\(\int \frac {d+e x^2}{\sqrt {a+b \arccos (c x)}} \, dx\) [698]

Optimal result

Integrand size = 20, antiderivative size = 329 \[ \int \frac {d+e x^2}{\sqrt {a+b \arccos (c x)}} \, dx=\frac {e \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}+\frac {d \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{\sqrt {b} c}-\frac {e \sqrt {\frac {\pi }{6}} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{2 \sqrt {b} c^3}+\frac {e \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{2 \sqrt {b} c^3}+\frac {d \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{\sqrt {b} c}-\frac {e \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{2 \sqrt {b} c^3} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.74 \[ \int \frac {d+e x^2}{\sqrt {a+b \arccos (c x)}} \, dx=\frac {e^{-\frac {3 i a}{b}} \left (3 \left (4 c^2 d+e\right ) e^{\frac {2 i a}{b}} \sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {i (a+b \arccos (c x))}{b}\right )+3 \left (4 c^2 d+e\right ) e^{\frac {4 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {i (a+b \arccos (c x))}{b}\right )+\sqrt {3} e \left (\sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i (a+b \arccos (c x))}{b}\right )+e^{\frac {6 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {3 i (a+b \arccos (c x))}{b}\right )\right )\right )}{24 c^3 \sqrt {a+b \arccos (c x)}} \] Input:

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

Output:

(3*(4*c^2*d + e)*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[ 
1/2, ((-I)*(a + b*ArcCos[c*x]))/b] + 3*(4*c^2*d + e)*E^(((4*I)*a)/b)*Sqrt[ 
(I*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, (I*(a + b*ArcCos[c*x]))/b] + Sqrt[3] 
*e*(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]))/(24*c^3*E^(((3*I)*a)/b)*Sqrt[a + b*ArcCos[c*x] 
])
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5173, 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.

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

\(\Big \downarrow \) 5173

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(e*Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/ 
Sqrt[b]])/(Sqrt[b]*c^3) - (d*Sqrt[2*Pi]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt 
[a + b*ArcCos[c*x]])/Sqrt[b]])/(Sqrt[b]*c) - (e*Sqrt[Pi/6]*Cos[(3*a)/b]*Fr 
esnelS[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/(2*Sqrt[b]*c^3) + (e 
*Sqrt[Pi/2]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b 
])/(2*Sqrt[b]*c^3) + (d*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[ 
c*x]])/Sqrt[b]]*Sin[a/b])/(Sqrt[b]*c) + (e*Sqrt[Pi/6]*FresnelC[(Sqrt[6/Pi] 
*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[(3*a)/b])/(2*Sqrt[b]*c^3)
 

Defintions of rubi rules used

rule 2009

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

rule 5173

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Int[ExpandIntegrand[(a + b*ArcCos[c*x])^n, (d + e*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (G 
tQ[p, 0] || IGtQ[n, 0])
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, \left (4 \,\operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sin \left (\frac {a}{b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {-\frac {3}{b}}\, b \,c^{2} d +4 \,\operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \cos \left (\frac {a}{b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {-\frac {3}{b}}\, b \,c^{2} d +\operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sin \left (\frac {a}{b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {-\frac {3}{b}}\, b e +\operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \cos \left (\frac {a}{b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {-\frac {3}{b}}\, b e -e \sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right )-e \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right )\right )}{12 c^{3}}\)	\(310\)

Input:

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

Output:

-1/12/c^3*2^(1/2)*Pi^(1/2)*(-3/b)^(1/2)*(4*FresnelC(2^(1/2)/Pi^(1/2)/(-1/b 
)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*sin(a/b)*(-1/b)^(1/2)*(-3/b)^(1/2)*b*c^ 
2*d+4*FresnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*co 
s(a/b)*(-1/b)^(1/2)*(-3/b)^(1/2)*b*c^2*d+FresnelC(2^(1/2)/Pi^(1/2)/(-1/b)^ 
(1/2)*(a+b*arccos(c*x))^(1/2)/b)*sin(a/b)*(-1/b)^(1/2)*(-3/b)^(1/2)*b*e+Fr 
esnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*cos(a/b)*( 
-1/b)^(1/2)*(-3/b)^(1/2)*b*e-e*sin(3*a/b)*FresnelC(3*2^(1/2)/Pi^(1/2)/(-3/ 
b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)-e*cos(3*a/b)*FresnelS(3*2^(1/2)/Pi^(1/ 
2)/(-3/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b))
 

Fricas [F(-2)]

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

integrate((e*x^2+d)/(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)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.54 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.46 \[ \int \frac {d+e x^2}{\sqrt {a+b \arccos (c x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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