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

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

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Rubi [A]  time = 0.20, 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 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]])

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 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.31, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)^2}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.12, size = 59, normalized size = 0.44 \[ \frac {8 \, x^{4} \operatorname {arcsec}\relax (x)^{3} + {\left (17 \, x^{4} - 40 \, x^{2} + 8\right )} \operatorname {arcsec}\relax (x) - {\left (8 \, {\left (5 \, x^{2} - 2\right )} \operatorname {arcsec}\relax (x)^{2} - 17 \, x^{2} + 2\right )} \sqrt {x^{2} - 1}}{64 \, x^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} - 1\right )}^{\frac {3}{2}} \operatorname {arcsec}\relax (x)^{2}}{x^{5}}\,{d x} \]

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)

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maple [C]  time = 0.70, size = 386, normalized size = 2.90

method	result	size
default	\(\frac {\sqrt {\frac {x^{2}-1}{x^{2}}}\, x \mathrm {arcsec}\relax (x )^{3}}{8 \sqrt {x^{2}-1}}+\frac {\left (-5 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}+x^{6}+20 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-13 x^{4}-16 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +28 x^{2}-16\right ) \left (4 i \mathrm {arcsec}\relax (x )+8 \mathrm {arcsec}\relax (x )^{2}-1\right )}{1024 \sqrt {x^{2}-1}\, x^{4}}-\frac {\left (-i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +x^{2}-1\right ) \left (2 \mathrm {arcsec}\relax (x )^{2}-1+2 i \mathrm {arcsec}\relax (x )\right )}{32 \sqrt {x^{2}-1}}+\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -2 x^{2}+2\right ) \left (2 \mathrm {arcsec}\relax (x )^{2}-1-2 i \mathrm {arcsec}\relax (x )\right )}{16 \sqrt {x^{2}-1}\, x^{2}}-\frac {\left (-5 x^{2}+4+3 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}+x^{4}-4 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \right ) \left (-4 i \mathrm {arcsec}\relax (x )+8 \mathrm {arcsec}\relax (x )^{2}-1\right )}{1024 \sqrt {x^{2}-1}\, x^{2}}+\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +x^{2}-1\right ) \left (7 i \mathrm {arcsec}\relax (x )+8 \mathrm {arcsec}\relax (x )^{2}-4\right ) \cos \left (4 \,\mathrm {arcsec}\relax (x )\right )}{128 \sqrt {x^{2}-1}}+\frac {\left (i x^{2}-\sqrt {\frac {x^{2}-1}{x^{2}}}\, x -i\right ) \left (32 i \mathrm {arcsec}\relax (x )+24 \mathrm {arcsec}\relax (x )^{2}-15\right ) \sin \left (4 \,\mathrm {arcsec}\relax (x )\right )}{512 \sqrt {x^{2}-1}}\)	\(386\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*arcsec(x)^3+1/1024/(x^2-1)^(1/2)/x^4*(-5*I*((x^2-1)/x^2)^(1/2)*x^5+x^6
+20*I*((x^2-1)/x^2)^(1/2)*x^3-13*x^4-16*I*((x^2-1)/x^2)^(1/2)*x+28*x^2-16)*(4*I*arcsec(x)+8*arcsec(x)^2-1)-1/3
2/(x^2-1)^(1/2)*(-I*((x^2-1)/x^2)^(1/2)*x+x^2-1)*(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))-1/1024/(x^2-1)^(1/2)*(
-5*x^2+4+3*I*((x^2-1)/x^2)^(1/2)*x^3+x^4-4*I*((x^2-1)/x^2)^(1/2)*x)*(-4*I*arcsec(x)+8*arcsec(x)^2-1)/x^2+1/128
/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x+x^2-1)*(7*I*arcsec(x)+8*arcsec(x)^2-4)*cos(4*arcsec(x))+1/512/(x^2-1)^
(1/2)*(I*x^2-((x^2-1)/x^2)^(1/2)*x-I)*(32*I*arcsec(x)+24*arcsec(x)^2-15)*sin(4*arcsec(x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} - 1\right )}^{\frac {3}{2}} \operatorname {arcsec}\relax (x)^{2}}{x^{5}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {acos}\left (\frac {1}{x}\right )}^2\,{\left (x^2-1\right )}^{3/2}}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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