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

Optimal. Leaf size=133 \[ \frac{\sqrt{x^2-1} \left (17 x^2-2\right )}{64 x^4}+\frac{x \sec ^{-1}(x)^3}{8 \sqrt{x^2}}-\frac{\left (x^2-1\right )^{3/2} \sec ^{-1}(x)^2}{4 x^4}-\frac{3 \sqrt{x^2-1} \sec ^{-1}(x)^2}{8 x^2}+\frac{\left (x^2-1\right )^2 \sec ^{-1}(x)}{8 x^3 \sqrt{x^2}}+\frac{9 x \sec ^{-1}(x)}{64 \sqrt{x^2}}-\frac{3 \sec ^{-1}(x)}{8 x \sqrt{x^2}} \]

[Out]

(Sqrt[-1 + x^2]*(-2 + 17*x^2))/(64*x^4) - (3*ArcSec[x])/(8*x*Sqrt[x^2]) + (9*x*ArcSec[x])/(64*Sqrt[x^2]) + ((-
1 + x^2)^2*ArcSec[x])/(8*x^3*Sqrt[x^2]) - (3*Sqrt[-1 + x^2]*ArcSec[x]^2)/(8*x^2) - ((-1 + x^2)^(3/2)*ArcSec[x]
^2)/(4*x^4) + (x*ArcSec[x]^3)/(8*Sqrt[x^2])

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Rubi [A]  time = 0.201347, antiderivative size = 172, normalized size of antiderivative = 1.29, number of steps used = 11, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {5242, 4650, 4648, 4642, 4628, 321, 216, 4678, 195} \[ \frac{\left (1-\frac{1}{x^2}\right )^{3/2}}{32 \sqrt{x^2}}+\frac{15 \sqrt{1-\frac{1}{x^2}}}{64 \sqrt{x^2}}-\frac{9 \sqrt{x^2} \csc ^{-1}(x)}{64 x}+\frac{\sqrt{x^2} \sec ^{-1}(x)^3}{8 x}-\frac{\left (1-\frac{1}{x^2}\right )^{3/2} \sec ^{-1}(x)^2}{4 \sqrt{x^2}}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^2}{8 \sqrt{x^2}}+\frac{\left (1-\frac{1}{x^2}\right )^2 \sqrt{x^2} \sec ^{-1}(x)}{8 x}-\frac{3 \sqrt{x^2} \sec ^{-1}(x)}{8 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[1 - x^(-2)])/(64*Sqrt[x^2]) + (1 - x^(-2))^(3/2)/(32*Sqrt[x^2]) - (9*Sqrt[x^2]*ArcCsc[x])/(64*x) - (3
*Sqrt[x^2]*ArcSec[x])/(8*x^3) + ((1 - x^(-2))^2*Sqrt[x^2]*ArcSec[x])/(8*x) - (3*Sqrt[1 - x^(-2)]*ArcSec[x]^2)/
(8*Sqrt[x^2]) - ((1 - x^(-2))^(3/2)*ArcSec[x]^2)/(4*Sqrt[x^2]) + (Sqrt[x^2]*ArcSec[x]^3)/(8*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 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 4648

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

Rule 4642

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

Rule 4628

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

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

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

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rubi steps

\begin{align*} \int \frac{\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)^2}{x^5} \, dx &=-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2 \, dx,x,\frac{1}{x}\right )}{x}\\ &=-\frac{\left (1-\frac{1}{x^2}\right )^{3/2} \sec ^{-1}(x)^2}{4 \sqrt{x^2}}-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int x \left (1-x^2\right ) \cos ^{-1}(x) \, dx,x,\frac{1}{x}\right )}{2 x}-\frac{\left (3 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \sqrt{1-x^2} \cos ^{-1}(x)^2 \, dx,x,\frac{1}{x}\right )}{4 x}\\ &=\frac{\left (1-\frac{1}{x^2}\right )^2 \sqrt{x^2} \sec ^{-1}(x)}{8 x}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^2}{8 \sqrt{x^2}}-\frac{\left (1-\frac{1}{x^2}\right )^{3/2} \sec ^{-1}(x)^2}{4 \sqrt{x^2}}+\frac{\sqrt{x^2} \operatorname{Subst}\left (\int \left (1-x^2\right )^{3/2} \, dx,x,\frac{1}{x}\right )}{8 x}-\frac{\left (3 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{\cos ^{-1}(x)^2}{\sqrt{1-x^2}} \, dx,x,\frac{1}{x}\right )}{8 x}-\frac{\left (3 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int x \cos ^{-1}(x) \, dx,x,\frac{1}{x}\right )}{4 x}\\ &=\frac{\left (1-\frac{1}{x^2}\right )^{3/2}}{32 \sqrt{x^2}}-\frac{3 \sqrt{x^2} \sec ^{-1}(x)}{8 x^3}+\frac{\left (1-\frac{1}{x^2}\right )^2 \sqrt{x^2} \sec ^{-1}(x)}{8 x}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^2}{8 \sqrt{x^2}}-\frac{\left (1-\frac{1}{x^2}\right )^{3/2} \sec ^{-1}(x)^2}{4 \sqrt{x^2}}+\frac{\sqrt{x^2} \sec ^{-1}(x)^3}{8 x}+\frac{\left (3 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \sqrt{1-x^2} \, dx,x,\frac{1}{x}\right )}{32 x}-\frac{\left (3 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2}} \, dx,x,\frac{1}{x}\right )}{8 x}\\ &=\frac{15 \sqrt{1-\frac{1}{x^2}}}{64 \sqrt{x^2}}+\frac{\left (1-\frac{1}{x^2}\right )^{3/2}}{32 \sqrt{x^2}}-\frac{3 \sqrt{x^2} \sec ^{-1}(x)}{8 x^3}+\frac{\left (1-\frac{1}{x^2}\right )^2 \sqrt{x^2} \sec ^{-1}(x)}{8 x}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^2}{8 \sqrt{x^2}}-\frac{\left (1-\frac{1}{x^2}\right )^{3/2} \sec ^{-1}(x)^2}{4 \sqrt{x^2}}+\frac{\sqrt{x^2} \sec ^{-1}(x)^3}{8 x}+\frac{\left (3 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,\frac{1}{x}\right )}{64 x}-\frac{\left (3 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,\frac{1}{x}\right )}{16 x}\\ &=\frac{15 \sqrt{1-\frac{1}{x^2}}}{64 \sqrt{x^2}}+\frac{\left (1-\frac{1}{x^2}\right )^{3/2}}{32 \sqrt{x^2}}-\frac{9 \sqrt{x^2} \csc ^{-1}(x)}{64 x}-\frac{3 \sqrt{x^2} \sec ^{-1}(x)}{8 x^3}+\frac{\left (1-\frac{1}{x^2}\right )^2 \sqrt{x^2} \sec ^{-1}(x)}{8 x}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^2}{8 \sqrt{x^2}}-\frac{\left (1-\frac{1}{x^2}\right )^{3/2} \sec ^{-1}(x)^2}{4 \sqrt{x^2}}+\frac{\sqrt{x^2} \sec ^{-1}(x)^3}{8 x}\\ \end{align*}

Mathematica [A]  time = 0.240602, size = 84, normalized size = 0.63 \[ \frac{\sqrt{x^2-1} \left (32 \sec ^{-1}(x)^3+4 \sec ^{-1}(x) \left (\cos \left (4 \sec ^{-1}(x)\right )-16 \cos \left (2 \sec ^{-1}(x)\right )\right )+8 \sec ^{-1}(x)^2 \left (\sin \left (4 \sec ^{-1}(x)\right )-8 \sin \left (2 \sec ^{-1}(x)\right )\right )+32 \sin \left (2 \sec ^{-1}(x)\right )-\sin \left (4 \sec ^{-1}(x)\right )\right )}{256 \sqrt{1-\frac{1}{x^2}} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-1 + x^2]*(32*ArcSec[x]^3 + 4*ArcSec[x]*(-16*Cos[2*ArcSec[x]] + Cos[4*ArcSec[x]]) + 32*Sin[2*ArcSec[x]]
- Sin[4*ArcSec[x]] + 8*ArcSec[x]^2*(-8*Sin[2*ArcSec[x]] + Sin[4*ArcSec[x]])))/(256*Sqrt[1 - x^(-2)]*x)

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Maple [C]  time = 0.355, size = 327, normalized size = 2.5 \begin{align*}{\frac{x \left ({\rm arcsec} \left (x\right ) \right ) ^{3}}{8}\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{\frac{1}{\sqrt{{x}^{2}-1}}}}-{\frac{4\,i{\rm arcsec} \left (x\right )+8\, \left ({\rm arcsec} \left (x\right ) \right ) ^{2}-1}{512\,{x}^{4}} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{5}-8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}+4\,{x}^{4}+8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-12\,{x}^{2}+8 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}}-{\frac{2\, \left ({\rm arcsec} \left (x\right ) \right ) ^{2}-1+2\,i{\rm arcsec} \left (x\right )}{16\,{x}^{2}} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-2\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+2\,{x}^{2}-2 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}}+{\frac{2\, \left ({\rm arcsec} \left (x\right ) \right ) ^{2}-1-2\,i{\rm arcsec} \left (x\right )}{16\,{x}^{2}} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-2\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-2\,{x}^{2}+2 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}}+{\frac{-4\,i{\rm arcsec} \left (x\right )+8\, \left ({\rm arcsec} \left (x\right ) \right ) ^{2}-1}{512\,{x}^{4}} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{5}-8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-4\,{x}^{4}+8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+12\,{x}^{2}-8 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*arcsec(x)^3-1/512/(x^2-1)^(1/2)/x^4*(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)*(4*I*arcsec(x)+8*arcsec(x)^2-1)-1/16/(x^2-1)^(1
/2)/x^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)*(2*arcsec(x)^2-1+2*I*arcsec(x))+1/16/(x^
2-1)^(1/2)/x^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)*(2*arcsec(x)^2-1-2*I*arcsec(x))+1
/512/(x^2-1)^(1/2)/x^4*(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)*(-4*I*arcsec(x)+8*arcsec(x)^2-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{2} - 1\right )}^{\frac{3}{2}} \operatorname{arcsec}\left (x\right )^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.39838, size = 163, normalized size = 1.23 \begin{align*} \frac{8 \, x^{4} \operatorname{arcsec}\left (x\right )^{3} +{\left (17 \, x^{4} - 40 \, x^{2} + 8\right )} \operatorname{arcsec}\left (x\right ) -{\left (8 \,{\left (5 \, x^{2} - 2\right )} \operatorname{arcsec}\left (x\right )^{2} - 17 \, x^{2} + 2\right )} \sqrt{x^{2} - 1}}{64 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(8*x^4*arcsec(x)^3 + (17*x^4 - 40*x^2 + 8)*arcsec(x) - (8*(5*x^2 - 2)*arcsec(x)^2 - 17*x^2 + 2)*sqrt(x^2
- 1))/x^4

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{2} - 1\right )}^{\frac{3}{2}} \operatorname{arcsec}\left (x\right )^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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