-1)/x^2)^(
1/2)*x^3-18*x^2*arcsec(x)^3-42*arcsec(x)*x^4-9*arcsec(x)^2*((x^2-1)/x^2)^(1/2)*x-42*((x^2-1)/x^2)^(1/2)*x^3+9*
arcsec(x)^3+48*arcsec(x)*x^2+2*((x^2-1)/x^2)^(1/2)*x-6*arcsec(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=\frac {9 \, {\left (3 \, x^{2} - 1\right )} \operatorname {arcsec}\left (x\right )^{2} - 42 \, x^{2} + 3 \, {\left (3 \, {\left (x^{2} - 1\right )} \operatorname {arcsec}\left (x\right )^{3} - 2 \, {\left (7 \, x^{2} - 1\right )} \operatorname {arcsec}\left (x\right )\right )} \sqrt {x^{2} - 1} + 2}{27 \, x^{3}} \]

[In]

integrate(arcsec(x)^3*(x^2-1)^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/27*(9*(3*x^2 - 1)*arcsec(x)^2 - 42*x^2 + 3*(3*(x^2 - 1)*arcsec(x)^3 - 2*(7*x^2 - 1)*arcsec(x))*sqrt(x^2 - 1)
+ 2)/x^3

Sympy [F]

\[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=\int \frac {\sqrt {\left (x - 1\right ) \left (x + 1\right )} \operatorname {asec}^{3}{\left (x \right )}}{x^{4}}\, dx \]

[In]

integrate(asec(x)**3*(x**2-1)**(1/2)/x**4,x)

[Out]

Integral(sqrt((x - 1)*(x + 1))*asec(x)**3/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=\frac {{\left (x^{2} - 1\right )}^{\frac {3}{2}} \operatorname {arcsec}\left (x\right )^{3}}{3 \, x^{3}} + \frac {{\left (3 \, x^{2} - 1\right )} \operatorname {arcsec}\left (x\right )^{2}}{3 \, x^{3}} - \frac {2 \, {\left ({\left (21 \, x^{2} - 1\right )} \sqrt {x + 1} \sqrt {x - 1} + 3 \, {\left (7 \, x^{4} - 8 \, x^{2} + 1\right )} \arctan \left (\sqrt {x + 1} \sqrt {x - 1}\right )\right )}}{27 \, \sqrt {x + 1} \sqrt {x - 1} x^{3}} \]

[In]

integrate(arcsec(x)^3*(x^2-1)^(1/2)/x^4,x, algorithm="maxima")

[Out]

1/3*(x^2 - 1)^(3/2)*arcsec(x)^3/x^3 + 1/3*(3*x^2 - 1)*arcsec(x)^2/x^3 - 2/27*((21*x^2 - 1)*sqrt(x + 1)*sqrt(x
- 1) + 3*(7*x^4 - 8*x^2 + 1)*arctan(sqrt(x + 1)*sqrt(x - 1)))/(sqrt(x + 1)*sqrt(x - 1)*x^3)

Giac [F]

\[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=\int { \frac {\sqrt {x^{2} - 1} \operatorname {arcsec}\left (x\right )^{3}}{x^{4}} \,d x } \]

[In]

integrate(arcsec(x)^3*(x^2-1)^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate(sqrt(x^2 - 1)*arcsec(x)^3/x^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx=\int \frac {{\mathrm {acos}\left (\frac {1}{x}\right )}^3\,\sqrt {x^2-1}}{x^4} \,d x \]

[In]

int((acos(1/x)^3*(x^2 - 1)^(1/2))/x^4,x)

[Out]

int((acos(1/x)^3*(x^2 - 1)^(1/2))/x^4, x)