3.37 \(\int \frac {1}{x^3 (a+b \sec ^{-1}(c x))} \, dx\)

Optimal. Leaf size=63 \[ \frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{2 b}-\frac {c^2 \sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{2 b} \]

[Out]

1/2*c^2*cos(2*a/b)*Si(2*a/b+2*arcsec(c*x))/b-1/2*c^2*Ci(2*a/b+2*arcsec(c*x))*sin(2*a/b)/b

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Rubi [A]  time = 0.14, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5222, 4406, 12, 3303, 3299, 3302} \[ \frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{2 b}-\frac {c^2 \sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*ArcSec[c*x])),x]

[Out]

-(c^2*CosIntegral[(2*a)/b + 2*ArcSec[c*x]]*Sin[(2*a)/b])/(2*b) + (c^2*Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*Arc
Sec[c*x]])/(2*b)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 5222

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 {1}{x^3 \left (a+b \sec ^{-1}(c x)\right )} \, dx &=c^2 \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )\\ &=c^2 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 (a+b x)} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {1}{2} c^2 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {1}{2} \left (c^2 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )-\frac {1}{2} \left (c^2 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac {c^2 \text {Ci}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right ) \sin \left (\frac {2 a}{b}\right )}{2 b}+\frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 56, normalized size = 0.89 \[ \frac {c^2 \left (\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )-\sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*ArcSec[c*x])),x]

[Out]

(c^2*(-(CosIntegral[(2*a)/b + 2*ArcSec[c*x]]*Sin[(2*a)/b]) + Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSec[c*x]]
))/(2*b)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b x^{3} \operatorname {arcsec}\left (c x\right ) + a x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsec(c*x)),x, algorithm="fricas")

[Out]

integral(1/(b*x^3*arcsec(c*x) + a*x^3), x)

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giac [A]  time = 0.13, size = 95, normalized size = 1.51 \[ -\frac {1}{2} \, {\left (\frac {2 \, c \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b} - \frac {2 \, c \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b} + \frac {c \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsec(c*x)),x, algorithm="giac")

[Out]

-1/2*(2*c*cos(a/b)*cos_integral(2*a/b + 2*arccos(1/(c*x)))*sin(a/b)/b - 2*c*cos(a/b)^2*sin_integral(2*a/b + 2*
arccos(1/(c*x)))/b + c*sin_integral(2*a/b + 2*arccos(1/(c*x)))/b)*c

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maple [A]  time = 0.10, size = 58, normalized size = 0.92 \[ c^{2} \left (\frac {\Si \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \cos \left (\frac {2 a}{b}\right )}{2 b}-\frac {\Ci \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \sin \left (\frac {2 a}{b}\right )}{2 b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b*arcsec(c*x)),x)

[Out]

c^2*(1/2*Si(2*a/b+2*arcsec(c*x))*cos(2*a/b)/b-1/2*Ci(2*a/b+2*arcsec(c*x))*sin(2*a/b)/b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsec(c*x)),x, algorithm="maxima")

[Out]

integrate(1/((b*arcsec(c*x) + a)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{x^3\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*acos(1/(c*x)))),x)

[Out]

int(1/(x^3*(a + b*acos(1/(c*x)))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \left (a + b \operatorname {asec}{\left (c x \right )}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a+b*asec(c*x)),x)

[Out]

Integral(1/(x**3*(a + b*asec(c*x))), x)

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