\(\int \frac {x}{(a+b \arccos (c x))^2} \, dx\) [164]

Optimal result

Integrand size = 12, antiderivative size = 91 \[ \int \frac {x}{(a+b \arccos (c x))^2} \, dx=\frac {x \sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{b^2 c^2}-\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{b^2 c^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {x}{(a+b \arccos (c x))^2} \, dx=\frac {\frac {b c x \sqrt {1-c^2 x^2}}{a+b \arccos (c x)}-\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{b^2 c^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5143, 25, 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.

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

\(\Big \downarrow \) 5143

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

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

Output:

(x*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcCos[c*x])) + (-(Cos[(2*a)/b]*CosInteg 
ral[(2*(a + b*ArcCos[c*x]))/b]) - Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcCo 
s[c*x]))/b])/(b^2*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 5143

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[( 
-x^m)*Sqrt[1 - c^2*x^2]*((a + b*ArcCos[c*x])^(n + 1)/(b*c*(n + 1))), x] - S 
imp[1/(b^2*c^(m + 1)*(n + 1))   Subst[Int[ExpandTrigReduce[x^(n + 1), Cos[- 
a/b + x/b]^(m - 1)*(m - (m + 1)*Cos[-a/b + x/b]^2), x], x], x, a + b*ArcCos 
[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (2 \arccos \left (c x \right )\right )}{2 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {Ci}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{b^{2}}}{c^{2}}\)	\(78\)
default	\(\frac {\frac {\sin \left (2 \arccos \left (c x \right )\right )}{2 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {Ci}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{b^{2}}}{c^{2}}\)	\(78\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

(sqrt(c*x + 1)*sqrt(-c*x + 1)*x - (b^2*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 
 1), c*x) + a*b*c)*integrate((2*c^2*x^2 - 1)*sqrt(c*x + 1)*sqrt(-c*x + 1)/ 
(a*b*c^3*x^2 - a*b*c + (b^2*c^3*x^2 - b^2*c)*arctan2(sqrt(c*x + 1)*sqrt(-c 
*x + 1), c*x)), x))/(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. 323 vs. \(2 (89) = 178\).

Time = 0.15 (sec) , antiderivative size = 323, normalized size of antiderivative = 3.55 \[ \int \frac {x}{(a+b \arccos (c x))^2} \, dx=-\frac {2 \, b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {2 \, b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {2 \, a \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {2 \, a \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} b c x}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac {b \arccos \left (c x\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac {a \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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