3.2.73 \(\int x^2 \sqrt {a+b \arccos (c x)} \, dx\) [173]

3.2.73.1 Optimal result

Integrand size = 16, antiderivative size = 242 \[ \int x^2 \sqrt {a+b \arccos (c x)} \, dx=\frac {1}{3} x^3 \sqrt {a+b \arccos (c x)}-\frac {\sqrt {b} \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{4 c^3}-\frac {\sqrt {b} \sqrt {\frac {\pi }{6}} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{12 c^3}-\frac {\sqrt {b} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{4 c^3}-\frac {\sqrt {b} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{12 c^3} \]

output

-1/72*cos(3*a/b)*FresnelC(6^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2) 
)*b^(1/2)*6^(1/2)*Pi^(1/2)/c^3-1/72*FresnelS(6^(1/2)/Pi^(1/2)*(a+b*arccos( 
c*x))^(1/2)/b^(1/2))*sin(3*a/b)*b^(1/2)*6^(1/2)*Pi^(1/2)/c^3-1/8*cos(a/b)* 
FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*b^(1/2)*2^(1/2) 
*Pi^(1/2)/c^3-1/8*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2 
))*sin(a/b)*b^(1/2)*2^(1/2)*Pi^(1/2)/c^3+1/3*x^3*(a+b*arccos(c*x))^(1/2)
 

3.2.73.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.95 \[ \int x^2 \sqrt {a+b \arccos (c x)} \, dx=-\frac {i b e^{-\frac {3 i a}{b}} \left (-9 e^{\frac {2 i a}{b}} \sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {3}{2},-\frac {i (a+b \arccos (c x))}{b}\right )+9 e^{\frac {4 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {3}{2},\frac {i (a+b \arccos (c x))}{b}\right )+\sqrt {3} \left (-\sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {3}{2},-\frac {3 i (a+b \arccos (c x))}{b}\right )+e^{\frac {6 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {3}{2},\frac {3 i (a+b \arccos (c x))}{b}\right )\right )\right )}{72 c^3 \sqrt {a+b \arccos (c x)}} \]

input

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

output

((-1/72*I)*b*(-9*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[ 
3/2, ((-I)*(a + b*ArcCos[c*x]))/b] + 9*E^(((4*I)*a)/b)*Sqrt[(I*(a + b*ArcC 
os[c*x]))/b]*Gamma[3/2, (I*(a + b*ArcCos[c*x]))/b] + Sqrt[3]*(-(Sqrt[((-I) 
*(a + b*ArcCos[c*x]))/b]*Gamma[3/2, ((-3*I)*(a + b*ArcCos[c*x]))/b]) + E^( 
((6*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[3/2, ((3*I)*(a + b*ArcC 
os[c*x]))/b])))/(c^3*E^(((3*I)*a)/b)*Sqrt[a + b*ArcCos[c*x]])
 

3.2.73.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5141, 5225, 3042, 3793, 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.

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

\(\Big \downarrow \) 5141

\(\displaystyle \frac {1}{6} b c \int \frac {x^3}{\sqrt {1-c^2 x^2} \sqrt {a+b \arccos (c x)}}dx+\frac {1}{3} x^3 \sqrt {a+b \arccos (c x)}\)

\(\Big \downarrow \) 5225

\(\displaystyle \frac {1}{3} x^3 \sqrt {a+b \arccos (c x)}-\frac {\int \frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{\sqrt {a+b \arccos (c x)}}d(a+b \arccos (c x))}{6 c^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

\(\displaystyle \frac {1}{3} x^3 \sqrt {a+b \arccos (c x)}-\frac {\int \left (\frac {\cos \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{4 \sqrt {a+b \arccos (c x)}}+\frac {3 \cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{4 \sqrt {a+b \arccos (c x)}}\right )d(a+b \arccos (c x))}{6 c^3}\)

\(\Big \downarrow \) 2009

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

input

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

output

(x^3*Sqrt[a + b*ArcCos[c*x]])/3 - ((3*Sqrt[b]*Sqrt[Pi/2]*Cos[a/b]*FresnelC 
[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/2 + (Sqrt[b]*Sqrt[Pi/6]*Co 
s[(3*a)/b]*FresnelC[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/2 + (3* 
Sqrt[b]*Sqrt[Pi/2]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]* 
Sin[a/b])/2 + (Sqrt[b]*Sqrt[Pi/6]*FresnelS[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c 
*x]])/Sqrt[b]]*Sin[(3*a)/b])/2)/(6*c^3)
 

3.2.73.3.1 Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 5141

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

3.2.73.4 Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {9 \sin \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, b -\cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, b -9 \,\operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \cos \left (\frac {a}{b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, b +\sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, b +18 \arccos \left (c x \right ) \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) b +18 \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a +6 \arccos \left (c x \right ) \cos \left (-\frac {3 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {3 a}{b}\right ) b +6 \cos \left (-\frac {3 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {3 a}{b}\right ) a}{72 c^{3} \sqrt {a +b \arccos \left (c x \right )}}\)	\(362\)

input

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

output

1/72/c^3/(a+b*arccos(c*x))^(1/2)*(9*sin(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(-1 
/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*2^(1/2)*Pi^(1/2)*(-1/b)^(1/2)*(a+b*ar 
ccos(c*x))^(1/2)*b-cos(3*a/b)*FresnelC(3*2^(1/2)/Pi^(1/2)/(-3/b)^(1/2)*(a+ 
b*arccos(c*x))^(1/2)/b)*2^(1/2)*Pi^(1/2)*(-3/b)^(1/2)*(a+b*arccos(c*x))^(1 
/2)*b-9*FresnelC(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)* 
cos(a/b)*2^(1/2)*Pi^(1/2)*(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)*b+sin(3*a/b 
)*FresnelS(3*2^(1/2)/Pi^(1/2)/(-3/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*2^(1 
/2)*Pi^(1/2)*(-3/b)^(1/2)*(a+b*arccos(c*x))^(1/2)*b+18*arccos(c*x)*cos(-(a 
+b*arccos(c*x))/b+a/b)*b+18*cos(-(a+b*arccos(c*x))/b+a/b)*a+6*arccos(c*x)* 
cos(-3*(a+b*arccos(c*x))/b+3*a/b)*b+6*cos(-3*(a+b*arccos(c*x))/b+3*a/b)*a)
 

3.2.73.5 Fricas [F(-2)]

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

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)
 

3.2.73.6 Sympy [F]

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

input

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

output

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

3.2.73.7 Maxima [F]

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

input

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

output

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

3.2.73.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.96 (sec) , antiderivative size = 1057, normalized size of antiderivative = 4.37 \[ \int x^2 \sqrt {a+b \arccos (c x)} \, dx=\text {Too large to display} \]

input

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

output

-1/8*I*sqrt(2)*sqrt(pi)*a*b*erf(-1/2*I*sqrt(2)*sqrt(b*arccos(c*x) + a)/sqr 
t(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(c*x) + a)*sqrt(abs(b))/b)*e^(I*a/b)/ 
((I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c^3) + 1/16*sqrt(2)*sqrt(pi)*b^2*er 
f(-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)/((I*b^2/sqrt(abs(b)) + b*sqrt( 
abs(b)))*c^3) + 1/8*I*sqrt(2)*sqrt(pi)*a*b*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)/((-I*b^2/sqrt(abs(b)) + b*sqrt(abs(b)))*c^3) + 1/16*sqrt(2) 
*sqrt(pi)*b^2*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)/((-I*b^2/sqrt( 
abs(b)) + b*sqrt(abs(b)))*c^3) - 1/4*I*sqrt(pi)*a*sqrt(b)*erf(-1/2*sqrt(6) 
*sqrt(b*arccos(c*x) + a)/sqrt(b) - 1/2*I*sqrt(6)*sqrt(b*arccos(c*x) + a)*s 
qrt(b)/abs(b))*e^(3*I*a/b)/((sqrt(6)*b + I*sqrt(6)*b^2/abs(b))*c^3) + 1/24 
*sqrt(pi)*b^(3/2)*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)*b + 
 I*sqrt(6)*b^2/abs(b))*c^3) + 1/4*I*sqrt(pi)*a*sqrt(b)*erf(-1/2*sqrt(6)*sq 
rt(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)*b - I*sqrt(6)*b^2/abs(b))*c^3) + 1/24*s 
qrt(pi)*b^(3/2)*erf(-1/2*sqrt(6)*sqrt(b*arccos(c*x) + a)/sqrt(b) + 1/2*I*s 
qrt(6)*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-3*I*a/b)/((sqrt(6)*b...
 

3.2.73.9 Mupad [F(-1)]

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

input

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

output

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