\(\int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))} \, dx\) [355]

Optimal result

Integrand size = 26, antiderivative size = 54 \[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))} \, dx=-\frac {\operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{b c^2}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b c^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))} \, dx=-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )}{b c^2} \] Input:

Integrate[x/(Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {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.

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

\(\Big \downarrow \) 5225

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

Int[x/(Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])),x]
 

Output:

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

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 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 = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{c^{2} b}\)	\(46\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x/(sqrt(-(c*x - 1)*(c*x + 1))*(a + b*acos(c*x))), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))} \, dx=-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b c^{2}} - \frac {\sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b c^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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