3.11 \(\int \frac {a+b \sec ^{-1}(c x)}{x^4} \, dx\)

Optimal. Leaf size=60 \[ -\frac {a+b \sec ^{-1}(c x)}{3 x^3}-\frac {1}{9} b c^3 \left (1-\frac {1}{c^2 x^2}\right )^{3/2}+\frac {1}{3} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \]

[Out]

-1/9*b*c^3*(1-1/c^2/x^2)^(3/2)+1/3*(-a-b*arcsec(c*x))/x^3+1/3*b*c^3*(1-1/c^2/x^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5220, 266, 43} \[ -\frac {a+b \sec ^{-1}(c x)}{3 x^3}-\frac {1}{9} b c^3 \left (1-\frac {1}{c^2 x^2}\right )^{3/2}+\frac {1}{3} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/x^4,x]

[Out]

(b*c^3*Sqrt[1 - 1/(c^2*x^2)])/3 - (b*c^3*(1 - 1/(c^2*x^2))^(3/2))/9 - (a + b*ArcSec[c*x])/(3*x^3)

Rule 43

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 266

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 5220

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*} \int \frac {a+b \sec ^{-1}(c x)}{x^4} \, dx &=-\frac {a+b \sec ^{-1}(c x)}{3 x^3}+\frac {b \int \frac {1}{\sqrt {1-\frac {1}{c^2 x^2}} x^5} \, dx}{3 c}\\ &=-\frac {a+b \sec ^{-1}(c x)}{3 x^3}-\frac {b \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{6 c}\\ &=-\frac {a+b \sec ^{-1}(c x)}{3 x^3}-\frac {b \operatorname {Subst}\left (\int \left (\frac {c^2}{\sqrt {1-\frac {x}{c^2}}}-c^2 \sqrt {1-\frac {x}{c^2}}\right ) \, dx,x,\frac {1}{x^2}\right )}{6 c}\\ &=\frac {1}{3} b c^3 \sqrt {1-\frac {1}{c^2 x^2}}-\frac {1}{9} b c^3 \left (1-\frac {1}{c^2 x^2}\right )^{3/2}-\frac {a+b \sec ^{-1}(c x)}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 59, normalized size = 0.98 \[ -\frac {a}{3 x^3}+b \left (\frac {2 c^3}{9}+\frac {c}{9 x^2}\right ) \sqrt {\frac {c^2 x^2-1}{c^2 x^2}}-\frac {b \sec ^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/x^4,x]

[Out]

-1/3*a/x^3 + b*((2*c^3)/9 + c/(9*x^2))*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)] - (b*ArcSec[c*x])/(3*x^3)

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fricas [A]  time = 0.68, size = 40, normalized size = 0.67 \[ -\frac {3 \, b \operatorname {arcsec}\left (c x\right ) - {\left (2 \, b c^{2} x^{2} + b\right )} \sqrt {c^{2} x^{2} - 1} + 3 \, a}{9 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/9*(3*b*arcsec(c*x) - (2*b*c^2*x^2 + b)*sqrt(c^2*x^2 - 1) + 3*a)/x^3

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giac [A]  time = 0.15, size = 65, normalized size = 1.08 \[ \frac {1}{9} \, {\left (2 \, b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + \frac {b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{2}} - \frac {3 \, b \arccos \left (\frac {1}{c x}\right )}{c x^{3}} - \frac {3 \, a}{c x^{3}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^4,x, algorithm="giac")

[Out]

1/9*(2*b*c^2*sqrt(-1/(c^2*x^2) + 1) + b*sqrt(-1/(c^2*x^2) + 1)/x^2 - 3*b*arccos(1/(c*x))/(c*x^3) - 3*a/(c*x^3)
)*c

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maple [A]  time = 0.05, size = 75, normalized size = 1.25 \[ c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\mathrm {arcsec}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\left (c^{2} x^{2}-1\right ) \left (2 c^{2} x^{2}+1\right )}{9 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))/x^4,x)

[Out]

c^3*(-1/3*a/c^3/x^3+b*(-1/3/c^3/x^3*arcsec(c*x)+1/9*(c^2*x^2-1)*(2*c^2*x^2+1)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c^4/
x^4))

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maxima [A]  time = 0.32, size = 58, normalized size = 0.97 \[ -\frac {1}{9} \, b {\left (\frac {c^{4} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} + \frac {3 \, \operatorname {arcsec}\left (c x\right )}{x^{3}}\right )} - \frac {a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^4,x, algorithm="maxima")

[Out]

-1/9*b*((c^4*(-1/(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(-1/(c^2*x^2) + 1))/c + 3*arcsec(c*x)/x^3) - 1/3*a/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))/x^4,x)

[Out]

int((a + b*acos(1/(c*x)))/x^4, x)

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sympy [A]  time = 2.93, size = 110, normalized size = 1.83 \[ - \frac {a}{3 x^{3}} - \frac {b \operatorname {asec}{\left (c x \right )}}{3 x^{3}} + \frac {b \left (\begin {cases} \frac {2 c^{3} \sqrt {c^{2} x^{2} - 1}}{3 x} + \frac {c \sqrt {c^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {2 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{3 x} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right )}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))/x**4,x)

[Out]

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

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