3.7.93 \(\int \frac {(-1+x^2)^{3/2} \sec ^{-1}(x)^2}{x^5} \, dx\) [693]

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

[Out]

-1/4*(x^2-1)^(3/2)*arcsec(x)^2/x^4-3/8*arcsec(x)/x/(x^2)^(1/2)+9/64*x*arcsec(x)/(x^2)^(1/2)+1/8*(x^2-1)^2*arcs
ec(x)/x^3/(x^2)^(1/2)+1/8*x*arcsec(x)^3/(x^2)^(1/2)+1/64*(17*x^2-2)*(x^2-1)^(1/2)/x^4-3/8*arcsec(x)^2*(x^2-1)^
(1/2)/x^2

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Rubi [A]
time = 0.14, 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 = {5350, 4744, 4742, 4738, 4724, 327, 222, 4768, 201} \begin {gather*} \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} \end {gather*}

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 201

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 222

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 327

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 4724

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 4738

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

Rule 4742

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[(1/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(a + b*ArcCos[c*x])^n/S
qrt[1 - c^2*x^2], x], x] + Dist[b*c*(n/2)*Simp[Sqrt[d + e*x^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 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*} \int \frac {\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)^2}{x^5} \, dx &=-\frac {\sqrt {x^2} \text {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} \text {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 ) \text {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} \text {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 ) \text {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 ) \text {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 ) \text {Subst}\left (\int \sqrt {1-x^2} \, dx,x,\frac {1}{x}\right )}{32 x}-\frac {\left (3 \sqrt {x^2}\right ) \text {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 ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,\frac {1}{x}\right )}{64 x}-\frac {\left (3 \sqrt {x^2}\right ) \text {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 = 0.19, size = 84, normalized size = 0.63 \begin {gather*} \frac {\sqrt {-1+x^2} \left (32 \sec ^{-1}(x)^3+4 \sec ^{-1}(x) \left (-16 \cos \left (2 \sec ^{-1}(x)\right )+\cos \left (4 \sec ^{-1}(x)\right )\right )+32 \sin \left (2 \sec ^{-1}(x)\right )-\sin \left (4 \sec ^{-1}(x)\right )+8 \sec ^{-1}(x)^2 \left (-8 \sin \left (2 \sec ^{-1}(x)\right )+\sin \left (4 \sec ^{-1}(x)\right )\right )\right )}{256 \sqrt {1-\frac {1}{x^2}} x} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.58, size = 386, normalized size = 2.90

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

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^4/(x^2-1)^(1/2)*(-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/(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)*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.60, size = 59, normalized size = 0.44 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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