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

Optimal. Leaf size=75 \[ \frac{c \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b^2}+\frac{c \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b^2}-\frac{c \sqrt{1-\frac{1}{c^2 x^2}}}{b \left (a+b \sec ^{-1}(c x)\right )} \]

[Out]

-((c*Sqrt[1 - 1/(c^2*x^2)])/(b*(a + b*ArcSec[c*x]))) + (c*Cos[a/b]*CosIntegral[a/b + ArcSec[c*x]])/b^2 + (c*Si
n[a/b]*SinIntegral[a/b + ArcSec[c*x]])/b^2

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Rubi [A]  time = 0.128973, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5222, 3297, 3303, 3299, 3302} \[ \frac{c \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b^2}+\frac{c \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b^2}-\frac{c \sqrt{1-\frac{1}{c^2 x^2}}}{b \left (a+b \sec ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*ArcSec[c*x])^2),x]

[Out]

-((c*Sqrt[1 - 1/(c^2*x^2)])/(b*(a + b*ArcSec[c*x]))) + (c*Cos[a/b]*CosIntegral[a/b + ArcSec[c*x]])/b^2 + (c*Si
n[a/b]*SinIntegral[a/b + ArcSec[c*x]])/b^2

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 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && 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 )^2} \, dx &=c \operatorname{Subst}\left (\int \frac{\sin (x)}{(a+b x)^2} \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{c \sqrt{1-\frac{1}{c^2 x^2}}}{b \left (a+b \sec ^{-1}(c x)\right )}+\frac{c \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{b}\\ &=-\frac{c \sqrt{1-\frac{1}{c^2 x^2}}}{b \left (a+b \sec ^{-1}(c x)\right )}+\frac{\left (c \cos \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 )}{b}+\frac{\left (c \sin \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 )}{b}\\ &=-\frac{c \sqrt{1-\frac{1}{c^2 x^2}}}{b \left (a+b \sec ^{-1}(c x)\right )}+\frac{c \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b^2}+\frac{c \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.256383, size = 69, normalized size = 0.92 \[ \frac{c \left (-\frac{b \sqrt{1-\frac{1}{c^2 x^2}}}{a+b \sec ^{-1}(c x)}+\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )+\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(-((b*Sqrt[1 - 1/(c^2*x^2)])/(a + b*ArcSec[c*x])) + Cos[a/b]*CosIntegral[a/b + ArcSec[c*x]] + Sin[a/b]*SinI
ntegral[a/b + ArcSec[c*x]]))/b^2

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Maple [A]  time = 0.241, size = 78, normalized size = 1. \begin{align*} c \left ( -{\frac{1}{ \left ( a+b{\rm arcsec} \left (cx\right ) \right ) b}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{1}{{b}^{2}} \left ({\it Si} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \sin \left ({\frac{a}{b}} \right ) +{\it Ci} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \cos \left ({\frac{a}{b}} \right ) \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{4 \, \sqrt{c x + 1} \sqrt{c x - 1}{\left (b \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) + a\right )} - 4 \,{\left (4 \, b^{3} x \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{2} + b^{3} x \log \left (c^{2} x^{2}\right )^{2} + 8 \, b^{3} x \log \left (c\right ) \log \left (x\right ) + 4 \, b^{3} x \log \left (x\right )^{2} + 8 \, a b^{2} x \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) + 4 \,{\left (b^{3} \log \left (c\right )^{2} + a^{2} b\right )} x - 4 \,{\left (b^{3} x \log \left (c\right ) + b^{3} x \log \left (x\right )\right )} \log \left (c^{2} x^{2}\right )\right )} \int \frac{\sqrt{c x + 1} \sqrt{c x - 1}{\left (b \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) + a\right )}}{4 \,{\left (b^{3} c^{2} \log \left (c\right )^{2} + a^{2} b c^{2}\right )} x^{4} - 4 \,{\left (b^{3} \log \left (c\right )^{2} + a^{2} b\right )} x^{2} + 4 \,{\left (b^{3} c^{2} x^{4} - b^{3} x^{2}\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{2} +{\left (b^{3} c^{2} x^{4} - b^{3} x^{2}\right )} \log \left (c^{2} x^{2}\right )^{2} + 4 \,{\left (b^{3} c^{2} x^{4} - b^{3} x^{2}\right )} \log \left (x\right )^{2} + 8 \,{\left (a b^{2} c^{2} x^{4} - a b^{2} x^{2}\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) - 4 \,{\left (b^{3} c^{2} x^{4} \log \left (c\right ) - b^{3} x^{2} \log \left (c\right ) +{\left (b^{3} c^{2} x^{4} - b^{3} x^{2}\right )} \log \left (x\right )\right )} \log \left (c^{2} x^{2}\right ) + 8 \,{\left (b^{3} c^{2} x^{4} \log \left (c\right ) - b^{3} x^{2} \log \left (c\right )\right )} \log \left (x\right )}\,{d x}}{4 \, b^{3} x \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{2} + b^{3} x \log \left (c^{2} x^{2}\right )^{2} + 8 \, b^{3} x \log \left (c\right ) \log \left (x\right ) + 4 \, b^{3} x \log \left (x\right )^{2} + 8 \, a b^{2} x \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) + 4 \,{\left (b^{3} \log \left (c\right )^{2} + a^{2} b\right )} x - 4 \,{\left (b^{3} x \log \left (c\right ) + b^{3} x \log \left (x\right )\right )} \log \left (c^{2} x^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*sqrt(c*x + 1)*sqrt(c*x - 1)*(b*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + a) - (4*b^3*x*arctan(sqrt(c*x + 1)*sq
rt(c*x - 1))^2 + b^3*x*log(c^2*x^2)^2 + 8*b^3*x*log(c)*log(x) + 4*b^3*x*log(x)^2 + 8*a*b^2*x*arctan(sqrt(c*x +
1)*sqrt(c*x - 1)) + 4*(b^3*log(c)^2 + a^2*b)*x - 4*(b^3*x*log(c) + b^3*x*log(x))*log(c^2*x^2))*integrate(4*sq
rt(c*x + 1)*sqrt(c*x - 1)*(b*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + a)/(4*(b^3*c^2*log(c)^2 + a^2*b*c^2)*x^4 -
4*(b^3*log(c)^2 + a^2*b)*x^2 + 4*(b^3*c^2*x^4 - b^3*x^2)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + (b^3*c^2*x^4
- b^3*x^2)*log(c^2*x^2)^2 + 4*(b^3*c^2*x^4 - b^3*x^2)*log(x)^2 + 8*(a*b^2*c^2*x^4 - a*b^2*x^2)*arctan(sqrt(c*x
+ 1)*sqrt(c*x - 1)) - 4*(b^3*c^2*x^4*log(c) - b^3*x^2*log(c) + (b^3*c^2*x^4 - b^3*x^2)*log(x))*log(c^2*x^2) +
8*(b^3*c^2*x^4*log(c) - b^3*x^2*log(c))*log(x)), x))/(4*b^3*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + b^3*x*l
og(c^2*x^2)^2 + 8*b^3*x*log(c)*log(x) + 4*b^3*x*log(x)^2 + 8*a*b^2*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + 4*(
b^3*log(c)^2 + a^2*b)*x - 4*(b^3*x*log(c) + b^3*x*log(x))*log(c^2*x^2))

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

[Out]

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

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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 )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*(a + b*asec(c*x))**2), 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 )}^{2} 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))^2,x, algorithm="giac")

[Out]

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