\(\int \frac {(d-c^2 d x^2)^2}{(a+b \arccos (c x))^{3/2}} \, dx\) [441]

Optimal result

Integrand size = 26, antiderivative size = 390 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=-\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {5 d^2 \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{2 b^{3/2} c}-\frac {5 d^2 \sqrt {\frac {3 \pi }{2}} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}-\frac {d^2 \sqrt {\frac {5 \pi }{2}} \cos \left (\frac {5 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c}+\frac {5 d^2 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{2 b^{3/2} c}+\frac {5 d^2 \sqrt {\frac {3 \pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b^{3/2} c}+\frac {d^2 \sqrt {\frac {5 \pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {5 a}{b}\right )}{4 b^{3/2} c} \] Output:

-2*d^2*(-c^2*x^2+1)^(5/2)/b/c/(a+b*arccos(c*x))^(1/2)-5/4*d^2*2^(1/2)*Pi^( 
1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))/b 
^(3/2)/c-5/8*d^2*6^(1/2)*Pi^(1/2)*cos(3*a/b)*FresnelS(6^(1/2)/Pi^(1/2)*(a+ 
b*arccos(c*x))^(1/2)/b^(1/2))/b^(3/2)/c-1/8*d^2*10^(1/2)*Pi^(1/2)*cos(5*a/ 
b)*FresnelS(10^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))/b^(3/2)/c+5 
/4*d^2*2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/ 
b^(1/2))*sin(a/b)/b^(3/2)/c+5/8*d^2*6^(1/2)*Pi^(1/2)*FresnelC(6^(1/2)/Pi^( 
1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*sin(3*a/b)/b^(3/2)/c+1/8*d^2*10^(1/2 
)*Pi^(1/2)*FresnelC(10^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*sin 
(5*a/b)/b^(3/2)/c
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.10 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.08 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=\frac {d^2 e^{-\frac {5 i a}{b}} \left (20 e^{\frac {5 i a}{b}} \sqrt {1-c^2 x^2}+10 i e^{\frac {4 i a}{b}} \sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {i (a+b \arccos (c x))}{b}\right )-10 i e^{\frac {6 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {i (a+b \arccos (c x))}{b}\right )-5 i \sqrt {3} e^{\frac {2 i a}{b}} \sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i (a+b \arccos (c x))}{b}\right )+5 i \sqrt {3} e^{\frac {8 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {3 i (a+b \arccos (c x))}{b}\right )+i \sqrt {5} \sqrt {-\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {5 i (a+b \arccos (c x))}{b}\right )-i \sqrt {5} e^{\frac {10 i a}{b}} \sqrt {\frac {i (a+b \arccos (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {5 i (a+b \arccos (c x))}{b}\right )-10 e^{\frac {5 i a}{b}} \sin (3 \arccos (c x))+2 e^{\frac {5 i a}{b}} \sin (5 \arccos (c x))\right )}{16 b c \sqrt {a+b \arccos (c x)}} \] Input:

Integrate[(d - c^2*d*x^2)^2/(a + b*ArcCos[c*x])^(3/2),x]
 

Output:

(d^2*(20*E^(((5*I)*a)/b)*Sqrt[1 - c^2*x^2] + (10*I)*E^(((4*I)*a)/b)*Sqrt[( 
(-I)*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, ((-I)*(a + b*ArcCos[c*x]))/b] - (1 
0*I)*E^(((6*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, (I*(a + b* 
ArcCos[c*x]))/b] - (5*I)*Sqrt[3]*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcCos[ 
c*x]))/b]*Gamma[1/2, ((-3*I)*(a + b*ArcCos[c*x]))/b] + (5*I)*Sqrt[3]*E^((( 
8*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, ((3*I)*(a + b*ArcCos 
[c*x]))/b] + I*Sqrt[5]*Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, ((-5* 
I)*(a + b*ArcCos[c*x]))/b] - I*Sqrt[5]*E^(((10*I)*a)/b)*Sqrt[(I*(a + b*Arc 
Cos[c*x]))/b]*Gamma[1/2, ((5*I)*(a + b*ArcCos[c*x]))/b] - 10*E^(((5*I)*a)/ 
b)*Sin[3*ArcCos[c*x]] + 2*E^(((5*I)*a)/b)*Sin[5*ArcCos[c*x]]))/(16*b*c*E^( 
((5*I)*a)/b)*Sqrt[a + b*ArcCos[c*x]])
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5167, 5225, 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.

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

\(\Big \downarrow \) 5167

\(\displaystyle \frac {10 c d^2 \int \frac {x \left (1-c^2 x^2\right )^{3/2}}{\sqrt {a+b \arccos (c x)}}dx}{b}+\frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \arccos (c x)}}\)

\(\Big \downarrow \) 5225

\(\displaystyle \frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {10 d^2 \int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^4\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{\sqrt {a+b \arccos (c x)}}d(a+b \arccos (c x))}{b^2 c}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {10 d^2 \int \left (\frac {\cos \left (\frac {5 a}{b}-\frac {5 (a+b \arccos (c x))}{b}\right )}{16 \sqrt {a+b \arccos (c x)}}-\frac {3 \cos \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{16 \sqrt {a+b \arccos (c x)}}+\frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{8 \sqrt {a+b \arccos (c x)}}\right )d(a+b \arccos (c x))}{b^2 c}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 d^2 \left (1-c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {10 d^2 \left (\frac {1}{4} \sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )-\frac {1}{8} \sqrt {\frac {3 \pi }{2}} \sqrt {b} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )+\frac {1}{8} \sqrt {\frac {\pi }{10}} \sqrt {b} \cos \left (\frac {5 a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )+\frac {1}{4} \sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )-\frac {1}{8} \sqrt {\frac {3 \pi }{2}} \sqrt {b} \sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )+\frac {1}{8} \sqrt {\frac {\pi }{10}} \sqrt {b} \sin \left (\frac {5 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )\right )}{b^2 c}\)

Input:

Int[(d - c^2*d*x^2)^2/(a + b*ArcCos[c*x])^(3/2),x]
 

Output:

(2*d^2*(1 - c^2*x^2)^(5/2))/(b*c*Sqrt[a + b*ArcCos[c*x]]) - (10*d^2*((Sqrt 
[b]*Sqrt[Pi/2]*Cos[a/b]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt 
[b]])/4 - (Sqrt[b]*Sqrt[(3*Pi)/2]*Cos[(3*a)/b]*FresnelC[(Sqrt[6/Pi]*Sqrt[a 
 + b*ArcCos[c*x]])/Sqrt[b]])/8 + (Sqrt[b]*Sqrt[Pi/10]*Cos[(5*a)/b]*Fresnel 
C[(Sqrt[10/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/8 + (Sqrt[b]*Sqrt[Pi/2]* 
FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/4 - (Sqrt 
[b]*Sqrt[(3*Pi)/2]*FresnelS[(Sqrt[6/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]* 
Sin[(3*a)/b])/8 + (Sqrt[b]*Sqrt[Pi/10]*FresnelS[(Sqrt[10/Pi]*Sqrt[a + b*Ar 
cCos[c*x]])/Sqrt[b]]*Sin[(5*a)/b])/8))/(b^2*c)
 

Defintions of rubi rules used

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 5167

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[(-Sqrt[1 - c^2*x^2])*(d + e*x^2)^p*((a + b*ArcCos[c*x])^(n 
+ 1)/(b*c*(n + 1))), x] - Simp[c*((2*p + 1)/(b*(n + 1)))*Simp[(d + e*x^2)^p 
/(1 - c^2*x^2)^p]   Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcCos[c*x])^(n + 
1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, - 
1]
 

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]
 

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 450, normalized size of antiderivative = 1.15

method	result	size
default	\(-\frac {d^{2} \left (10 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )-10 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )-5 \sqrt {-\frac {3}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) \sqrt {a +b \arccos \left (c x \right )}+5 \sqrt {-\frac {3}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) \sqrt {a +b \arccos \left (c x \right )}+\sqrt {-\frac {5}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {5 a}{b}\right ) \operatorname {FresnelC}\left (\frac {5 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {5}{b}}\, b}\right )-\sqrt {-\frac {5}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {5 a}{b}\right ) \operatorname {FresnelS}\left (\frac {5 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {5}{b}}\, b}\right )+10 \sin \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right )-5 \sin \left (-\frac {3 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {3 a}{b}\right )+\sin \left (-\frac {5 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {5 a}{b}\right )\right )}{8 c b \sqrt {a +b \arccos \left (c x \right )}}\)	\(450\)

Input:

int((-c^2*d*x^2+d)^2/(a+b*arccos(c*x))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*d^2/c/b*(10*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1/2)*cos 
(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)-10 
*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1/2)*sin(a/b)*FresnelS(2 
^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)-5*(-3/b)^(1/2)*Pi^ 
(1/2)*2^(1/2)*cos(3*a/b)*FresnelC(3*2^(1/2)/Pi^(1/2)/(-3/b)^(1/2)*(a+b*arc 
cos(c*x))^(1/2)/b)*(a+b*arccos(c*x))^(1/2)+5*(-3/b)^(1/2)*Pi^(1/2)*2^(1/2) 
*sin(3*a/b)*FresnelS(3*2^(1/2)/Pi^(1/2)/(-3/b)^(1/2)*(a+b*arccos(c*x))^(1/ 
2)/b)*(a+b*arccos(c*x))^(1/2)+(-5/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(c* 
x))^(1/2)*cos(5*a/b)*FresnelC(5*2^(1/2)/Pi^(1/2)/(-5/b)^(1/2)*(a+b*arccos( 
c*x))^(1/2)/b)-(-5/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1/2)*sin(5 
*a/b)*FresnelS(5*2^(1/2)/Pi^(1/2)/(-5/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)+ 
10*sin(-(a+b*arccos(c*x))/b+a/b)-5*sin(-3*(a+b*arccos(c*x))/b+3*a/b)+sin(- 
5*(a+b*arccos(c*x))/b+5*a/b))/(a+b*arccos(c*x))^(1/2)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((-c^2*d*x^2+d)^2/(a+b*arccos(c*x))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=d^{2} \left (\int \left (- \frac {2 c^{2} x^{2}}{a \sqrt {a + b \operatorname {acos}{\left (c x \right )}} + b \sqrt {a + b \operatorname {acos}{\left (c x \right )}} \operatorname {acos}{\left (c x \right )}}\right )\, dx + \int \frac {c^{4} x^{4}}{a \sqrt {a + b \operatorname {acos}{\left (c x \right )}} + b \sqrt {a + b \operatorname {acos}{\left (c x \right )}} \operatorname {acos}{\left (c x \right )}}\, dx + \int \frac {1}{a \sqrt {a + b \operatorname {acos}{\left (c x \right )}} + b \sqrt {a + b \operatorname {acos}{\left (c x \right )}} \operatorname {acos}{\left (c x \right )}}\, dx\right ) \] Input:

integrate((-c**2*d*x**2+d)**2/(a+b*acos(c*x))**(3/2),x)
 

Output:

d**2*(Integral(-2*c**2*x**2/(a*sqrt(a + b*acos(c*x)) + b*sqrt(a + b*acos(c 
*x))*acos(c*x)), x) + Integral(c**4*x**4/(a*sqrt(a + b*acos(c*x)) + b*sqrt 
(a + b*acos(c*x))*acos(c*x)), x) + Integral(1/(a*sqrt(a + b*acos(c*x)) + b 
*sqrt(a + b*acos(c*x))*acos(c*x)), x))
 

Maxima [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=\int { \frac {{\left (c^{2} d x^{2} - d\right )}^{2}}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-c^2*d*x^2+d)^2/(a+b*arccos(c*x))^(3/2),x, algorithm="maxima")
 

Output:

integrate((c^2*d*x^2 - d)^2/(b*arccos(c*x) + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=\int { \frac {{\left (c^{2} d x^{2} - d\right )}^{2}}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-c^2*d*x^2+d)^2/(a+b*arccos(c*x))^(3/2),x, algorithm="giac")
 

Output:

integrate((c^2*d*x^2 - d)^2/(b*arccos(c*x) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=\int \frac {{\left (d-c^2\,d\,x^2\right )}^2}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^{3/2}} \,d x \] Input:

int((d - c^2*d*x^2)^2/(a + b*acos(c*x))^(3/2),x)
 

Output:

int((d - c^2*d*x^2)^2/(a + b*acos(c*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^2}{(a+b \arccos (c x))^{3/2}} \, dx=\text {too large to display} \] Input:

int((-c^2*d*x^2+d)^2/(a+b*acos(c*x))^(3/2),x)
 

Output:

(d**2*( - 20*sqrt(acos(c*x)*b + a)*sqrt( - c**2*x**2 + 1)*acos(c*x)*b + 20 
*acos(c*x)*int(sqrt(acos(c*x)*b + a)/(acos(c*x)**2*b**2*c**2*x**2 - acos(c 
*x)**2*b**2 + 2*acos(c*x)*a*b*c**2*x**2 - 2*acos(c*x)*a*b + a**2*c**2*x**2 
 - a**2),x)*a*b**2*c + 8*acos(c*x)*int((sqrt(acos(c*x)*b + a)*x**6)/(acos( 
c*x)**2*b**2*c**2*x**2 - acos(c*x)**2*b**2 + 2*acos(c*x)*a*b*c**2*x**2 - 2 
*acos(c*x)*a*b + a**2*c**2*x**2 - a**2),x)*a*b**2*c**7 - 28*acos(c*x)*int( 
(sqrt(acos(c*x)*b + a)*x**4)/(acos(c*x)**2*b**2*c**2*x**2 - acos(c*x)**2*b 
**2 + 2*acos(c*x)*a*b*c**2*x**2 - 2*acos(c*x)*a*b + a**2*c**2*x**2 - a**2) 
,x)*a*b**2*c**5 - 24*acos(c*x)*int((sqrt(acos(c*x)*b + a)*sqrt( - c**2*x** 
2 + 1)*acos(c*x)*x**3)/(acos(c*x)**2*b**2*c**2*x**2 - acos(c*x)**2*b**2 + 
2*acos(c*x)*a*b*c**2*x**2 - 2*acos(c*x)*a*b + a**2*c**2*x**2 - a**2),x)*a* 
b**2*c**4 + 20*acos(c*x)*int((sqrt(acos(c*x)*b + a)*sqrt( - c**2*x**2 + 1) 
*acos(c*x)**2*x)/(acos(c*x)**2*b**2*c**2*x**2 - acos(c*x)**2*b**2 + 2*acos 
(c*x)*a*b*c**2*x**2 - 2*acos(c*x)*a*b + a**2*c**2*x**2 - a**2),x)*b**3*c** 
2 - 24*acos(c*x)*int((sqrt(acos(c*x)*b + a)*sqrt( - c**2*x**2 + 1)*x**3)/( 
acos(c*x)**2*b**2*c**2*x**2 - acos(c*x)**2*b**2 + 2*acos(c*x)*a*b*c**2*x** 
2 - 2*acos(c*x)*a*b + a**2*c**2*x**2 - a**2),x)*a**2*b*c**4 - 20*acos(c*x) 
*int((sqrt(acos(c*x)*b + a)*sqrt( - c**2*x**2 + 1)*x)/(acos(c*x)**2*b**2*c 
**2*x**2 - acos(c*x)**2*b**2 + 2*acos(c*x)*a*b*c**2*x**2 - 2*acos(c*x)*a*b 
 + a**2*c**2*x**2 - a**2),x)*a**2*b*c**2 - 5*acos(c*x)*int((sqrt(acos(c...