3.12 \(\int \frac {a+b \sec ^{-1}(c x)}{x^5} \, dx\)

Optimal. Leaf size=76 \[ -\frac {a+b \sec ^{-1}(c x)}{4 x^4}-\frac {3}{32} b c^4 \csc ^{-1}(c x)+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}}}{16 x^3}+\frac {3 b c^3 \sqrt {1-\frac {1}{c^2 x^2}}}{32 x} \]

[Out]

-3/32*b*c^4*arccsc(c*x)+1/4*(-a-b*arcsec(c*x))/x^4+1/16*b*c*(1-1/c^2/x^2)^(1/2)/x^3+3/32*b*c^3*(1-1/c^2/x^2)^(
1/2)/x

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Rubi [A]  time = 0.05, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5220, 335, 321, 216} \[ -\frac {a+b \sec ^{-1}(c x)}{4 x^4}+\frac {3 b c^3 \sqrt {1-\frac {1}{c^2 x^2}}}{32 x}+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}}}{16 x^3}-\frac {3}{32} b c^4 \csc ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/x^5,x]

[Out]

(b*c*Sqrt[1 - 1/(c^2*x^2)])/(16*x^3) + (3*b*c^3*Sqrt[1 - 1/(c^2*x^2)])/(32*x) - (3*b*c^4*ArcCsc[c*x])/32 - (a
+ b*ArcSec[c*x])/(4*x^4)

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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^5} \, dx &=-\frac {a+b \sec ^{-1}(c x)}{4 x^4}+\frac {b \int \frac {1}{\sqrt {1-\frac {1}{c^2 x^2}} x^6} \, dx}{4 c}\\ &=-\frac {a+b \sec ^{-1}(c x)}{4 x^4}-\frac {b \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{4 c}\\ &=\frac {b c \sqrt {1-\frac {1}{c^2 x^2}}}{16 x^3}-\frac {a+b \sec ^{-1}(c x)}{4 x^4}-\frac {1}{16} (3 b c) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {b c \sqrt {1-\frac {1}{c^2 x^2}}}{16 x^3}+\frac {3 b c^3 \sqrt {1-\frac {1}{c^2 x^2}}}{32 x}-\frac {a+b \sec ^{-1}(c x)}{4 x^4}-\frac {1}{32} \left (3 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {b c \sqrt {1-\frac {1}{c^2 x^2}}}{16 x^3}+\frac {3 b c^3 \sqrt {1-\frac {1}{c^2 x^2}}}{32 x}-\frac {3}{32} b c^4 \csc ^{-1}(c x)-\frac {a+b \sec ^{-1}(c x)}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 78, normalized size = 1.03 \[ -\frac {a}{4 x^4}-\frac {3}{32} b c^4 \sin ^{-1}\left (\frac {1}{c x}\right )+b \left (\frac {3 c^3}{32 x}+\frac {c}{16 x^3}\right ) \sqrt {\frac {c^2 x^2-1}{c^2 x^2}}-\frac {b \sec ^{-1}(c x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/x^5,x]

[Out]

-1/4*a/x^4 + b*(c/(16*x^3) + (3*c^3)/(32*x))*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)] - (b*ArcSec[c*x])/(4*x^4) - (3*b*c
^4*ArcSin[1/(c*x)])/32

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fricas [A]  time = 0.71, size = 52, normalized size = 0.68 \[ \frac {{\left (3 \, b c^{4} x^{4} - 8 \, b\right )} \operatorname {arcsec}\left (c x\right ) + {\left (3 \, b c^{2} x^{2} + 2 \, b\right )} \sqrt {c^{2} x^{2} - 1} - 8 \, a}{32 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^5,x, algorithm="fricas")

[Out]

1/32*((3*b*c^4*x^4 - 8*b)*arcsec(c*x) + (3*b*c^2*x^2 + 2*b)*sqrt(c^2*x^2 - 1) - 8*a)/x^4

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giac [A]  time = 0.15, size = 83, normalized size = 1.09 \[ \frac {1}{32} \, {\left (3 \, b c^{3} \arccos \left (\frac {1}{c x}\right ) + \frac {3 \, b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x} + \frac {2 \, b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{3}} - \frac {8 \, b \arccos \left (\frac {1}{c x}\right )}{c x^{4}} - \frac {8 \, a}{c x^{4}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^5,x, algorithm="giac")

[Out]

1/32*(3*b*c^3*arccos(1/(c*x)) + 3*b*c^2*sqrt(-1/(c^2*x^2) + 1)/x + 2*b*sqrt(-1/(c^2*x^2) + 1)/x^3 - 8*b*arccos
(1/(c*x))/(c*x^4) - 8*a/(c*x^4))*c

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maple [B]  time = 0.05, size = 147, normalized size = 1.93 \[ -\frac {a}{4 x^{4}}-\frac {b \,\mathrm {arcsec}\left (c x \right )}{4 x^{4}}-\frac {3 c^{3} b \sqrt {c^{2} x^{2}-1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{32 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}+\frac {3 c^{3} b}{32 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}-\frac {c b}{32 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x^{3}}-\frac {b}{16 c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))/x^5,x)

[Out]

-1/4*a/x^4-1/4*b/x^4*arcsec(c*x)-3/32*c^3*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*arctan(1/(c^2*x^2-
1)^(1/2))+3/32*c^3*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x-1/32*c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x^3-1/16/c*b/((c^2*x^2
-1)/c^2/x^2)^(1/2)/x^5

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maxima [A]  time = 0.42, size = 125, normalized size = 1.64 \[ \frac {1}{32} \, b {\left (\frac {3 \, c^{5} \arctan \left (c x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}\right ) + \frac {3 \, c^{8} x^{3} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 5 \, c^{6} x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{4} x^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{2} - 2 \, c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + 1}}{c} - \frac {8 \, \operatorname {arcsec}\left (c x\right )}{x^{4}}\right )} - \frac {a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^5,x, algorithm="maxima")

[Out]

1/32*b*((3*c^5*arctan(c*x*sqrt(-1/(c^2*x^2) + 1)) + (3*c^8*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 5*c^6*x*sqrt(-1/(c^2
*x^2) + 1))/(c^4*x^4*(1/(c^2*x^2) - 1)^2 - 2*c^2*x^2*(1/(c^2*x^2) - 1) + 1))/c - 8*arcsec(c*x)/x^4) - 1/4*a/x^
4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))/x^5,x)

[Out]

int((a + b*acos(1/(c*x)))/x^5, x)

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sympy [A]  time = 5.23, size = 192, normalized size = 2.53 \[ - \frac {a}{4 x^{4}} - \frac {b \operatorname {asec}{\left (c x \right )}}{4 x^{4}} + \frac {b \left (\begin {cases} \frac {3 i c^{5} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{8} - \frac {3 i c^{4}}{8 x \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} + \frac {i c^{2}}{8 x^{3} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} + \frac {i}{4 x^{5} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\- \frac {3 c^{5} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{8} + \frac {3 c^{4}}{8 x \sqrt {1 - \frac {1}{c^{2} x^{2}}}} - \frac {c^{2}}{8 x^{3} \sqrt {1 - \frac {1}{c^{2} x^{2}}}} - \frac {1}{4 x^{5} \sqrt {1 - \frac {1}{c^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))/x**5,x)

[Out]

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

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