\(\int \frac {1}{x^2 (a+b \sec ^{-1}(c x))} \, dx\) [36]

Optimal result

Integrand size = 14, antiderivative size = 46 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=-\frac {c \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )}{b}+\frac {c \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5330, 3384, 3380, 3383} \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=\frac {c \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{b}-\frac {c \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{b} \]

[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 3380

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 3383

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 3384

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 5330

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*} \text {integral}& = c \text {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 ) \text {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 ) \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right ) \\ & = -\frac {c \operatorname {CosIntegral}\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] (verified)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=\frac {c \left (-\operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )}{b} \]

[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] (verified)

Time = 0.45 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(c \left (\frac {\operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{b}-\frac {\operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )}{b}\right )\)	\(47\)
default	\(c \left (\frac {\operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{b}-\frac {\operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )}{b}\right )\)	\(47\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=\int { \frac {1}{{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{2}} \,d x } \]

[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]

\[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=\int \frac {1}{x^{2} \left (a + b \operatorname {asec}{\left (c x \right )}\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=\int { \frac {1}{{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=-c {\left (\frac {\operatorname {Ci}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b} - \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right )}{b}\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )} \, dx=\int \frac {1}{x^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )} \,d x \]

[In]

int(1/(x^2*(a + b*acos(1/(c*x)))),x)

[Out]

int(1/(x^2*(a + b*acos(1/(c*x)))), x)