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

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

[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)

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Rubi [A]  time = 0.135768, antiderivative size = 63, normalized size of antiderivative = 1., 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 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])

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 12

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

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 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]

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*}

Mathematica [A]  time = 0.0716536, size = 56, normalized size = 0.89 \[ \frac{c^2 \left (\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sec ^{-1}(c x)\right )-\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sec ^{-1}(c x)\right )\right )}{2 b} \]

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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Maple [A]  time = 0.24, size = 58, normalized size = 0.9 \begin{align*}{c}^{2} \left ({\frac{1}{2\,b}{\it Si} \left ( 2\,{\frac{a}{b}}+2\,{\rm arcsec} \left (cx\right ) \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) }-{\frac{1}{2\,b}{\it Ci} \left ( 2\,{\frac{a}{b}}+2\,{\rm arcsec} \left (cx\right ) \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) } \right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )} x^{3}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b x^{3} \operatorname{arcsec}\left (c x\right ) + a x^{3}}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b \operatorname{asec}{\left (c x \right )}\right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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