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

3.7.50.1 Optimal result

Integrand size = 8, antiderivative size = 148 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=-\frac {3}{128 x^4}-\frac {45}{128 x^2}-\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac {45 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac {45}{128} \sec ^{-1}(x)^2+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {9 \sec ^{-1}(x)^2}{16 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac {3 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac {3}{32} \sec ^{-1}(x)^4-\frac {\sec ^{-1}(x)^4}{4 x^4} \]

output

-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
 

3.7.50.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\frac {-3-45 x^2-6 \sqrt {1-\frac {1}{x^2}} x \left (2+15 x^2\right ) \sec ^{-1}(x)+\left (24+72 x^2-45 x^4\right ) \sec ^{-1}(x)^2+16 \sqrt {1-\frac {1}{x^2}} x \left (2+3 x^2\right ) \sec ^{-1}(x)^3+4 \left (-8+3 x^4\right ) \sec ^{-1}(x)^4}{128 x^4} \]

input

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

output

(-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)
 

3.7.50.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.37, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.750, Rules used = {5745, 3925, 3042, 3792, 3042, 3791, 3042, 3791, 15, 3792, 15, 3042, 3791, 15}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx\)

\(\Big \downarrow \) 5745

\(\displaystyle \int \frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^4}{x^3}d\sec ^{-1}(x)\)

\(\Big \downarrow \) 3925

\(\displaystyle \int \frac {\sec ^{-1}(x)^3}{x^4}d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec ^{-1}(x)^3 \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^4d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}\)

\(\Big \downarrow \) 3792

\(\displaystyle -\frac {3}{8} \int \frac {\sec ^{-1}(x)}{x^4}d\sec ^{-1}(x)+\frac {3}{4} \int \frac {\sec ^{-1}(x)^3}{x^2}d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{4} \int \sec ^{-1}(x)^3 \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(x)-\frac {3}{8} \int \sec ^{-1}(x) \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^4d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}\)

\(\Big \downarrow \) 3791

\(\displaystyle -\frac {3}{8} \left (\frac {3}{4} \int \frac {\sec ^{-1}(x)}{x^2}d\sec ^{-1}(x)+\frac {1}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )+\frac {3}{4} \int \sec ^{-1}(x)^3 \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3}{8} \left (\frac {3}{4} \int \sec ^{-1}(x) \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(x)+\frac {1}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )+\frac {3}{4} \int \sec ^{-1}(x)^3 \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}\)

\(\Big \downarrow \) 3791

\(\displaystyle -\frac {3}{8} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec ^{-1}(x)d\sec ^{-1}(x)+\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}\right )+\frac {1}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )+\frac {3}{4} \int \sec ^{-1}(x)^3 \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {3}{4} \int \sec ^{-1}(x)^3 \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(x)-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3}{8} \left (\frac {1}{16 x^4}+\frac {3}{4} \left (\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}+\frac {1}{4} \sec ^{-1}(x)^2\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )\)

\(\Big \downarrow \) 3792

\(\displaystyle \frac {3}{4} \left (-\frac {3}{2} \int \frac {\sec ^{-1}(x)}{x^2}d\sec ^{-1}(x)+\frac {1}{2} \int \sec ^{-1}(x)^3d\sec ^{-1}(x)+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{2 x}+\frac {3 \sec ^{-1}(x)^2}{4 x^2}\right )-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3}{8} \left (\frac {1}{16 x^4}+\frac {3}{4} \left (\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}+\frac {1}{4} \sec ^{-1}(x)^2\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {3}{4} \left (-\frac {3}{2} \int \frac {\sec ^{-1}(x)}{x^2}d\sec ^{-1}(x)+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{2 x}+\frac {3 \sec ^{-1}(x)^2}{4 x^2}+\frac {1}{8} \sec ^{-1}(x)^4\right )-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3}{8} \left (\frac {1}{16 x^4}+\frac {3}{4} \left (\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}+\frac {1}{4} \sec ^{-1}(x)^2\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{4} \left (-\frac {3}{2} \int \sec ^{-1}(x) \sin \left (\sec ^{-1}(x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(x)+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{2 x}+\frac {3 \sec ^{-1}(x)^2}{4 x^2}+\frac {1}{8} \sec ^{-1}(x)^4\right )-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3}{8} \left (\frac {1}{16 x^4}+\frac {3}{4} \left (\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}+\frac {1}{4} \sec ^{-1}(x)^2\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )\)

\(\Big \downarrow \) 3791

\(\displaystyle \frac {3}{4} \left (-\frac {3}{2} \left (\frac {1}{2} \int \sec ^{-1}(x)d\sec ^{-1}(x)+\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{2 x}+\frac {3 \sec ^{-1}(x)^2}{4 x^2}+\frac {1}{8} \sec ^{-1}(x)^4\right )-\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3}{8} \left (\frac {1}{16 x^4}+\frac {3}{4} \left (\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}+\frac {1}{4} \sec ^{-1}(x)^2\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {\sec ^{-1}(x)^4}{4 x^4}+\frac {3 \sec ^{-1}(x)^2}{16 x^4}+\frac {3}{4} \left (\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{2 x}+\frac {3 \sec ^{-1}(x)^2}{4 x^2}-\frac {3}{2} \left (\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}+\frac {1}{4} \sec ^{-1}(x)^2\right )+\frac {1}{8} \sec ^{-1}(x)^4\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac {3}{8} \left (\frac {1}{16 x^4}+\frac {3}{4} \left (\frac {1}{4 x^2}+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{2 x}+\frac {1}{4} \sec ^{-1}(x)^2\right )+\frac {\sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x)}{4 x^3}\right )\)

input

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

output

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

3.7.50.3.1 Defintions of rubi rules used

rule 15

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3791

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] + (-Simp[b*(c + d*x)*Cos[e + f*x 
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n)   Int[(c + d* 
x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 
 1]
 

rule 3792

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

rule 3925

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] + Simp[(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 5745

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[1 
/c^(m + 1)   Subst[Int[(a + b*x)^n*Sec[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])
 

3.7.50.4 Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.18

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

input

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

output

-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+3/16*arcsec(x)^2/x^4-3/64*arcsec(x)*(3*ar 
csec(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*ar 
csec(x)^2-3/512*(3*x^2+2)^2/x^4+9/16*arcsec(x)^2/x^2-9/16*arcsec(x)*(x*arc 
sec(x)+((x^2-1)/x^2)^(1/2))/x+9/32-9/32/x^2-9/32*arcsec(x)^4
 

3.7.50.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.52 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\frac {4 \, {\left (3 \, x^{4} - 8\right )} \operatorname {arcsec}\left (x\right )^{4} - 3 \, {\left (15 \, x^{4} - 24 \, x^{2} - 8\right )} \operatorname {arcsec}\left (x\right )^{2} - 45 \, x^{2} + 2 \, {\left (8 \, {\left (3 \, x^{2} + 2\right )} \operatorname {arcsec}\left (x\right )^{3} - 3 \, {\left (15 \, x^{2} + 2\right )} \operatorname {arcsec}\left (x\right )\right )} \sqrt {x^{2} - 1} - 3}{128 \, x^{4}} \]

input

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

output

1/128*(4*(3*x^4 - 8)*arcsec(x)^4 - 3*(15*x^4 - 24*x^2 - 8)*arcsec(x)^2 - 4 
5*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
 

3.7.50.6 Sympy [F]

\[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\int \frac {\operatorname {asec}^{4}{\left (x \right )}}{x^{5}}\, dx \]

input

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

output

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

3.7.50.7 Maxima [F]

\[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\int { \frac {\operatorname {arcsec}\left (x\right )^{4}}{x^{5}} \,d x } \]

input

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

output

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)*log(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)*l 
og(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)*sqrt(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*arcta 
n(sqrt(x + 1)*sqrt(x - 1))^2*log(x^2)^2 - log(x^2)^4)/x^4
 

3.7.50.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\frac {3}{32} \, \arccos \left (\frac {1}{x}\right )^{4} + \frac {3 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )^{3}}{8 \, x} - \frac {45}{128} \, \arccos \left (\frac {1}{x}\right )^{2} - \frac {45 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )}{64 \, x} + \frac {\sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )^{3}}{4 \, x^{3}} + \frac {9 \, \arccos \left (\frac {1}{x}\right )^{2}}{16 \, x^{2}} - \frac {\arccos \left (\frac {1}{x}\right )^{4}}{4 \, x^{4}} - \frac {3 \, \sqrt {-\frac {1}{x^{2}} + 1} \arccos \left (\frac {1}{x}\right )}{32 \, x^{3}} - \frac {45}{128 \, x^{2}} + \frac {3 \, \arccos \left (\frac {1}{x}\right )^{2}}{16 \, x^{4}} - \frac {3}{128 \, x^{4}} + \frac {189}{1024} \]

input

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

output

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)*arccos(1/x)/x + 1/4*sqrt(-1/x^2 + 1)*arcco 
s(1/x)^3/x^3 + 9/16*arccos(1/x)^2/x^2 - 1/4*arccos(1/x)^4/x^4 - 3/32*sqrt( 
-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
 

3.7.50.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\int \frac {{\mathrm {acos}\left (\frac {1}{x}\right )}^4}{x^5} \,d x \]

input

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

output

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

3.7.50.10 Reduce [F]

\[ \int \frac {\sec ^{-1}(x)^4}{x^5} \, dx=\int \frac {\mathit {asec} \left (x \right )^{4}}{x^{5}}d x \]

input

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

output

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