\(\int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx\) [13]

Optimal result

Integrand size = 12, antiderivative size = 82 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=\frac {1}{5} b c^5 \sqrt {1-\frac {1}{c^2 x^2}}-\frac {2}{15} b c^5 \left (1-\frac {1}{c^2 x^2}\right )^{3/2}+\frac {1}{25} b c^5 \left (1-\frac {1}{c^2 x^2}\right )^{5/2}-\frac {a+b \sec ^{-1}(c x)}{5 x^5} \]

[Out]

-2/15*b*c^5*(1-1/c^2/x^2)^(3/2)+1/25*b*c^5*(1-1/c^2/x^2)^(5/2)+1/5*(-a-b*arcsec(c*x))/x^5+1/5*b*c^5*(1-1/c^2/x
^2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5328, 272, 45} \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=-\frac {a+b \sec ^{-1}(c x)}{5 x^5}+\frac {1}{25} b c^5 \left (1-\frac {1}{c^2 x^2}\right )^{5/2}-\frac {2}{15} b c^5 \left (1-\frac {1}{c^2 x^2}\right )^{3/2}+\frac {1}{5} b c^5 \sqrt {1-\frac {1}{c^2 x^2}} \]

[In]

Int[(a + b*ArcSec[c*x])/x^6,x]

[Out]

(b*c^5*Sqrt[1 - 1/(c^2*x^2)])/5 - (2*b*c^5*(1 - 1/(c^2*x^2))^(3/2))/15 + (b*c^5*(1 - 1/(c^2*x^2))^(5/2))/25 -
(a + b*ArcSec[c*x])/(5*x^5)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5328

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSec[c*x]
)/(d*(m + 1))), x] - Dist[b*(d/(c*(m + 1))), Int[(d*x)^(m - 1)/Sqrt[1 - 1/(c^2*x^2)], x], x] /; FreeQ[{a, b, c
, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \sec ^{-1}(c x)}{5 x^5}+\frac {b \int \frac {1}{\sqrt {1-\frac {1}{c^2 x^2}} x^7} \, dx}{5 c} \\ & = -\frac {a+b \sec ^{-1}(c x)}{5 x^5}-\frac {b \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{10 c} \\ & = -\frac {a+b \sec ^{-1}(c x)}{5 x^5}-\frac {b \text {Subst}\left (\int \left (\frac {c^4}{\sqrt {1-\frac {x}{c^2}}}-2 c^4 \sqrt {1-\frac {x}{c^2}}+c^4 \left (1-\frac {x}{c^2}\right )^{3/2}\right ) \, dx,x,\frac {1}{x^2}\right )}{10 c} \\ & = \frac {1}{5} b c^5 \sqrt {1-\frac {1}{c^2 x^2}}-\frac {2}{15} b c^5 \left (1-\frac {1}{c^2 x^2}\right )^{3/2}+\frac {1}{25} b c^5 \left (1-\frac {1}{c^2 x^2}\right )^{5/2}-\frac {a+b \sec ^{-1}(c x)}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=-\frac {a}{5 x^5}+b \left (\frac {8 c^5}{75}+\frac {c}{25 x^4}+\frac {4 c^3}{75 x^2}\right ) \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}-\frac {b \sec ^{-1}(c x)}{5 x^5} \]

[In]

Integrate[(a + b*ArcSec[c*x])/x^6,x]

[Out]

-1/5*a/x^5 + b*((8*c^5)/75 + c/(25*x^4) + (4*c^3)/(75*x^2))*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)] - (b*ArcSec[c*x])/(
5*x^5)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96

method	result	size
parts	\(-\frac {a}{5 x^{5}}+b \,c^{5} \left (-\frac {\operatorname {arcsec}\left (c x \right )}{5 c^{5} x^{5}}+\frac {\left (c^{2} x^{2}-1\right ) \left (8 c^{4} x^{4}+4 c^{2} x^{2}+3\right )}{75 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{6} x^{6}}\right )\)	\(79\)
derivativedivides	\(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}+b \left (-\frac {\operatorname {arcsec}\left (c x \right )}{5 c^{5} x^{5}}+\frac {\left (c^{2} x^{2}-1\right ) \left (8 c^{4} x^{4}+4 c^{2} x^{2}+3\right )}{75 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{6} x^{6}}\right )\right )\)	\(83\)
default	\(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}+b \left (-\frac {\operatorname {arcsec}\left (c x \right )}{5 c^{5} x^{5}}+\frac {\left (c^{2} x^{2}-1\right ) \left (8 c^{4} x^{4}+4 c^{2} x^{2}+3\right )}{75 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{6} x^{6}}\right )\right )\)	\(83\)

[In]

int((a+b*arcsec(c*x))/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/5*a/x^5+b*c^5*(-1/5/c^5/x^5*arcsec(c*x)+1/75*(c^2*x^2-1)*(8*c^4*x^4+4*c^2*x^2+3)/((c^2*x^2-1)/c^2/x^2)^(1/2
)/c^6/x^6)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.62 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=-\frac {15 \, b \operatorname {arcsec}\left (c x\right ) - {\left (8 \, b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 3 \, b\right )} \sqrt {c^{2} x^{2} - 1} + 15 \, a}{75 \, x^{5}} \]

[In]

integrate((a+b*arcsec(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/75*(15*b*arcsec(c*x) - (8*b*c^4*x^4 + 4*b*c^2*x^2 + 3*b)*sqrt(c^2*x^2 - 1) + 15*a)/x^5

Sympy [A] (verification not implemented)

Time = 3.60 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.90 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=- \frac {a}{5 x^{5}} - \frac {b \operatorname {asec}{\left (c x \right )}}{5 x^{5}} + \frac {b \left (\begin {cases} \frac {8 c^{5} \sqrt {c^{2} x^{2} - 1}}{15 x} + \frac {4 c^{3} \sqrt {c^{2} x^{2} - 1}}{15 x^{3}} + \frac {c \sqrt {c^{2} x^{2} - 1}}{5 x^{5}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {8 i c^{5} \sqrt {- c^{2} x^{2} + 1}}{15 x} + \frac {4 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{15 x^{3}} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{5 x^{5}} & \text {otherwise} \end {cases}\right )}{5 c} \]

[In]

integrate((a+b*asec(c*x))/x**6,x)

[Out]

-a/(5*x**5) - b*asec(c*x)/(5*x**5) + b*Piecewise((8*c**5*sqrt(c**2*x**2 - 1)/(15*x) + 4*c**3*sqrt(c**2*x**2 -
1)/(15*x**3) + c*sqrt(c**2*x**2 - 1)/(5*x**5), Abs(c**2*x**2) > 1), (8*I*c**5*sqrt(-c**2*x**2 + 1)/(15*x) + 4*
I*c**3*sqrt(-c**2*x**2 + 1)/(15*x**3) + I*c*sqrt(-c**2*x**2 + 1)/(5*x**5), True))/(5*c)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=\frac {1}{75} \, b {\left (\frac {3 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 10 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, c^{6} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {15 \, \operatorname {arcsec}\left (c x\right )}{x^{5}}\right )} - \frac {a}{5 \, x^{5}} \]

[In]

integrate((a+b*arcsec(c*x))/x^6,x, algorithm="maxima")

[Out]

1/75*b*((3*c^6*(-1/(c^2*x^2) + 1)^(5/2) - 10*c^6*(-1/(c^2*x^2) + 1)^(3/2) + 15*c^6*sqrt(-1/(c^2*x^2) + 1))/c -
15*arcsec(c*x)/x^5) - 1/5*a/x^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=\frac {1}{75} \, {\left (8 \, b c^{4} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + \frac {4 \, b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{2}} + \frac {3 \, b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{4}} - \frac {15 \, b \arccos \left (\frac {1}{c x}\right )}{c x^{5}} - \frac {15 \, a}{c x^{5}}\right )} c \]

[In]

integrate((a+b*arcsec(c*x))/x^6,x, algorithm="giac")

[Out]

1/75*(8*b*c^4*sqrt(-1/(c^2*x^2) + 1) + 4*b*c^2*sqrt(-1/(c^2*x^2) + 1)/x^2 + 3*b*sqrt(-1/(c^2*x^2) + 1)/x^4 - 1
5*b*arccos(1/(c*x))/(c*x^5) - 15*a/(c*x^5))*c

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \sec ^{-1}(c x)}{x^6} \, dx=\int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x^6} \,d x \]

[In]

int((a + b*acos(1/(c*x)))/x^6,x)

[Out]

int((a + b*acos(1/(c*x)))/x^6, x)