3.7.50 \(\int \frac {\sec ^{-1}(x)^4}{x^5} \, dx\) [650]

Optimal. Leaf size=148 \[ -\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac {45}{128} \sec ^{-1}(x)^2+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4} \]

[Out]

-3/128/x^4-45/128/x^2-45/128*arcsec(x)^2+3/16*arcsec(x)^2/x^4+9/16*arcsec(x)^2/x^2+3/32*arcsec(x)^4-1/4*arcsec
(x)^4/x^4-3/32*arcsec(x)*(1-1/x^2)^(1/2)/x^3-45/64*arcsec(x)*(1-1/x^2)^(1/2)/x+1/4*arcsec(x)^3*(1-1/x^2)^(1/2)
/x^3+3/8*arcsec(x)^3*(1-1/x^2)^(1/2)/x

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Rubi [A]
time = 0.10, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5330, 3525, 3392, 30, 3391} \begin {gather*} -\frac {3}{128 x^4}-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}-\frac {45}{128 x^2}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {45}{128} \sec ^{-1}(x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSec[x]^4/x^5,x]

[Out]

-3/(128*x^4) - 45/(128*x^2) - (3*Sqrt[1 - x^(-2)]*ArcSec[x])/(32*x^3) - (45*Sqrt[1 - x^(-2)]*ArcSec[x])/(64*x)
- (45*ArcSec[x]^2)/128 + (3*ArcSec[x]^2)/(16*x^4) + (9*ArcSec[x]^2)/(16*x^2) + (Sqrt[1 - x^(-2)]*ArcSec[x]^3)
/(4*x^3) + (3*Sqrt[1 - x^(-2)]*ArcSec[x]^3)/(8*x) + (3*ArcSec[x]^4)/32 - ArcSec[x]^4/(4*x^4)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3525

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[(-x^(m - n
+ 1))*(Cos[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] + Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cos[a + b*x^
n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 5330

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -1])

Rubi steps

\begin {align*} \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx &=\text {Subst}\left (\int x^4 \cos ^3(x) \sin (x) \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac {\sec ^{-1}(x)^4}{4 x^4}+\text {Subst}\left (\int x^3 \cos ^4(x) \, dx,x,\sec ^{-1}(x)\right )\\ &=\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {\sec ^{-1}(x)^4}{4 x^4}-\frac {3}{8} \text {Subst}\left (\int x \cos ^4(x) \, dx,x,\sec ^{-1}(x)\right )+\frac {3}{4} \text {Subst}\left (\int x^3 \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac {3}{128 x^4}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}-\frac {\sec ^{-1}(x)^4}{4 x^4}-\frac {9}{32} \text {Subst}\left (\int x \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right )+\frac {3}{8} \text {Subst}\left (\int x^3 \, dx,x,\sec ^{-1}(x)\right )-\frac {9}{8} \text {Subst}\left (\int x \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4}-\frac {9}{64} \text {Subst}\left (\int x \, dx,x,\sec ^{-1}(x)\right )-\frac {9}{16} \text {Subst}\left (\int x \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac {45}{128} \sec ^{-1}(x)^2+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 92, normalized size = 0.62 \begin {gather*} \frac {-3-45 x^2-6 \sqrt {1-\frac {1}{x^2}} x \left (2+15 x^2\right ) \sec ^{-1}(x)+\left (24+72 x^2-45 x^4\right ) \sec ^{-1}(x)^2+16 \sqrt {1-\frac {1}{x^2}} x \left (2+3 x^2\right ) \sec ^{-1}(x)^3+4 \left (-8+3 x^4\right ) \sec ^{-1}(x)^4}{128 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[x]^4/x^5,x]

[Out]

(-3 - 45*x^2 - 6*Sqrt[1 - x^(-2)]*x*(2 + 15*x^2)*ArcSec[x] + (24 + 72*x^2 - 45*x^4)*ArcSec[x]^2 + 16*Sqrt[1 -
x^(-2)]*x*(2 + 3*x^2)*ArcSec[x]^3 + 4*(-8 + 3*x^4)*ArcSec[x]^4)/(128*x^4)

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Maple [A]
time = 0.15, size = 174, normalized size = 1.18

method	result	size
default	\(-\frac {\mathrm {arcsec}\left (x \right )^{4}}{4 x^{4}}+\frac {\mathrm {arcsec}\left (x \right )^{3} \left (3 \,\mathrm {arcsec}\left (x \right ) x^{3}+3 x^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{8 x^{3}}+\frac {3 \mathrm {arcsec}\left (x \right )^{2}}{16 x^{4}}-\frac {3 \,\mathrm {arcsec}\left (x \right ) \left (3 \,\mathrm {arcsec}\left (x \right ) x^{3}+3 x^{2} \sqrt {\frac {x^{2}-1}{x^{2}}}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{64 x^{3}}+\frac {45 \mathrm {arcsec}\left (x \right )^{2}}{128}-\frac {3 \left (3 x^{2}+2\right )^{2}}{512 x^{4}}+\frac {9 \mathrm {arcsec}\left (x \right )^{2}}{16 x^{2}}-\frac {9 \,\mathrm {arcsec}\left (x \right ) \left (\mathrm {arcsec}\left (x \right ) x +\sqrt {\frac {x^{2}-1}{x^{2}}}\right )}{16 x}+\frac {9}{32}-\frac {9}{32 x^{2}}-\frac {9 \mathrm {arcsec}\left (x \right )^{4}}{32}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(x)^4/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*arcsec(x)^4/x^4+1/8*arcsec(x)^3*(3*arcsec(x)*x^3+3*x^2*((x^2-1)/x^2)^(1/2)+2*((x^2-1)/x^2)^(1/2))/x^3+3/1
6*arcsec(x)^2/x^4-3/64*arcsec(x)*(3*arcsec(x)*x^3+3*x^2*((x^2-1)/x^2)^(1/2)+2*((x^2-1)/x^2)^(1/2))/x^3+45/128*
arcsec(x)^2-3/512*(3*x^2+2)^2/x^4+9/16*arcsec(x)^2/x^2-9/16*arcsec(x)*(arcsec(x)*x+((x^2-1)/x^2)^(1/2))/x+9/32
-9/32/x^2-9/32*arcsec(x)^4

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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(arcsec(x)^4/x^5,x, algorithm="maxima")

[Out]

1/64*(64*x^4*integrate(1/8*(12*(x^2 - 1)*log(x^2)^2*log(x)^2 - 16*(x^2 - 1)*log(x^2)*log(x)^3 + 8*(x^2 - 1)*lo
g(x)^4 + (x^2 - 4*(x^2 - 1)*log(x) - 1)*log(x^2)^3 - 12*(4*(x^2 - 1)*log(x)^2 + (x^2 - 4*(x^2 - 1)*log(x) - 1)
*log(x^2))*arctan(sqrt(x + 1)*sqrt(x - 1))^2 + 2*(4*arctan(sqrt(x + 1)*sqrt(x - 1))^3 - 3*arctan(sqrt(x + 1)*s
qrt(x - 1))*log(x^2)^2)*sqrt(x + 1)*sqrt(x - 1))/(x^7 - x^5), x) - 16*arctan(sqrt(x + 1)*sqrt(x - 1))^4 + 24*a
rctan(sqrt(x + 1)*sqrt(x - 1))^2*log(x^2)^2 - log(x^2)^4)/x^4

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Fricas [A]
time = 0.64, size = 77, normalized size = 0.52 \begin {gather*} \frac {4 \, {\left (3 \, x^{4} - 8\right )} \operatorname {arcsec}\left (x\right )^{4} - 3 \, {\left (15 \, x^{4} - 24 \, x^{2} - 8\right )} \operatorname {arcsec}\left (x\right )^{2} - 45 \, x^{2} + 2 \, {\left (8 \, {\left (3 \, x^{2} + 2\right )} \operatorname {arcsec}\left (x\right )^{3} - 3 \, {\left (15 \, x^{2} + 2\right )} \operatorname {arcsec}\left (x\right )\right )} \sqrt {x^{2} - 1} - 3}{128 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)^4/x^5,x, algorithm="fricas")

[Out]

1/128*(4*(3*x^4 - 8)*arcsec(x)^4 - 3*(15*x^4 - 24*x^2 - 8)*arcsec(x)^2 - 45*x^2 + 2*(8*(3*x^2 + 2)*arcsec(x)^3
- 3*(15*x^2 + 2)*arcsec(x))*sqrt(x^2 - 1) - 3)/x^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asec}^{4}{\left (x \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(x)**4/x**5,x)

[Out]

Integral(asec(x)**4/x**5, x)

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Giac [A]
time = 1.57, size = 137, normalized size = 0.93 \begin {gather*} \frac {3}{32} \, \arccos \left (\frac {1}{x}\right )^{4} + \frac {3 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )^{3}}{8 \, x} - \frac {45}{128} \, \arccos \left (\frac {1}{x}\right )^{2} - \frac {45 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )}{64 \, x} + \frac {\sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )^{3}}{4 \, x^{3}} + \frac {9 \, \arccos \left (\frac {1}{x}\right )^{2}}{16 \, x^{2}} - \frac {\arccos \left (\frac {1}{x}\right )^{4}}{4 \, x^{4}} - \frac {3 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )}{32 \, x^{3}} - \frac {45}{128 \, x^{2}} + \frac {3 \, \arccos \left (\frac {1}{x}\right )^{2}}{16 \, x^{4}} - \frac {3}{128 \, x^{4}} + \frac {189}{1024} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)^4/x^5,x, algorithm="giac")

[Out]

3/32*arccos(1/x)^4 + 3/8*sqrt(-1/x^2 + 1)*arccos(1/x)^3/x - 45/128*arccos(1/x)^2 - 45/64*sqrt(-1/x^2 + 1)*arcc
os(1/x)/x + 1/4*sqrt(-1/x^2 + 1)*arccos(1/x)^3/x^3 + 9/16*arccos(1/x)^2/x^2 - 1/4*arccos(1/x)^4/x^4 - 3/32*sqr
t(-1/x^2 + 1)*arccos(1/x)/x^3 - 45/128/x^2 + 3/16*arccos(1/x)^2/x^4 - 3/128/x^4 + 189/1024

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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 )}^4}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/x)^4/x^5,x)

[Out]

int(acos(1/x)^4/x^5, x)

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