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

Optimal. Leaf size=117 \[ -\frac {c^3 \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{4 b}-\frac {c^3 \sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \sec ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sec ^{-1}(c x)\right )}{4 b} \]

[Out]

1/4*c^3*cos(a/b)*Si(a/b+arcsec(c*x))/b+1/4*c^3*cos(3*a/b)*Si(3*a/b+3*arcsec(c*x))/b-1/4*c^3*Ci(a/b+arcsec(c*x)
)*sin(a/b)/b-1/4*c^3*Ci(3*a/b+3*arcsec(c*x))*sin(3*a/b)/b

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Rubi [A]  time = 0.22, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5222, 4406, 3303, 3299, 3302} \[ -\frac {c^3 \sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{4 b}-\frac {c^3 \sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sec ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sec ^{-1}(c x)\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*ArcSec[c*x])),x]

[Out]

-(c^3*CosIntegral[a/b + ArcSec[c*x]]*Sin[a/b])/(4*b) - (c^3*CosIntegral[(3*a)/b + 3*ArcSec[c*x]]*Sin[(3*a)/b])
/(4*b) + (c^3*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]])/(4*b) + (c^3*Cos[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSe
c[c*x]])/(4*b)

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^4 \left (a+b \sec ^{-1}(c x)\right )} \, dx &=c^3 \operatorname {Subst}\left (\int \frac {\cos ^2(x) \sin (x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )\\ &=c^3 \operatorname {Subst}\left (\int \left (\frac {\sin (x)}{4 (a+b x)}+\frac {\sin (3 x)}{4 (a+b x)}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {1}{4} c^3 \operatorname {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )+\frac {1}{4} c^3 \operatorname {Subst}\left (\int \frac {\sin (3 x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {1}{4} \left (c^3 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )+\frac {1}{4} \left (c^3 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )-\frac {1}{4} \left (c^3 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )-\frac {1}{4} \left (c^3 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac {c^3 \text {Ci}\left (\frac {a}{b}+\sec ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )}{4 b}-\frac {c^3 \text {Ci}\left (\frac {3 a}{b}+3 \sec ^{-1}(c x)\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}+\frac {c^3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sec ^{-1}(c x)\right )}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 91, normalized size = 0.78 \[ \frac {c^3 \left (\sin \left (\frac {a}{b}\right ) \left (-\text {Ci}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )-\sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*ArcSec[c*x])),x]

[Out]

(c^3*(-(CosIntegral[a/b + ArcSec[c*x]]*Sin[a/b]) - CosIntegral[3*(a/b + ArcSec[c*x])]*Sin[(3*a)/b] + Cos[a/b]*
SinIntegral[a/b + ArcSec[c*x]] + Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSec[c*x])]))/(4*b)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b x^{4} \operatorname {arcsec}\left (c x\right ) + a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsec(c*x)),x, algorithm="fricas")

[Out]

integral(1/(b*x^4*arcsec(c*x) + a*x^4), x)

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giac [A]  time = 0.14, size = 199, normalized size = 1.70 \[ -\frac {1}{4} \, {\left (\frac {4 \, c^{2} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b} - \frac {4 \, c^{2} \cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b} + \frac {c^{2} \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b} + \frac {3 \, c^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right )}{b}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsec(c*x)),x, algorithm="giac")

[Out]

-1/4*(4*c^2*cos(a/b)^2*cos_integral(3*a/b + 3*arccos(1/(c*x)))*sin(a/b)/b - 4*c^2*cos(a/b)^3*sin_integral(3*a/
b + 3*arccos(1/(c*x)))/b - c^2*cos_integral(3*a/b + 3*arccos(1/(c*x)))*sin(a/b)/b + c^2*cos_integral(a/b + arc
cos(1/(c*x)))*sin(a/b)/b + 3*c^2*cos(a/b)*sin_integral(3*a/b + 3*arccos(1/(c*x)))/b - c^2*cos(a/b)*sin_integra
l(a/b + arccos(1/(c*x)))/b)*c

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maple [A]  time = 0.11, size = 102, normalized size = 0.87 \[ c^{3} \left (\frac {\Si \left (\frac {3 a}{b}+3 \,\mathrm {arcsec}\left (c x \right )\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}-\frac {\Ci \left (\frac {3 a}{b}+3 \,\mathrm {arcsec}\left (c x \right )\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}+\frac {\Si \left (\frac {a}{b}+\mathrm {arcsec}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{4 b}-\frac {\Ci \left (\frac {a}{b}+\mathrm {arcsec}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )}{4 b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+b*arcsec(c*x)),x)

[Out]

c^3*(1/4*Si(3*a/b+3*arcsec(c*x))*cos(3*a/b)/b-1/4*Ci(3*a/b+3*arcsec(c*x))*sin(3*a/b)/b+1/4*Si(a/b+arcsec(c*x))
*cos(a/b)/b-1/4*Ci(a/b+arcsec(c*x))*sin(a/b)/b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsec(c*x)),x, algorithm="maxima")

[Out]

integrate(1/((b*arcsec(c*x) + a)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^4\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*acos(1/(c*x)))),x)

[Out]

int(1/(x^4*(a + b*acos(1/(c*x)))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \left (a + b \operatorname {asec}{\left (c x \right )}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b*asec(c*x)),x)

[Out]

Integral(1/(x**4*(a + b*asec(c*x))), x)

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