\(\int \frac {x^2}{a+b \arccos (c x)} \, dx\) [158]

Optimal result

Integrand size = 14, antiderivative size = 121 \[ \int \frac {x^2}{a+b \arccos (c x)} \, dx=\frac {\operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b c^3}+\frac {\operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b c^3}-\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b c^3}-\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b c^3} \] Output:

1/4*Ci((a+b*arccos(c*x))/b)*sin(a/b)/b/c^3+1/4*Ci(3*(a+b*arccos(c*x))/b)*s 
in(3*a/b)/b/c^3-1/4*cos(a/b)*Si((a+b*arccos(c*x))/b)/b/c^3-1/4*cos(3*a/b)* 
Si(3*(a+b*arccos(c*x))/b)/b/c^3
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{a+b \arccos (c x)} \, dx=-\frac {-\operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right ) \sin \left (\frac {a}{b}\right )-\operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{4 b c^3} \] Input:

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

Output:

-1/4*(-(CosIntegral[a/b + ArcCos[c*x]]*Sin[a/b]) - CosIntegral[3*(a/b + Ar 
cCos[c*x])]*Sin[(3*a)/b] + Cos[a/b]*SinIntegral[a/b + ArcCos[c*x]] + Cos[( 
3*a)/b]*SinIntegral[3*(a/b + ArcCos[c*x])])/(b*c^3)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5147, 25, 4906, 2009}

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.

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

\(\Big \downarrow \) 5147

\(\displaystyle -\frac {\int -\frac {\cos ^2\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\cos ^2\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c^3}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {\int \left (\frac {\sin \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{4 (a+b \arccos (c x))}+\frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{4 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b c^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {1}{4} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {1}{4} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )+\frac {1}{4} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{b c^3}\)

Input:

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

Output:

-((-1/4*(CosIntegral[(a + b*ArcCos[c*x])/b]*Sin[a/b]) - (CosIntegral[(3*(a 
 + b*ArcCos[c*x]))/b]*Sin[(3*a)/b])/4 + (Cos[a/b]*SinIntegral[(a + b*ArcCo 
s[c*x])/b])/4 + (Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/4)/( 
b*c^3))
 

Defintions of rubi rules used

rule 25

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG 
tQ[p, 0]
 

rule 5147

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}+\frac {\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b}+\frac {\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b}}{c^{3}}\)	\(102\)
default	\(\frac {-\frac {\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}+\frac {\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b}+\frac {\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b}}{c^{3}}\)	\(102\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.42 \[ \int \frac {x^2}{a+b \arccos (c x)} \, dx=\frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b c^{3}} - \frac {\operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c^{3}} + \frac {\operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c^{3}} + \frac {3 \, \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, b c^{3}} - \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{4 \, b c^{3}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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