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

Optimal. Leaf size=110 \[ \frac{2 \left (1-21 x^2\right )}{27 \left (x^2\right )^{3/2}}+\frac{\left (x^2-1\right )^{3/2} \sec ^{-1}(x)^3}{3 x^3}+\frac{\left (x^2-1\right ) \sec ^{-1}(x)^2}{3 \left (x^2\right )^{3/2}}+\frac{2 \sec ^{-1}(x)^2}{3 \sqrt{x^2}}-\frac{2 \left (x^2-1\right )^{3/2} \sec ^{-1}(x)}{9 x^3}-\frac{4 \sqrt{x^2-1} \sec ^{-1}(x)}{3 x} \]

[Out]

(2*(1 - 21*x^2))/(27*(x^2)^(3/2)) - (4*Sqrt[-1 + x^2]*ArcSec[x])/(3*x) - (2*(-1 + x^2)^(3/2)*ArcSec[x])/(9*x^3
) + (2*ArcSec[x]^2)/(3*Sqrt[x^2]) + ((-1 + x^2)*ArcSec[x]^2)/(3*(x^2)^(3/2)) + ((-1 + x^2)^(3/2)*ArcSec[x]^3)/
(3*x^3)

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Rubi [A]  time = 0.199997, antiderivative size = 146, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {5242, 4678, 4650, 4620, 8} \[ \frac{2 \sqrt{x^2}}{27 x^4}-\frac{14}{9 \sqrt{x^2}}+\frac{\left (1-\frac{1}{x^2}\right )^{3/2} \sqrt{x^2} \sec ^{-1}(x)^3}{3 x}+\frac{\left (1-\frac{1}{x^2}\right ) \sec ^{-1}(x)^2}{3 \sqrt{x^2}}+\frac{2 \sec ^{-1}(x)^2}{3 \sqrt{x^2}}-\frac{2 \left (1-\frac{1}{x^2}\right )^{3/2} \sqrt{x^2} \sec ^{-1}(x)}{9 x}-\frac{4 \sqrt{1-\frac{1}{x^2}} \sqrt{x^2} \sec ^{-1}(x)}{3 x} \]

Antiderivative was successfully verified.

[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 5242

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]

Rule 4678

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[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 4650

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c
^2*x^2)^(p - 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && Gt
Q[n, 0] && GtQ[p, 0]

Rule 4620

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 8

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

Rubi steps

\begin{align*} \int \frac{\sqrt{-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx &=-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int x \sqrt{1-x^2} \cos ^{-1}(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} \operatorname{Subst}\left (\int \left (1-x^2\right ) \cos ^{-1}(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 ) \operatorname{Subst}\left (\int x \sqrt{1-x^2} \cos ^{-1}(x) \, dx,x,\frac{1}{x}\right )}{3 x}+\frac{\left (2 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \cos ^{-1}(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 ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\frac{1}{x}\right )}{9 x}+\frac{\left (4 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{x \cos ^{-1}(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 ) \operatorname{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]  time = 0.0896488, size = 92, normalized size = 0.84 \[ \frac{2 \sqrt{1-\frac{1}{x^2}} x \left (1-21 x^2\right )+9 \left (x^2-1\right )^2 \sec ^{-1}(x)^3+9 \sqrt{1-\frac{1}{x^2}} x \left (3 x^2-1\right ) \sec ^{-1}(x)^2-6 \left (7 x^4-8 x^2+1\right ) \sec ^{-1}(x)}{27 x^3 \sqrt{x^2-1}} \]

Antiderivative was successfully verified.

[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])

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Maple [C]  time = 0.588, size = 1153, normalized size = 10.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48/(x^2-1)^(1/2)/x^3*(I*((x^2-1)/x^2)^(1/2)*x^5-8*I*((x^2-1)/x^2)^(1/2)*x^3+4*x^4+8*I*((x^2-1)/x^2)^(1/2)*x-
12*x^2+8)*arcsec(x)^3+1/24/(x^2-1)^(1/2)/x*(I*((x^2-1)/x^2)^(1/2)*x^3-2*I*((x^2-1)/x^2)^(1/2)*x+2*x^2-2)*arcse
c(x)^3+10/27*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x^3-3*I*x^2-4*((x^2-1)/x^2)^(1/2)*x+4*I)/(-I*((x^2-1)/x^2)^(1/
2)*x+x^2-1)/x-3/4*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x-I)*x/(-I*((x^2-1)/x^2)^(1/2)*x+x^2-1)-1/72/(x^2-1)^(1/2
)/x^3*(I*((x^2-1)/x^2)^(1/2)*x^5-8*I*((x^2-1)/x^2)^(1/2)*x^3+4*x^4+8*I*((x^2-1)/x^2)^(1/2)*x-12*x^2+8)*arcsec(
x)-13/36/(x^2-1)^(1/2)/x*(I*((x^2-1)/x^2)^(1/2)*x^3-2*I*((x^2-1)/x^2)^(1/2)*x+2*x^2-2)*arcsec(x)+I*(x^2-1)^(1/
2)*x^5/(216*I*((x^2-1)/x^2)^(1/2)*x^5-1728*I*((x^2-1)/x^2)^(1/2)*x^3+864*x^4+1728*I*((x^2-1)/x^2)^(1/2)*x-2592
*x^2+1728)-1/24/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x^3-2*I*((x^2-1)/x^2)^(1/2)*x-2*x^2+2)*arcsec(x)^3/x+10*I
*(x^2-1)^(1/2)*x^3/(27*I*((x^2-1)/x^2)^(1/2)*x^3-54*I*((x^2-1)/x^2)^(1/2)*x+54*x^2-54)+13/36/(x^2-1)^(1/2)*(I*
((x^2-1)/x^2)^(1/2)*x^3-2*I*((x^2-1)/x^2)^(1/2)*x-2*x^2+2)*arcsec(x)/x-1/48/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/
2)*x^5-8*I*((x^2-1)/x^2)^(1/2)*x^3-4*x^4+8*I*((x^2-1)/x^2)^(1/2)*x+12*x^2-8)*arcsec(x)^3/x^3-1/6/(x^2-1)^(1/2)
*(((x^2-1)/x^2)^(1/2)*x^3+2*I*x^2-2*((x^2-1)/x^2)^(1/2)*x-2*I)*arcsec(x)^2/x+1/72/(x^2-1)^(1/2)*(I*((x^2-1)/x^
2)^(1/2)*x^5-8*I*((x^2-1)/x^2)^(1/2)*x^3-4*x^4+8*I*((x^2-1)/x^2)^(1/2)*x+12*x^2-8)*arcsec(x)/x^3-1/48/(x^2-1)^
(1/2)*(((x^2-1)/x^2)^(1/2)*x^5+4*I*x^4-8*((x^2-1)/x^2)^(1/2)*x^3-12*I*x^2+8*((x^2-1)/x^2)^(1/2)*x+8*I)*arcsec(
x)^2/x^3-1/48/(x^2-1)^(1/2)/x^3*(((x^2-1)/x^2)^(1/2)*x^5-4*I*x^4-8*((x^2-1)/x^2)^(1/2)*x^3+12*I*x^2+8*((x^2-1)
/x^2)^(1/2)*x-8*I)*arcsec(x)^2-1/6/(x^2-1)^(1/2)/x*(((x^2-1)/x^2)^(1/2)*x^3-2*I*x^2-2*((x^2-1)/x^2)^(1/2)*x+2*
I)*arcsec(x)^2+3/8/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x^2*arcsec(x)^2+1/216*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*
x^5-5*I*x^4-12*((x^2-1)/x^2)^(1/2)*x^3+20*I*x^2+16*((x^2-1)/x^2)^(1/2)*x-16*I)/(-I*((x^2-1)/x^2)^(1/2)*x+x^2-1
)/x^3

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Maxima [A]  time = 2.53848, size = 126, normalized size = 1.15 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.43214, size = 163, normalized size = 1.48 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - 1} \operatorname{arcsec}\left (x\right )^{3}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)