3.36 \(\int \frac{1}{x^2 (a+b \sec ^{-1}(c x))} \, dx\)

Optimal. Leaf size=46 \[ \frac{c \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b}-\frac{c \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.105162, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5222, 3303, 3299, 3302} \[ \frac{c \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b}-\frac{c \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*ArcSec[c*x])),x]

[Out]

-((c*CosIntegral[a/b + ArcSec[c*x]]*Sin[a/b])/b) + (c*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]])/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 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^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx &=c \operatorname{Subst}\left (\int \frac{\sin (x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\left (c \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 )-\left (c \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{c \text{Ci}\left (\frac{a}{b}+\sec ^{-1}(c x)\right ) \sin \left (\frac{a}{b}\right )}{b}+\frac{c \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0741781, size = 43, normalized size = 0.93 \[ \frac{c \left (\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )-\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*ArcSec[c*x])),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.247, size = 47, normalized size = 1. \begin{align*} c \left ({\frac{1}{b}{\it Si} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \cos \left ({\frac{a}{b}} \right ) }-{\frac{1}{b}{\it Ci} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \sin \left ({\frac{a}{b}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(a+b*arcsec(c*x)),x)

[Out]

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

________________________________________________________________________________________

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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*arcsec(c*x)),x, algorithm="maxima")

[Out]

integrate(1/((b*arcsec(c*x) + a)*x^2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b x^{2} \operatorname{arcsec}\left (c x\right ) + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*arcsec(c*x)),x, algorithm="fricas")

[Out]

integral(1/(b*x^2*arcsec(c*x) + a*x^2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \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**2/(a+b*asec(c*x)),x)

[Out]

Integral(1/(x**2*(a + b*asec(c*x))), x)

________________________________________________________________________________________

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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*arcsec(c*x)),x, algorithm="giac")

[Out]

integrate(1/((b*arcsec(c*x) + a)*x^2), x)