3.2.88 \(\int \frac {x^2}{\sqrt {a+b \arccos (c x)}} \, dx\) [188]

3.2.88.1 Optimal result

Integrand size = 16, antiderivative size = 223 \[ \int \frac {x^2}{\sqrt {a+b \arccos (c x)}} \, dx=-\frac {\sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}-\frac {\sqrt {\frac {\pi }{6}} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}+\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{2 \sqrt {b} c^3}+\frac {\sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{2 \sqrt {b} c^3} \]

output

-1/12*cos(3*a/b)*FresnelS(6^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2) 
)*6^(1/2)*Pi^(1/2)/c^3/b^(1/2)+1/12*FresnelC(6^(1/2)/Pi^(1/2)*(a+b*arccos( 
c*x))^(1/2)/b^(1/2))*sin(3*a/b)*6^(1/2)*Pi^(1/2)/c^3/b^(1/2)-1/4*cos(a/b)* 
FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2 
)/c^3/b^(1/2)+1/4*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2 
))*sin(a/b)*2^(1/2)*Pi^(1/2)/c^3/b^(1/2)
 

3.2.88.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.01 \[ \int \frac {x^2}{\sqrt {a+b \arccos (c x)}} \, dx=\frac {e^{-\frac {3 i a}{b}} \left (3 e^{\frac {2 i a}{b}} \sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {i (a+b \arccos (c x))}{b}\right )+3 e^{\frac {4 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {i (a+b \arccos (c x))}{b}\right )+\sqrt {3} \left (\sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i (a+b \arccos (c x))}{b}\right )+e^{\frac {6 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {3 i (a+b \arccos (c x))}{b}\right )\right )\right )}{24 c^3 \sqrt {a+b \arccos (c x)}} \]

input

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

output

(3*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, ((-I)*(a 
+ b*ArcCos[c*x]))/b] + 3*E^(((4*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*G 
amma[1/2, (I*(a + b*ArcCos[c*x]))/b] + Sqrt[3]*(Sqrt[((-I)*(a + b*ArcCos[c 
*x]))/b]*Gamma[1/2, ((-3*I)*(a + b*ArcCos[c*x]))/b] + E^(((6*I)*a)/b)*Sqrt 
[(I*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, ((3*I)*(a + b*ArcCos[c*x]))/b]))/(2 
4*c^3*E^(((3*I)*a)/b)*Sqrt[a + b*ArcCos[c*x]])
 

3.2.88.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.

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

\(\Big \downarrow \) 5147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {1}{2} \sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )-\frac {1}{2} \sqrt {\frac {\pi }{6}} \sqrt {b} \sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \sqrt {b} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{b c^3}\)

input

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

output

-(((Sqrt[b]*Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x 
]])/Sqrt[b]])/2 + (Sqrt[b]*Sqrt[Pi/6]*Cos[(3*a)/b]*FresnelS[(Sqrt[6/Pi]*Sq 
rt[a + b*ArcCos[c*x]])/Sqrt[b]])/2 - (Sqrt[b]*Sqrt[Pi/2]*FresnelC[(Sqrt[2/ 
Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/2 - (Sqrt[b]*Sqrt[Pi/6]*Fr 
esnelC[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[(3*a)/b])/2)/(b*c 
^3))
 

3.2.88.3.1 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]
 

3.2.88.4 Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\sqrt {\pi }\, \sqrt {2}\, \sqrt {-\frac {1}{b}}\, \left (3 \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )+3 \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )-\sqrt {-\frac {1}{b}}\, \sqrt {-\frac {3}{b}}\, \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) b -\sqrt {-\frac {1}{b}}\, \sqrt {-\frac {3}{b}}\, \sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) b \right )}{12 c^{3}}\)	\(198\)

input

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

output

1/12/c^3*Pi^(1/2)*2^(1/2)*(-1/b)^(1/2)*(3*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/ 
2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)+3*cos(a/b)*FresnelS(2^(1/2)/Pi^ 
(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)-(-1/b)^(1/2)*(-3/b)^(1/2)*co 
s(3*a/b)*FresnelS(3*2^(1/2)/Pi^(1/2)/(-3/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/ 
b)*b-(-1/b)^(1/2)*(-3/b)^(1/2)*sin(3*a/b)*FresnelC(3*2^(1/2)/Pi^(1/2)/(-3/ 
b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b)
 

3.2.88.5 Fricas [F(-2)]

Exception generated. \[ \int \frac {x^2}{\sqrt {a+b \arccos (c x)}} \, dx=\text {Exception raised: TypeError} \]

input

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

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

3.2.88.6 Sympy [F]

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

input

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

output

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

3.2.88.7 Maxima [F]

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

input

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

output

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

3.2.88.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.42 \[ \int \frac {x^2}{\sqrt {a+b \arccos (c x)}} \, dx=\frac {i \, \sqrt {\pi } \operatorname {erf}\left (-\frac {\sqrt {6} \sqrt {b \arccos \left (c x\right ) + a}}{2 \, \sqrt {b}} - \frac {i \, \sqrt {6} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{2 \, {\left | b \right |}}\right ) e^{\left (\frac {3 i \, a}{b}\right )}}{4 \, {\left (\sqrt {6} \sqrt {b} + \frac {i \, \sqrt {6} b^{\frac {3}{2}}}{{\left | b \right |}}\right )} c^{3}} + \frac {i \, \sqrt {\pi } \operatorname {erf}\left (-\frac {i \, \sqrt {2} \sqrt {b \arccos \left (c x\right ) + a}}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (\frac {i \, a}{b}\right )}}{4 \, c^{3} {\left (\frac {i \, \sqrt {2} b}{\sqrt {{\left | b \right |}}} + \sqrt {2} \sqrt {{\left | b \right |}}\right )}} - \frac {i \, \sqrt {\pi } \operatorname {erf}\left (\frac {i \, \sqrt {2} \sqrt {b \arccos \left (c x\right ) + a}}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (-\frac {i \, a}{b}\right )}}{4 \, c^{3} {\left (-\frac {i \, \sqrt {2} b}{\sqrt {{\left | b \right |}}} + \sqrt {2} \sqrt {{\left | b \right |}}\right )}} - \frac {i \, \sqrt {\pi } \operatorname {erf}\left (-\frac {\sqrt {6} \sqrt {b \arccos \left (c x\right ) + a}}{2 \, \sqrt {b}} + \frac {i \, \sqrt {6} \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{2 \, {\left | b \right |}}\right ) e^{\left (-\frac {3 i \, a}{b}\right )}}{4 \, {\left (\sqrt {6} \sqrt {b} - \frac {i \, \sqrt {6} b^{\frac {3}{2}}}{{\left | b \right |}}\right )} c^{3}} \]

input

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

output

1/4*I*sqrt(pi)*erf(-1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b) - 1/2*I*sq 
rt(6)*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(3*I*a/b)/((sqrt(6)*sqrt(b 
) + I*sqrt(6)*b^(3/2)/abs(b))*c^3) + 1/4*I*sqrt(pi)*erf(-1/2*I*sqrt(2)*sqr 
t(b*arccos(c*x) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sq 
rt(abs(b))/b)*e^(I*a/b)/(c^3*(I*sqrt(2)*b/sqrt(abs(b)) + sqrt(2)*sqrt(abs( 
b)))) - 1/4*I*sqrt(pi)*erf(1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqrt(abs( 
b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/(c^3* 
(-I*sqrt(2)*b/sqrt(abs(b)) + sqrt(2)*sqrt(abs(b)))) - 1/4*I*sqrt(pi)*erf(- 
1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b) + 1/2*I*sqrt(6)*sqrt(b*arccos( 
c*x) + a)*sqrt(b)/abs(b))*e^(-3*I*a/b)/((sqrt(6)*sqrt(b) - I*sqrt(6)*b^(3/ 
2)/abs(b))*c^3)
 

3.2.88.9 Mupad [F(-1)]

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

input

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

output

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