\(\int \frac {1}{(a+b \arccos (c x))^2} \, dx\) [681]

Optimal result

Integrand size = 10, antiderivative size = 86 \[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=-\frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}+\frac {\operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{b^2 c}-\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=\frac {\frac {b \sqrt {1-c^2 x^2}}{a+b \arccos (c x)}-\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right )-\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )}{b^2 c} \] Input:

Integrate[(a + b*ArcCos[c*x])^(-2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5133, 5225, 3042, 3784, 25, 3042, 3780, 3783}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(a+b \arccos (c x))^2} \, dx\)

\(\Big \downarrow \) 5133

\(\displaystyle \frac {c \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))}dx}{b}+\frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 5225

\(\displaystyle \frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c}\)

\(\Big \downarrow \) 3784

\(\displaystyle \frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {a}{b}\right ) \int \frac {\cos \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\sin \left (\frac {a}{b}\right ) \int -\frac {\sin \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\sin \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))+\cos \left (\frac {a}{b}\right ) \int \frac {\cos \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\sin \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))+\cos \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c}\)

\(\Big \downarrow \) 3780

\(\displaystyle \frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c}\)

\(\Big \downarrow \) 3783

\(\displaystyle \frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c}\)

Input:

Int[(a + b*ArcCos[c*x])^(-2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* 
e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[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 5133

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-Sqrt[1 - c 
^2*x^2])*((a + b*ArcCos[c*x])^(n + 1)/(b*c*(n + 1))), x] - Simp[c/(b*(n + 1 
))   Int[x*((a + b*ArcCos[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ 
[{a, b, c}, x] && LtQ[n, -1]
 

rule 5225

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(-(b*c^(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c 
^2*x^2)^p]   Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b]^(2*p + 1), x], 
 x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e 
, 0] && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{\left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{b^{2}}}{c}\)	\(74\)
default	\(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{\left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{b^{2}}}{c}\)	\(74\)

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2}}\, dx \] Input:

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

Output:

Integral((a + b*acos(c*x))**(-2), x)
 

Maxima [F]

\[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

-((b^2*c^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c^2)*integrate 
(sqrt(c*x + 1)*sqrt(-c*x + 1)*x/(a*b*c^2*x^2 - a*b + (b^2*c^2*x^2 - b^2)*a 
rctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)), x) - sqrt(c*x + 1)*sqrt(-c*x + 
 1))/(b^2*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (84) = 168\).

Time = 0.14 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.24 \[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=-\frac {b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} - \frac {b \arccos \left (c x\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} - \frac {a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} - \frac {a \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} + \frac {\sqrt {-c^{2} x^{2} + 1} b}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=\int \frac {1}{\mathit {acos} \left (c x \right )^{2} b^{2}+2 \mathit {acos} \left (c x \right ) a b +a^{2}}d x \] Input:

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

Output:

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