3.48 \(\int \frac{1}{x^2 (a+b \sec ^{-1}(c x))^3} \, dx\)

Optimal. Leaf size=103 \[ \frac{c \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{2 b^3}-\frac{c \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{2 b^3}-\frac{1}{2 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac{c \sqrt{1-\frac{1}{c^2 x^2}}}{2 b \left (a+b \sec ^{-1}(c x)\right )^2} \]

[Out]

-(c*Sqrt[1 - 1/(c^2*x^2)])/(2*b*(a + b*ArcSec[c*x])^2) - 1/(2*b^2*x*(a + b*ArcSec[c*x])) + (c*CosIntegral[a/b
+ ArcSec[c*x]]*Sin[a/b])/(2*b^3) - (c*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]])/(2*b^3)

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Rubi [A]  time = 0.147265, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{c \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{2 b^3}-\frac{c \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{2 b^3}-\frac{1}{2 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac{c \sqrt{1-\frac{1}{c^2 x^2}}}{2 b \left (a+b \sec ^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*ArcSec[c*x])^3),x]

[Out]

-(c*Sqrt[1 - 1/(c^2*x^2)])/(2*b*(a + b*ArcSec[c*x])^2) - 1/(2*b^2*x*(a + b*ArcSec[c*x])) + (c*CosIntegral[a/b
+ ArcSec[c*x]]*Sin[a/b])/(2*b^3) - (c*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]])/(2*b^3)

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

Mathematica [A]  time = 0.371504, size = 88, normalized size = 0.85 \[ -\frac{\frac{b \left (a+b c x \sqrt{1-\frac{1}{c^2 x^2}}+b \sec ^{-1}(c x)\right )}{x \left (a+b \sec ^{-1}(c x)\right )^2}-c \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )+c \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*ArcSec[c*x])^3),x]

[Out]

-((b*(a + b*c*Sqrt[1 - 1/(c^2*x^2)]*x + b*ArcSec[c*x]))/(x*(a + b*ArcSec[c*x])^2) - c*CosIntegral[a/b + ArcSec
[c*x]]*Sin[a/b] + c*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]])/(2*b^3)

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Maple [A]  time = 0.27, size = 154, normalized size = 1.5 \begin{align*} c \left ( -{\frac{1}{2\, \left ( a+b{\rm arcsec} \left (cx\right ) \right ) ^{2}b}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{1}{2\,cx \left ( a+b{\rm arcsec} \left (cx\right ) \right ){b}^{3}} \left ({\it Si} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \cos \left ({\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )cxb-{\it Ci} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \sin \left ({\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )cxb+{\it Si} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \cos \left ({\frac{a}{b}} \right ) cxa-{\it Ci} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \sin \left ({\frac{a}{b}} \right ) cxa+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))^3,x)

[Out]

c*(-1/2*((c^2*x^2-1)/c^2/x^2)^(1/2)/(a+b*arcsec(c*x))^2/b-1/2*(Si(a/b+arcsec(c*x))*cos(a/b)*arcsec(c*x)*c*x*b-
Ci(a/b+arcsec(c*x))*sin(a/b)*arcsec(c*x)*c*x*b+Si(a/b+arcsec(c*x))*cos(a/b)*c*x*a-Ci(a/b+arcsec(c*x))*sin(a/b)
*c*x*a+b)/c/x/(a+b*arcsec(c*x))/b^3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*arcsec(c*x))^3,x, algorithm="maxima")

[Out]

-(8*b^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 + 24*a*b^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + 2*a*b^2*log(c
^2*x^2)^2 + 8*a*b^2*log(c)^2 + 16*a*b^2*log(c)*log(x) + 8*a*b^2*log(x)^2 + 8*a^3 + 2*(4*b^3*arctan(sqrt(c*x +
1)*sqrt(c*x - 1))^2 - b^3*log(c^2*x^2)^2 - 4*b^3*log(c)^2 - 8*b^3*log(c)*log(x) - 4*b^3*log(x)^2 + 8*a*b^2*arc
tan(sqrt(c*x + 1)*sqrt(c*x - 1)) + 4*a^2*b + 4*(b^3*log(c) + b^3*log(x))*log(c^2*x^2))*sqrt(c*x + 1)*sqrt(c*x
- 1) + 2*(b^3*log(c^2*x^2)^2 + 4*b^3*log(c)^2 + 8*b^3*log(c)*log(x) + 4*b^3*log(x)^2 + 12*a^2*b - 4*(b^3*log(c
) + b^3*log(x))*log(c^2*x^2))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + (16*b^6*x*arctan(sqrt(c*x + 1)*sqrt(c*x -
1))^4 + b^6*x*log(c^2*x^2)^4 + 64*b^6*x*log(c)*log(x)^3 + 16*b^6*x*log(x)^4 + 64*a*b^5*x*arctan(sqrt(c*x + 1)*
sqrt(c*x - 1))^3 - 8*(b^6*x*log(c) + b^6*x*log(x))*log(c^2*x^2)^3 + 32*(3*b^6*log(c)^2 + a^2*b^4)*x*log(x)^2 +
8*(b^6*x*log(c^2*x^2)^2 + 8*b^6*x*log(c)*log(x) + 4*b^6*x*log(x)^2 + 4*(b^6*log(c)^2 + 3*a^2*b^4)*x - 4*(b^6*
x*log(c) + b^6*x*log(x))*log(c^2*x^2))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + 8*(6*b^6*x*log(c)*log(x) + 3*b^
6*x*log(x)^2 + (3*b^6*log(c)^2 + a^2*b^4)*x)*log(c^2*x^2)^2 + 64*(b^6*log(c)^3 + a^2*b^4*log(c))*x*log(x) + 16
*(b^6*log(c)^4 + 2*a^2*b^4*log(c)^2 + a^4*b^2)*x + 16*(a*b^5*x*log(c^2*x^2)^2 + 8*a*b^5*x*log(c)*log(x) + 4*a*
b^5*x*log(x)^2 + 4*(a*b^5*log(c)^2 + a^3*b^3)*x - 4*(a*b^5*x*log(c) + a*b^5*x*log(x))*log(c^2*x^2))*arctan(sqr
t(c*x + 1)*sqrt(c*x - 1)) - 32*(3*b^6*x*log(c)*log(x)^2 + b^6*x*log(x)^3 + (3*b^6*log(c)^2 + a^2*b^4)*x*log(x)
+ (b^6*log(c)^3 + a^2*b^4*log(c))*x)*log(c^2*x^2))*integrate(2*(b*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + a)/(4
*b^4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + b^4*x^2*log(c^2*x^2)^2 + 8*b^4*x^2*log(c)*log(x) + 4*b^4*x^2*
log(x)^2 + 8*a*b^3*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + 4*(b^4*log(c)^2 + a^2*b^2)*x^2 - 4*(b^4*x^2*log(c
) + b^4*x^2*log(x))*log(c^2*x^2)), x) - 8*(a*b^2*log(c) + a*b^2*log(x))*log(c^2*x^2))/(16*b^6*x*arctan(sqrt(c*
x + 1)*sqrt(c*x - 1))^4 + b^6*x*log(c^2*x^2)^4 + 64*b^6*x*log(c)*log(x)^3 + 16*b^6*x*log(x)^4 + 64*a*b^5*x*arc
tan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 8*(b^6*x*log(c) + b^6*x*log(x))*log(c^2*x^2)^3 + 32*(3*b^6*log(c)^2 + a^2
*b^4)*x*log(x)^2 + 8*(b^6*x*log(c^2*x^2)^2 + 8*b^6*x*log(c)*log(x) + 4*b^6*x*log(x)^2 + 4*(b^6*log(c)^2 + 3*a^
2*b^4)*x - 4*(b^6*x*log(c) + b^6*x*log(x))*log(c^2*x^2))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + 8*(6*b^6*x*lo
g(c)*log(x) + 3*b^6*x*log(x)^2 + (3*b^6*log(c)^2 + a^2*b^4)*x)*log(c^2*x^2)^2 + 64*(b^6*log(c)^3 + a^2*b^4*log
(c))*x*log(x) + 16*(b^6*log(c)^4 + 2*a^2*b^4*log(c)^2 + a^4*b^2)*x + 16*(a*b^5*x*log(c^2*x^2)^2 + 8*a*b^5*x*lo
g(c)*log(x) + 4*a*b^5*x*log(x)^2 + 4*(a*b^5*log(c)^2 + a^3*b^3)*x - 4*(a*b^5*x*log(c) + a*b^5*x*log(x))*log(c^
2*x^2))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - 32*(3*b^6*x*log(c)*log(x)^2 + b^6*x*log(x)^3 + (3*b^6*log(c)^2 +
a^2*b^4)*x*log(x) + (b^6*log(c)^3 + a^2*b^4*log(c))*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^{3} x^{2} \operatorname{arcsec}\left (c x\right )^{3} + 3 \, a b^{2} x^{2} \operatorname{arcsec}\left (c x\right )^{2} + 3 \, a^{2} b x^{2} \operatorname{arcsec}\left (c x\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

integral(1/(b^3*x^2*arcsec(c*x)^3 + 3*a*b^2*x^2*arcsec(c*x)^2 + 3*a^2*b*x^2*arcsec(c*x) + a^3*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 )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(a+b*asec(c*x))**3,x)

[Out]

Integral(1/(x**2*(a + b*asec(c*x))**3), 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 )}^{3} 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))^3,x, algorithm="giac")

[Out]

integrate(1/((b*arcsec(c*x) + a)^3*x^2), x)