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

Optimal. Leaf size=112 \[ \frac{c^2 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^3}-\frac{c^2 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^3}-\frac{c^2 \cos \left (2 \sec ^{-1}(c x)\right )}{2 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{4 b \left (a+b \sec ^{-1}(c x)\right )^2} \]

[Out]

-(c^2*Cos[2*ArcSec[c*x]])/(2*b^2*(a + b*ArcSec[c*x])) + (c^2*CosIntegral[(2*a)/b + 2*ArcSec[c*x]]*Sin[(2*a)/b]
)/b^3 - (c^2*Sin[2*ArcSec[c*x]])/(4*b*(a + b*ArcSec[c*x])^2) - (c^2*Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSe
c[c*x]])/b^3

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Rubi [A]  time = 0.179502, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5222, 4406, 12, 3297, 3303, 3299, 3302} \[ \frac{c^2 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^3}-\frac{c^2 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^3}-\frac{c^2 \cos \left (2 \sec ^{-1}(c x)\right )}{2 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{4 b \left (a+b \sec ^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*ArcSec[c*x])^3),x]

[Out]

-(c^2*Cos[2*ArcSec[c*x]])/(2*b^2*(a + b*ArcSec[c*x])) + (c^2*CosIntegral[(2*a)/b + 2*ArcSec[c*x]]*Sin[(2*a)/b]
)/b^3 - (c^2*Sin[2*ArcSec[c*x]])/(4*b*(a + b*ArcSec[c*x])^2) - (c^2*Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSe
c[c*x]])/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 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 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^3 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx &=c^2 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{(a+b x)^3} \, dx,x,\sec ^{-1}(c x)\right )\\ &=c^2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 (a+b x)^3} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{1}{2} c^2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{(a+b x)^3} \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{4 b \left (a+b \sec ^{-1}(c x)\right )^2}+\frac{c^2 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{(a+b x)^2} \, dx,x,\sec ^{-1}(c x)\right )}{2 b}\\ &=-\frac{c^2 \cos \left (2 \sec ^{-1}(c x)\right )}{2 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{4 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{b^2}\\ &=-\frac{c^2 \cos \left (2 \sec ^{-1}(c x)\right )}{2 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{4 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{\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 )}{b^2}+\frac{\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 )}{b^2}\\ &=-\frac{c^2 \cos \left (2 \sec ^{-1}(c x)\right )}{2 b^2 \left (a+b \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 )}{b^3}-\frac{c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{4 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^2 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.400985, size = 114, normalized size = 1.02 \[ \frac{-\frac{b^2 c \sqrt{1-\frac{1}{c^2 x^2}}}{x \left (a+b \sec ^{-1}(c x)\right )^2}+2 c^2 \left (\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sec ^{-1}(c x)\right )\right )-\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sec ^{-1}(c x)\right )\right )\right )+\frac{b \left (c^2 x^2-2\right )}{x^2 \left (a+b \sec ^{-1}(c x)\right )}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.248, size = 157, normalized size = 1.4 \begin{align*}{c}^{2} \left ( -{\frac{\sin \left ( 2\,{\rm arcsec} \left (cx\right ) \right ) }{4\, \left ( a+b{\rm arcsec} \left (cx\right ) \right ) ^{2}b}}-{\frac{1}{ \left ( 2\,a+2\,b{\rm arcsec} \left (cx\right ) \right ){b}^{3}} \left ( 2\,{\it Si} \left ( 2\,{\frac{a}{b}}+2\,{\rm arcsec} \left (cx\right ) \right ) \cos \left ( 2\,{\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )b-2\,{\it Ci} \left ( 2\,{\frac{a}{b}}+2\,{\rm arcsec} \left (cx\right ) \right ) \sin \left ( 2\,{\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )b+2\,{\it Si} \left ( 2\,{\frac{a}{b}}+2\,{\rm arcsec} \left (cx\right ) \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) a-2\,{\it Ci} \left ( 2\,{\frac{a}{b}}+2\,{\rm arcsec} \left (cx\right ) \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) a+\cos \left ( 2\,{\rm arcsec} \left (cx\right ) \right ) 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))^3,x)

[Out]

c^2*(-1/4*sin(2*arcsec(c*x))/(a+b*arcsec(c*x))^2/b-1/2*(2*Si(2*a/b+2*arcsec(c*x))*cos(2*a/b)*arcsec(c*x)*b-2*C
i(2*a/b+2*arcsec(c*x))*sin(2*a/b)*arcsec(c*x)*b+2*Si(2*a/b+2*arcsec(c*x))*cos(2*a/b)*a-2*Ci(2*a/b+2*arcsec(c*x
))*sin(2*a/b)*a+cos(2*arcsec(c*x))*b)/(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^3/(a+b*arcsec(c*x))^3,x, algorithm="maxima")

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*(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^{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))^3,x, algorithm="giac")

[Out]

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