\(\int \frac {1}{x^2 (a+b \sec ^{-1}(c x))^3} \, dx\) [48]

Optimal result

Integrand size = 14, antiderivative size = 103 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=-\frac {c \sqrt {1-\frac {1}{c^2 x^2}}}{2 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac {1}{2 b^2 x \left (a+b \sec ^{-1}(c x)\right )}+\frac {c \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )}{2 b^3}-\frac {c \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{2 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5330, 3378, 3384, 3380, 3383} \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=\frac {c \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{2 b^3}-\frac {c \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{2 b^3}-\frac {1}{2 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac {c \sqrt {1-\frac {1}{c^2 x^2}}}{2 b \left (a+b \sec ^{-1}(c x)\right )^2} \]

[In]

Int[1/(x^2*(a + b*ArcSec[c*x])^3),x]

[Out]

-1/2*(c*Sqrt[1 - 1/(c^2*x^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 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 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)^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 \text {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 \text {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 ) \text {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 ) \text {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 \operatorname {CosIntegral}\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] (verified)

Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=-\frac {\frac {b \left (a+b c \sqrt {1-\frac {1}{c^2 x^2}} x+b \sec ^{-1}(c x)\right )}{x \left (a+b \sec ^{-1}(c x)\right )^2}-c \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )+c \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{2 b^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(c \left (-\frac {\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{2 \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )^{2} b}-\frac {\operatorname {arcsec}\left (c x \right ) \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) b c x -\operatorname {arcsec}\left (c x \right ) \sin \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) b c x +\cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) a c x -\sin \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) a c x +b}{2 c x \left (a +b \,\operatorname {arcsec}\left (c x \right )\right ) b^{3}}\right )\)	\(154\)
default	\(c \left (-\frac {\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{2 \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )^{2} b}-\frac {\operatorname {arcsec}\left (c x \right ) \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) b c x -\operatorname {arcsec}\left (c x \right ) \sin \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) b c x +\cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) a c x -\sin \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) a c x +b}{2 c x \left (a +b \,\operatorname {arcsec}\left (c x \right )\right ) b^{3}}\right )\)	\(154\)

[In]

int(1/x^2/(a+b*arcsec(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=\int { \frac {1}{{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} x^{2}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=\int \frac {1}{x^{2} \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=\int { \frac {1}{{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} x^{2}} \,d x } \]

[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))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (93) = 186\).

Time = 0.31 (sec) , antiderivative size = 580, normalized size of antiderivative = 5.63 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=\frac {1}{2} \, {\left (\frac {b^{2} \arccos \left (\frac {1}{c x}\right )^{2} \operatorname {Ci}\left (\frac {a}{b} + \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 {b^{2} \arccos \left (\frac {1}{c x}\right )^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \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 {2 \, a b \arccos \left (\frac {1}{c x}\right ) \operatorname {Ci}\left (\frac {a}{b} + \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 \, a b \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \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^{2} \operatorname {Ci}\left (\frac {a}{b} + \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 {a^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \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} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{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} \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} - \frac {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}\right )} c \]

[In]

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

[Out]

1/2*(b^2*arccos(1/(c*x))^2*cos_integral(a/b + arccos(1/(c*x)))*sin(a/b)/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arcco
s(1/(c*x)) + a^2*b^3) - b^2*arccos(1/(c*x))^2*cos(a/b)*sin_integral(a/b + 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*b*arccos(1/(c*x))*cos_integral(a/b + 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*a*b*arccos(1/(c*x))*cos(a/b)*sin_integral(a/b
+ arccos(1/(c*x)))/(b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3) + a^2*cos_integral(a/b + 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) - a^2*cos(a/b)*sin_integral(a/
b + 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*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) - 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) - a*b/((b^5*arccos(1/(c*x))^2 + 2*a*b^4*arccos(1/(c*x)) + a^2*b^3)*c*x))
*c

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx=\int \frac {1}{x^2\,{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3} \,d x \]

[In]

int(1/(x^2*(a + b*acos(1/(c*x)))^3),x)

[Out]

int(1/(x^2*(a + b*acos(1/(c*x)))^3), x)