\(\int \frac {(d-c^2 d x^2)^{5/2}}{(a+b \arccos (c x))^2} \, dx\) [74]

Optimal result

Integrand size = 26, antiderivative size = 428 \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=\frac {d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2}}{b c (a+b \arccos (c x))}+\frac {15 d^2 \sqrt {d-c^2 d x^2} \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{16 b^2 c \sqrt {1-c^2 x^2}}-\frac {3 d^2 \sqrt {d-c^2 d x^2} \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{4 b^2 c \sqrt {1-c^2 x^2}}+\frac {3 d^2 \sqrt {d-c^2 d x^2} \operatorname {CosIntegral}\left (\frac {6 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {6 a}{b}\right )}{16 b^2 c \sqrt {1-c^2 x^2}}-\frac {15 d^2 \sqrt {d-c^2 d x^2} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{16 b^2 c \sqrt {1-c^2 x^2}}+\frac {3 d^2 \sqrt {d-c^2 d x^2} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{4 b^2 c \sqrt {1-c^2 x^2}}-\frac {3 d^2 \sqrt {d-c^2 d x^2} \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arccos (c x))}{b}\right )}{16 b^2 c \sqrt {1-c^2 x^2}} \] Output:

d^2*(-c^2*x^2+1)^(5/2)*(-c^2*d*x^2+d)^(1/2)/b/c/(a+b*arccos(c*x))+15/16*d^ 
2*(-c^2*d*x^2+d)^(1/2)*Ci(2*(a+b*arccos(c*x))/b)*sin(2*a/b)/b^2/c/(-c^2*x^ 
2+1)^(1/2)-3/4*d^2*(-c^2*d*x^2+d)^(1/2)*Ci(4*(a+b*arccos(c*x))/b)*sin(4*a/ 
b)/b^2/c/(-c^2*x^2+1)^(1/2)+3/16*d^2*(-c^2*d*x^2+d)^(1/2)*Ci(6*(a+b*arccos 
(c*x))/b)*sin(6*a/b)/b^2/c/(-c^2*x^2+1)^(1/2)-15/16*d^2*(-c^2*d*x^2+d)^(1/ 
2)*cos(2*a/b)*Si(2*(a+b*arccos(c*x))/b)/b^2/c/(-c^2*x^2+1)^(1/2)+3/4*d^2*( 
-c^2*d*x^2+d)^(1/2)*cos(4*a/b)*Si(4*(a+b*arccos(c*x))/b)/b^2/c/(-c^2*x^2+1 
)^(1/2)-3/16*d^2*(-c^2*d*x^2+d)^(1/2)*cos(6*a/b)*Si(6*(a+b*arccos(c*x))/b) 
/b^2/c/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 343, normalized size of antiderivative = 0.80 \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=-\frac {d^2 \sqrt {d-c^2 d x^2} \left (-16 b+48 b c^2 x^2-48 b c^4 x^4+16 b c^6 x^6-15 (a+b \arccos (c x)) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+12 (a+b \arccos (c x)) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )-3 a \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )-3 b \arccos (c x) \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )+15 a \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )+15 b \arccos (c x) \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )-12 a \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )-12 b \arccos (c x) \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )+3 a \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right )+3 b \arccos (c x) \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{16 b^2 c \sqrt {1-c^2 x^2} (a+b \arccos (c x))} \] Input:

Integrate[(d - c^2*d*x^2)^(5/2)/(a + b*ArcCos[c*x])^2,x]
 

Output:

-1/16*(d^2*Sqrt[d - c^2*d*x^2]*(-16*b + 48*b*c^2*x^2 - 48*b*c^4*x^4 + 16*b 
*c^6*x^6 - 15*(a + b*ArcCos[c*x])*CosIntegral[2*(a/b + ArcCos[c*x])]*Sin[( 
2*a)/b] + 12*(a + b*ArcCos[c*x])*CosIntegral[4*(a/b + ArcCos[c*x])]*Sin[(4 
*a)/b] - 3*a*CosIntegral[6*(a/b + ArcCos[c*x])]*Sin[(6*a)/b] - 3*b*ArcCos[ 
c*x]*CosIntegral[6*(a/b + ArcCos[c*x])]*Sin[(6*a)/b] + 15*a*Cos[(2*a)/b]*S 
inIntegral[2*(a/b + ArcCos[c*x])] + 15*b*ArcCos[c*x]*Cos[(2*a)/b]*SinInteg 
ral[2*(a/b + ArcCos[c*x])] - 12*a*Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcCos 
[c*x])] - 12*b*ArcCos[c*x]*Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcCos[c*x])] 
 + 3*a*Cos[(6*a)/b]*SinIntegral[6*(a/b + ArcCos[c*x])] + 3*b*ArcCos[c*x]*C 
os[(6*a)/b]*SinIntegral[6*(a/b + ArcCos[c*x])]))/(b^2*c*Sqrt[1 - c^2*x^2]* 
(a + b*ArcCos[c*x]))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.56, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5167, 5225, 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.

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

\(\Big \downarrow \) 5167

\(\displaystyle \frac {6 c d^2 \sqrt {d-c^2 d x^2} \int \frac {x \left (1-c^2 x^2\right )^2}{a+b \arccos (c x)}dx}{b \sqrt {1-c^2 x^2}}+\frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{5/2}}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 5225

\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{5/2}}{b c (a+b \arccos (c x))}-\frac {6 d^2 \sqrt {d-c^2 d x^2} \int -\frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^5\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {6 d^2 \sqrt {d-c^2 d x^2} \int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^5\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c \sqrt {1-c^2 x^2}}+\frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{5/2}}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {6 d^2 \sqrt {d-c^2 d x^2} \int \left (\frac {\sin \left (\frac {6 a}{b}-\frac {6 (a+b \arccos (c x))}{b}\right )}{32 (a+b \arccos (c x))}-\frac {\sin \left (\frac {4 a}{b}-\frac {4 (a+b \arccos (c x))}{b}\right )}{8 (a+b \arccos (c x))}+\frac {5 \sin \left (\frac {2 a}{b}-\frac {2 (a+b \arccos (c x))}{b}\right )}{32 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b^2 c \sqrt {1-c^2 x^2}}+\frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{5/2}}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{5/2}}{b c (a+b \arccos (c x))}-\frac {6 d^2 \sqrt {d-c^2 d x^2} \left (-\frac {5}{32} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )+\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )-\frac {1}{32} \sin \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arccos (c x))}{b}\right )+\frac {5}{32} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )+\frac {1}{32} \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c \sqrt {1-c^2 x^2}}\)

Input:

Int[(d - c^2*d*x^2)^(5/2)/(a + b*ArcCos[c*x])^2,x]
 

Output:

(Sqrt[1 - c^2*x^2]*(d - c^2*d*x^2)^(5/2))/(b*c*(a + b*ArcCos[c*x])) - (6*d 
^2*Sqrt[d - c^2*d*x^2]*((-5*CosIntegral[(2*(a + b*ArcCos[c*x]))/b]*Sin[(2* 
a)/b])/32 + (CosIntegral[(4*(a + b*ArcCos[c*x]))/b]*Sin[(4*a)/b])/8 - (Cos 
Integral[(6*(a + b*ArcCos[c*x]))/b]*Sin[(6*a)/b])/32 + (5*Cos[(2*a)/b]*Sin 
Integral[(2*(a + b*ArcCos[c*x]))/b])/32 - (Cos[(4*a)/b]*SinIntegral[(4*(a 
+ b*ArcCos[c*x]))/b])/8 + (Cos[(6*a)/b]*SinIntegral[(6*(a + b*ArcCos[c*x]) 
)/b])/32))/(b^2*c*Sqrt[1 - c^2*x^2])
 

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 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 [C] (verified)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 717, normalized size of antiderivative = 1.68

method	result	size
default	\(\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (-7 i \sin \left (5 \arccos \left (c x \right )\right ) b +21 i \sin \left (3 \arccos \left (c x \right )\right ) b +24 i \operatorname {expIntegral}_{1}\left (4 i \arccos \left (c x \right )+\frac {4 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+4 a \right )}{b}} b \arccos \left (c x \right )+6 i \operatorname {expIntegral}_{1}\left (-6 i \arccos \left (c x \right )-\frac {6 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+6 a \right )}{b}} b \arccos \left (c x \right )+24 i \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} x^{2}-30 i \operatorname {expIntegral}_{1}\left (2 i \arccos \left (c x \right )+\frac {2 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+2 a \right )}{b}} b \arccos \left (c x \right )+30 i \operatorname {expIntegral}_{1}\left (-2 i \arccos \left (c x \right )-\frac {2 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+2 a \right )}{b}} b \arccos \left (c x \right )+30 i \operatorname {expIntegral}_{1}\left (-2 i \arccos \left (c x \right )-\frac {2 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+2 a \right )}{b}} a -6 i \operatorname {expIntegral}_{1}\left (6 i \arccos \left (c x \right )+\frac {6 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+6 a \right )}{b}} b \arccos \left (c x \right )-24 i \operatorname {expIntegral}_{1}\left (-4 i \arccos \left (c x \right )-\frac {4 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+4 a \right )}{b}} b \arccos \left (c x \right )-6 i \operatorname {expIntegral}_{1}\left (6 i \arccos \left (c x \right )+\frac {6 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+6 a \right )}{b}} a -24 i \operatorname {expIntegral}_{1}\left (-4 i \arccos \left (c x \right )-\frac {4 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+4 a \right )}{b}} a +64 i \sqrt {-c^{2} x^{2}+1}\, b \,c^{6} x^{6}-80 i \sqrt {-c^{2} x^{2}+1}\, b \,c^{4} x^{4}-5 \cos \left (5 \arccos \left (c x \right )\right ) b +9 \cos \left (3 \arccos \left (c x \right )\right ) b -12 c x b -36 i \sqrt {-c^{2} x^{2}+1}\, b +6 i \operatorname {expIntegral}_{1}\left (-6 i \arccos \left (c x \right )-\frac {6 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+6 a \right )}{b}} a +24 i \operatorname {expIntegral}_{1}\left (4 i \arccos \left (c x \right )+\frac {4 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+4 a \right )}{b}} a -30 i \operatorname {expIntegral}_{1}\left (2 i \arccos \left (c x \right )+\frac {2 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+2 a \right )}{b}} a +56 b \,c^{3} x^{3}+64 b \,c^{7} x^{7}-112 b \,c^{5} x^{5}\right ) d^{2}}{64 c \left (c^{2} x^{2}-1\right ) b^{2} \left (a +b \arccos \left (c x \right )\right )}\)	\(717\)

Input:

int((-c^2*d*x^2+d)^(5/2)/(a+b*arccos(c*x))^2,x,method=_RETURNVERBOSE)
 

Output:

1/64*(-d*(c^2*x^2-1))^(1/2)*(I*c^2*x^2+c*x*(-c^2*x^2+1)^(1/2)-I)*(-7*I*sin 
(5*arccos(c*x))*b+21*I*sin(3*arccos(c*x))*b+24*I*Ei(1,4*I*arccos(c*x)+4*I* 
a/b)*exp(I*(b*arccos(c*x)+4*a)/b)*b*arccos(c*x)+6*I*Ei(1,-6*I*arccos(c*x)- 
6*I*a/b)*exp(-I*(-b*arccos(c*x)+6*a)/b)*b*arccos(c*x)+24*I*(-c^2*x^2+1)^(1 
/2)*b*c^2*x^2-30*I*Ei(1,2*I*arccos(c*x)+2*I*a/b)*exp(I*(b*arccos(c*x)+2*a) 
/b)*b*arccos(c*x)+30*I*Ei(1,-2*I*arccos(c*x)-2*I*a/b)*exp(-I*(-b*arccos(c* 
x)+2*a)/b)*b*arccos(c*x)+30*I*Ei(1,-2*I*arccos(c*x)-2*I*a/b)*exp(-I*(-b*ar 
ccos(c*x)+2*a)/b)*a-6*I*Ei(1,6*I*arccos(c*x)+6*I*a/b)*exp(I*(b*arccos(c*x) 
+6*a)/b)*b*arccos(c*x)-24*I*Ei(1,-4*I*arccos(c*x)-4*I*a/b)*exp(-I*(-b*arcc 
os(c*x)+4*a)/b)*b*arccos(c*x)-6*I*Ei(1,6*I*arccos(c*x)+6*I*a/b)*exp(I*(b*a 
rccos(c*x)+6*a)/b)*a-24*I*Ei(1,-4*I*arccos(c*x)-4*I*a/b)*exp(-I*(-b*arccos 
(c*x)+4*a)/b)*a+64*I*(-c^2*x^2+1)^(1/2)*b*c^6*x^6-80*I*(-c^2*x^2+1)^(1/2)* 
b*c^4*x^4-5*cos(5*arccos(c*x))*b+9*cos(3*arccos(c*x))*b-12*c*x*b-36*I*(-c^ 
2*x^2+1)^(1/2)*b+6*I*Ei(1,-6*I*arccos(c*x)-6*I*a/b)*exp(-I*(-b*arccos(c*x) 
+6*a)/b)*a+24*I*Ei(1,4*I*arccos(c*x)+4*I*a/b)*exp(I*(b*arccos(c*x)+4*a)/b) 
*a-30*I*Ei(1,2*I*arccos(c*x)+2*I*a/b)*exp(I*(b*arccos(c*x)+2*a)/b)*a+56*b* 
c^3*x^3+64*b*c^7*x^7-112*b*c^5*x^5)*d^2/c/(c^2*x^2-1)/b^2/(a+b*arccos(c*x) 
)
 

Fricas [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=\int { \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{{\left (b \arccos \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((-c^2*d*x^2+d)^(5/2)/(a+b*arccos(c*x))^2,x, algorithm="fricas")
 

Output:

integral((c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2)*sqrt(-c^2*d*x^2 + d)/(b^2*arc 
cos(c*x)^2 + 2*a*b*arccos(c*x) + a^2), x)
 

Sympy [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=\int \frac {\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2}}\, dx \] Input:

integrate((-c**2*d*x**2+d)**(5/2)/(a+b*acos(c*x))**2,x)
 

Output:

Integral((-d*(c*x - 1)*(c*x + 1))**(5/2)/(a + b*acos(c*x))**2, x)
 

Maxima [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=\int { \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{{\left (b \arccos \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((-c^2*d*x^2+d)^(5/2)/(a+b*arccos(c*x))^2,x, algorithm="maxima")
 

Output:

-(c^6*d^2*x^6 - 3*c^4*d^2*x^4 + 3*c^2*d^2*x^2 - d^2 - (b^2*c*arctan2(sqrt( 
c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c)*integrate(6*(c^5*d^2*x^5 - 2*c^3*d^ 
2*x^3 + c*d^2*x)/(b^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b), x 
))*sqrt(d)/(b^2*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1706 vs. \(2 (388) = 776\).

Time = 1.06 (sec) , antiderivative size = 1706, normalized size of antiderivative = 3.99 \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=\text {Too large to display} \] Input:

integrate((-c^2*d*x^2+d)^(5/2)/(a+b*arccos(c*x))^2,x, algorithm="giac")
 

Output:

-1/16*(16*b*c^6*d^(5/2)*x^6/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 48*b*c^4*d 
^(5/2)*x^4/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 96*b*d^(5/2)*arccos(c*x)*co 
s(a/b)^5*cos_integral(6*a/b + 6*arccos(c*x))*sin(a/b)/(b^3*c^2*arccos(c*x) 
 + a*b^2*c^2) + 96*b*d^(5/2)*arccos(c*x)*cos(a/b)^6*sin_integral(6*a/b + 6 
*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 96*a*d^(5/2)*cos(a/b)^5* 
cos_integral(6*a/b + 6*arccos(c*x))*sin(a/b)/(b^3*c^2*arccos(c*x) + a*b^2* 
c^2) + 96*a*d^(5/2)*cos(a/b)^6*sin_integral(6*a/b + 6*arccos(c*x))/(b^3*c^ 
2*arccos(c*x) + a*b^2*c^2) + 96*b*d^(5/2)*arccos(c*x)*cos(a/b)^3*cos_integ 
ral(6*a/b + 6*arccos(c*x))*sin(a/b)/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + 96 
*b*d^(5/2)*arccos(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*arccos(c*x))*sin( 
a/b)/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 144*b*d^(5/2)*arccos(c*x)*cos(a/b 
)^4*sin_integral(6*a/b + 6*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) 
- 96*b*d^(5/2)*arccos(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arccos(c*x))/ 
(b^3*c^2*arccos(c*x) + a*b^2*c^2) + 96*a*d^(5/2)*cos(a/b)^3*cos_integral(6 
*a/b + 6*arccos(c*x))*sin(a/b)/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + 96*a*d^ 
(5/2)*cos(a/b)^3*cos_integral(4*a/b + 4*arccos(c*x))*sin(a/b)/(b^3*c^2*arc 
cos(c*x) + a*b^2*c^2) - 144*a*d^(5/2)*cos(a/b)^4*sin_integral(6*a/b + 6*ar 
ccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 96*a*d^(5/2)*cos(a/b)^4*sin 
_integral(4*a/b + 4*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + 48*b* 
c^2*d^(5/2)*x^2/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 18*b*d^(5/2)*arccos...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=\int \frac {{\left (d-c^2\,d\,x^2\right )}^{5/2}}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2} \,d x \] Input:

int((d - c^2*d*x^2)^(5/2)/(a + b*acos(c*x))^2,x)
 

Output:

int((d - c^2*d*x^2)^(5/2)/(a + b*acos(c*x))^2, x)
 

Reduce [F]

\[ \int \frac {\left (d-c^2 d x^2\right )^{5/2}}{(a+b \arccos (c x))^2} \, dx=\sqrt {d}\, d^{2} \left (\int \frac {\sqrt {-c^{2} x^{2}+1}}{\mathit {acos} \left (c x \right )^{2} b^{2}+2 \mathit {acos} \left (c x \right ) a b +a^{2}}d x +\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{4}}{\mathit {acos} \left (c x \right )^{2} b^{2}+2 \mathit {acos} \left (c x \right ) a b +a^{2}}d x \right ) c^{4}-2 \left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{2}}{\mathit {acos} \left (c x \right )^{2} b^{2}+2 \mathit {acos} \left (c x \right ) a b +a^{2}}d x \right ) c^{2}\right ) \] Input:

int((-c^2*d*x^2+d)^(5/2)/(a+b*acos(c*x))^2,x)
 

Output:

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