3.1.49 \(\int \frac {1}{x^3 (a+b \sec ^{-1}(c x))^3} \, dx\) [49]

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

[Out]

-1/2*c^2*cos(2*arcsec(c*x))/b^2/(a+b*arcsec(c*x))-c^2*cos(2*a/b)*Si(2*a/b+2*arcsec(c*x))/b^3+c^2*Ci(2*a/b+2*ar
csec(c*x))*sin(2*a/b)/b^3-1/4*c^2*sin(2*arcsec(c*x))/b/(a+b*arcsec(c*x))^2

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Rubi [A]
time = 0.14, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5330, 4491, 12, 3378, 3384, 3380, 3383} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*ArcSec[c*x])^3),x]

[Out]

-1/2*(c^2*Cos[2*ArcSec[c*x]])/(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*Arc
Sec[c*x]])/b^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3378

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 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 4491

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

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Mathematica [A]
time = 0.31, size = 114, normalized size = 1.02 \begin {gather*} \frac {-\frac {b^2 c \sqrt {1-\frac {1}{c^2 x^2}}}{x \left (a+b \sec ^{-1}(c x)\right )^2}+\frac {b \left (-2+c^2 x^2\right )}{x^2 \left (a+b \sec ^{-1}(c x)\right )}+2 c^2 \left (\text {CosIntegral}\left (2 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )-\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )\right )}{2 b^3} \end {gather*}

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 = 0.15, size = 157, normalized size = 1.40

method	result	size
derivativedivides	\(c^{2} \left (-\frac {\sin \left (2 \,\mathrm {arcsec}\left (c x \right )\right )}{4 \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )^{2} b}-\frac {2 \,\mathrm {arcsec}\left (c x \right ) \cos \left (\frac {2 a}{b}\right ) \sinIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) b -2 \,\mathrm {arcsec}\left (c x \right ) \sin \left (\frac {2 a}{b}\right ) \cosineIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) b +2 \cos \left (\frac {2 a}{b}\right ) \sinIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) a -2 \sin \left (\frac {2 a}{b}\right ) \cosineIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) a +\cos \left (2 \,\mathrm {arcsec}\left (c x \right )\right ) b}{2 \left (a +b \,\mathrm {arcsec}\left (c x \right )\right ) b^{3}}\right )\)	\(157\)
default	\(c^{2} \left (-\frac {\sin \left (2 \,\mathrm {arcsec}\left (c x \right )\right )}{4 \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )^{2} b}-\frac {2 \,\mathrm {arcsec}\left (c x \right ) \cos \left (\frac {2 a}{b}\right ) \sinIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) b -2 \,\mathrm {arcsec}\left (c x \right ) \sin \left (\frac {2 a}{b}\right ) \cosineIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) b +2 \cos \left (\frac {2 a}{b}\right ) \sinIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) a -2 \sin \left (\frac {2 a}{b}\right ) \cosineIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) a +\cos \left (2 \,\mathrm {arcsec}\left (c x \right )\right ) b}{2 \left (a +b \,\mathrm {arcsec}\left (c x \right )\right ) b^{3}}\right )\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b*arcsec(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

c^2*(-1/4*sin(2*arcsec(c*x))/(a+b*arcsec(c*x))^2/b-1/2*(2*arcsec(c*x)*cos(2*a/b)*Si(2*a/b+2*arcsec(c*x))*b-2*a
rcsec(c*x)*sin(2*a/b)*Ci(2*a/b+2*arcsec(c*x))*b+2*cos(2*a/b)*Si(2*a/b+2*arcsec(c*x))*a-2*sin(2*a/b)*Ci(2*a/b+2
*arcsec(c*x))*a+cos(2*arcsec(c*x))*b)/(a+b*arcsec(c*x))/b^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 929 vs. \(2 (108) = 216\).
time = 0.41, size = 929, normalized size = 8.29 \begin {gather*} \frac {1}{2} \, {\left (\frac {4 \, b^{2} c \arccos \left (\frac {1}{c x}\right )^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} - \frac {4 \, b^{2} c \arccos \left (\frac {1}{c x}\right )^{2} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} + \frac {8 \, a b c \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} - \frac {8 \, a b c \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} + \frac {4 \, a^{2} c \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} + \frac {2 \, b^{2} c \arccos \left (\frac {1}{c x}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} - \frac {4 \, a^{2} c \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} + \frac {4 \, a b c \arccos \left (\frac {1}{c x}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} + \frac {b^{2} c \arccos \left (\frac {1}{c x}\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} + \frac {2 \, a^{2} c \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} + \frac {a b c}{b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}} - \frac {b^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{{\left (b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}\right )} x} - \frac {2 \, b^{2} \arccos \left (\frac {1}{c x}\right )}{{\left (b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}\right )} c x^{2}} - \frac {2 \, a b}{{\left (b^{5} \arccos \left (\frac {1}{c x}\right )^{2} + 2 \, a b^{4} \arccos \left (\frac {1}{c x}\right ) + a^{2} b^{3}\right )} c x^{2}}\right )} c \end {gather*}

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]

1/2*(4*b^2*c*arccos(1/(c*x))^2*cos(a/b)*cos_integral(2*a/b + 2*arccos(1/(c*x)))*sin(a/b)/(b^5*arccos(1/(c*x))^
2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) - 4*b^2*c*arccos(1/(c*x))^2*cos(a/b)^2*sin_integral(2*a/b + 2*arccos(1/
(c*x)))/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) + 8*a*b*c*arccos(1/(c*x))*cos(a/b)*cos_int
egral(2*a/b + 2*arccos(1/(c*x)))*sin(a/b)/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) - 8*a*b*
c*arccos(1/(c*x))*cos(a/b)^2*sin_integral(2*a/b + 2*arccos(1/(c*x)))/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1
/(c*x)) + a^2*b^3) + 4*a^2*c*cos(a/b)*cos_integral(2*a/b + 2*arccos(1/(c*x)))*sin(a/b)/(b^5*arccos(1/(c*x))^2
+ 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) + 2*b^2*c*arccos(1/(c*x))^2*sin_integral(2*a/b + 2*arccos(1/(c*x)))/(b^5*
arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) - 4*a^2*c*cos(a/b)^2*sin_integral(2*a/b + 2*arccos(1/(c
*x)))/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) + 4*a*b*c*arccos(1/(c*x))*sin_integral(2*a/b
+ 2*arccos(1/(c*x)))/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) + b^2*c*arccos(1/(c*x))/(b^5
*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) + 2*a^2*c*sin_integral(2*a/b + 2*arccos(1/(c*x)))/(b^5
*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) + a*b*c/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x
)) + a^2*b^3) - b^2*sqrt(-1/(c^2*x^2) + 1)/((b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3)*x) - 2
*b^2*arccos(1/(c*x))/((b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3)*c*x^2) - 2*a*b/((b^5*arccos(
1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3)*c*x^2))*c

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*acos(1/(c*x)))^3),x)

[Out]

int(1/(x^3*(a + b*acos(1/(c*x)))^3), x)

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