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

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

[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

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Rubi [A]
time = 0.13, 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 = {5350, 4768, 4744, 4716, 8} \begin {gather*} -\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} \end {gather*}

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 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*} \int \frac {\sqrt {-1+x^2} \sec ^{-1}(x)^3}{x^4} \, dx &=-\frac {\sqrt {x^2} \text {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} \text {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 ) \text {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 ) \text {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 ) \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 \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 ) \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*}

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Mathematica [A]
time = 0.05, size = 92, normalized size = 0.84 \begin {gather*} \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}} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.54, size = 536, normalized size = 4.87

method	result	size
default	\(\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}-8 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}+4 x^{4}+8 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -12 x^{2}+8\right ) \mathrm {arcsec}\left (x \right )^{3}}{48 x^{3} \sqrt {x^{2}-1}}-\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}-8 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}+4 x^{4}+8 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -12 x^{2}+8\right ) \mathrm {arcsec}\left (x \right )}{72 x^{3} \sqrt {x^{2}-1}}-\frac {\left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}-4 i x^{4}-8 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}+12 i x^{2}+8 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -8 i\right ) \mathrm {arcsec}\left (x \right )^{2}}{48 x^{3} \sqrt {x^{2}-1}}+\frac {\sqrt {x^{2}-1}\, \left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{5}-5 i x^{4}-12 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}+20 i x^{2}+16 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -16 i\right )}{216 \left (-i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +x^{2}-1\right ) x^{3}}+\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +x^{2}-1\right ) \left (9 \mathrm {arcsec}\left (x \right )^{3}-117 i \mathrm {arcsec}\left (x \right )^{2}-78 \,\mathrm {arcsec}\left (x \right )+242 i\right )}{216 \sqrt {x^{2}-1}\, x}-\frac {\left (45 i \mathrm {arcsec}\left (x \right )^{2}+9 \mathrm {arcsec}\left (x \right )^{3}-82 i-78 \,\mathrm {arcsec}\left (x \right )\right ) \left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -1\right ) \sqrt {x^{2}-1}}{216 x}-\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +x^{2}-1\right ) \left (63 i \mathrm {arcsec}\left (x \right )^{2}+27 \mathrm {arcsec}\left (x \right )^{3}-158 i-162 \,\mathrm {arcsec}\left (x \right )\right ) \cos \left (3 \,\mathrm {arcsec}\left (x \right )\right )}{432 \sqrt {x^{2}-1}}-\frac {\left (i x^{2}-\sqrt {\frac {x^{2}-1}{x^{2}}}\, x -i\right ) \left (27 i \mathrm {arcsec}\left (x \right )^{2}+3 \mathrm {arcsec}\left (x \right )^{3}-54 i-50 \,\mathrm {arcsec}\left (x \right )\right ) \sin \left (3 \,\mathrm {arcsec}\left (x \right )\right )}{144 \sqrt {x^{2}-1}}\)	\(536\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48/x^3/(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)^3-1/72/x^3/(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)-1/48/x^3/(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+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
+1/216/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x+x^2-1)*(9*arcsec(x)^3-117*I*arcsec(x)^2-78*arcsec(x)+242*I)/x-1/
216*(45*I*arcsec(x)^2+9*arcsec(x)^3-82*I-78*arcsec(x))*(I*((x^2-1)/x^2)^(1/2)*x-1)*(x^2-1)^(1/2)/x-1/432/(x^2-
1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x+x^2-1)*(63*I*arcsec(x)^2+27*arcsec(x)^3-158*I-162*arcsec(x))*cos(3*arcsec(x)
)-1/144/(x^2-1)^(1/2)*(I*x^2-((x^2-1)/x^2)^(1/2)*x-I)*(27*I*arcsec(x)^2+3*arcsec(x)^3-54*I-50*arcsec(x))*sin(3
*arcsec(x))

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Maxima [A]
time = 3.20, size = 93, normalized size = 0.85 \begin {gather*} \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 {gather*}

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 = 0.78, size = 57, normalized size = 0.52 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\left (x - 1\right ) \left (x + 1\right )} \operatorname {asec}^{3}{\left (x \right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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(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)

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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 )}^3\,\sqrt {x^2-1}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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