\(\int \frac {(d-c^2 d x^2)^{3/2}}{(a+b \arccos (c x))^2} \, dx\) [75]

Optimal result

Integrand size = 26, antiderivative size = 287 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=\frac {d \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2}}{b c (a+b \arccos (c x))}+\frac {d \sqrt {d-c^2 d x^2} \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c \sqrt {1-c^2 x^2}}-\frac {d \sqrt {d-c^2 d x^2} \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c \sqrt {1-c^2 x^2}}-\frac {d \sqrt {d-c^2 d x^2} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{b^2 c \sqrt {1-c^2 x^2}}+\frac {d \sqrt {d-c^2 d x^2} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{2 b^2 c \sqrt {1-c^2 x^2}} \] Output:

d*(-c^2*x^2+1)^(3/2)*(-c^2*d*x^2+d)^(1/2)/b/c/(a+b*arccos(c*x))+d*(-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) 
-1/2*d*(-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)-d*(-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)+1/2*d*(-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)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.80 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=-\frac {d \sqrt {1-c^2 x^2} \left (-1+c^2 x^2\right ) \sqrt {-d \left (-1+c^2 x^2\right )}}{b c (a+b \arccos (c x))}+\frac {d \sqrt {-d \left (1-c^2 x^2\right )^2} \sqrt {d-c^2 d x^2} \left (2 \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )-\operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )-2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )+\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{2 b^2 c \sqrt {1-c^2 x^2} \sqrt {\left (-1+c^2 x^2\right ) \left (d-c^2 d x^2\right )}} \] Input:

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

Output:

-((d*Sqrt[1 - c^2*x^2]*(-1 + c^2*x^2)*Sqrt[-(d*(-1 + c^2*x^2))])/(b*c*(a + 
 b*ArcCos[c*x]))) + (d*Sqrt[-(d*(1 - c^2*x^2)^2)]*Sqrt[d - c^2*d*x^2]*(2*C 
osIntegral[2*(a/b + ArcCos[c*x])]*Sin[(2*a)/b] - CosIntegral[4*(a/b + ArcC 
os[c*x])]*Sin[(4*a)/b] - 2*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcCos[c*x])] 
 + Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcCos[c*x])]))/(2*b^2*c*Sqrt[1 - c^2 
*x^2]*Sqrt[(-1 + c^2*x^2)*(d - c^2*d*x^2)])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.65, 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.

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

\(\Big \downarrow \) 5167

\(\displaystyle \frac {4 c d \sqrt {d-c^2 d x^2} \int \frac {x \left (1-c^2 x^2\right )}{a+b \arccos (c x)}dx}{b \sqrt {1-c^2 x^2}}+\frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 5225

\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}}{b c (a+b \arccos (c x))}-\frac {4 d \sqrt {d-c^2 d x^2} \int -\frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 d \sqrt {d-c^2 d x^2} \int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c \sqrt {1-c^2 x^2}}+\frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {4 d \sqrt {d-c^2 d x^2} \int \left (\frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arccos (c x))}{b}\right )}{4 (a+b \arccos (c x))}-\frac {\sin \left (\frac {4 a}{b}-\frac {4 (a+b \arccos (c x))}{b}\right )}{8 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b^2 c \sqrt {1-c^2 x^2}}+\frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}}{b c (a+b \arccos (c x))}-\frac {4 d \sqrt {d-c^2 d x^2} \left (-\frac {1}{4} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )+\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c \sqrt {1-c^2 x^2}}\)

Input:

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

Output:

(Sqrt[1 - c^2*x^2]*(d - c^2*d*x^2)^(3/2))/(b*c*(a + b*ArcCos[c*x])) - (4*d 
*Sqrt[d - c^2*d*x^2]*(-1/4*(CosIntegral[(2*(a + b*ArcCos[c*x]))/b]*Sin[(2* 
a)/b]) + (CosIntegral[(4*(a + b*ArcCos[c*x]))/b]*Sin[(4*a)/b])/8 + (Cos[(2 
*a)/b]*SinIntegral[(2*(a + b*ArcCos[c*x]))/b])/4 - (Cos[(4*a)/b]*SinIntegr 
al[(4*(a + b*ArcCos[c*x]))/b])/8))/(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 = 0.32 (sec) , antiderivative size = 501, normalized size of antiderivative = 1.75

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 (5 i \sin \left (3 \arccos \left (c x \right )\right ) b -16 b \,c^{5} x^{5}-16 i \sqrt {-c^{2} x^{2}+1}\, b \,c^{4} x^{4}+20 b \,c^{3} x^{3}-11 i \sqrt {-c^{2} x^{2}+1}\, b +8 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 )+8 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 +4 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 )-8 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 )+4 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 -4 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 )-8 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 -4 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 +12 i \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} x^{2}-7 c x b +3 \cos \left (3 \arccos \left (c x \right )\right ) b \right ) d}{16 c \left (c^{2} x^{2}-1\right ) b^{2} \left (a +b \arccos \left (c x \right )\right )}\)	\(501\)

Input:

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

Output:

1/16*(-d*(c^2*x^2-1))^(1/2)*(I*c^2*x^2+c*x*(-c^2*x^2+1)^(1/2)-I)*(5*I*sin( 
3*arccos(c*x))*b-16*b*c^5*x^5-16*I*(-c^2*x^2+1)^(1/2)*b*c^4*x^4+20*b*c^3*x 
^3-11*I*(-c^2*x^2+1)^(1/2)*b+8*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)+8*I*Ei(1,-2*I*arccos(c*x)-2*I*a/b)*exp( 
-I*(-b*arccos(c*x)+2*a)/b)*a+4*I*Ei(1,4*I*arccos(c*x)+4*I*a/b)*exp(I*(b*ar 
ccos(c*x)+4*a)/b)*b*arccos(c*x)-8*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)+4*I*Ei(1,4*I*arccos(c*x)+4*I*a/b)*exp(I 
*(b*arccos(c*x)+4*a)/b)*a-4*I*exp(-I*(-b*arccos(c*x)+4*a)/b)*Ei(1,-4*I*arc 
cos(c*x)-4*I*a/b)*b*arccos(c*x)-8*I*Ei(1,2*I*arccos(c*x)+2*I*a/b)*exp(I*(b 
*arccos(c*x)+2*a)/b)*a-4*I*exp(-I*(-b*arccos(c*x)+4*a)/b)*Ei(1,-4*I*arccos 
(c*x)-4*I*a/b)*a+12*I*(-c^2*x^2+1)^(1/2)*b*c^2*x^2-7*c*x*b+3*cos(3*arccos( 
c*x))*b)*d/c/(c^2*x^2-1)/b^2/(a+b*arccos(c*x))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((-d*(c*x - 1)*(c*x + 1))**(3/2)/(a + b*acos(c*x))**2, x)
 

Maxima [F]

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

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

Output:

(c^4*d*x^4 - 2*c^2*d*x^2 - (b^2*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c* 
x) + a*b*c)*integrate(4*(c^3*d*x^3 - c*d*x)/(b^2*arctan2(sqrt(c*x + 1)*sqr 
t(-c*x + 1), c*x) + a*b), x) + d)*sqrt(d)/(b^2*c*arctan2(sqrt(c*x + 1)*sqr 
t(-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. 926 vs. \(2 (263) = 526\).

Time = 0.92 (sec) , antiderivative size = 926, normalized size of antiderivative = 3.23 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2}}{(a+b \arccos (c x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(2*b*c^4*d^(3/2)*x^4/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 8*b*d^(3/2)*a 
rccos(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) + 8*b*d^(3/2)*arccos(c*x)*cos(a/b)^4*sin_integr 
al(4*a/b + 4*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 8*a*d^(3/2)* 
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) + 8*a*d^(3/2)*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) - 4*b*c^2*d^(3/2)*x^2/(b^3*c^2*arccos 
(c*x) + a*b^2*c^2) + 4*b*d^(3/2)*arccos(c*x)*cos(a/b)*cos_integral(4*a/b + 
 4*arccos(c*x))*sin(a/b)/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + 4*b*d^(3/2)*a 
rccos(c*x)*cos(a/b)*cos_integral(2*a/b + 2*arccos(c*x))*sin(a/b)/(b^3*c^2* 
arccos(c*x) + a*b^2*c^2) - 8*b*d^(3/2)*arccos(c*x)*cos(a/b)^2*sin_integral 
(4*a/b + 4*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 4*b*d^(3/2)*ar 
ccos(c*x)*cos(a/b)^2*sin_integral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c 
*x) + a*b^2*c^2) + 4*a*d^(3/2)*cos(a/b)*cos_integral(4*a/b + 4*arccos(c*x) 
)*sin(a/b)/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + 4*a*d^(3/2)*cos(a/b)*cos_in 
tegral(2*a/b + 2*arccos(c*x))*sin(a/b)/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 
 8*a*d^(3/2)*cos(a/b)^2*sin_integral(4*a/b + 4*arccos(c*x))/(b^3*c^2*arcco 
s(c*x) + a*b^2*c^2) - 4*a*d^(3/2)*cos(a/b)^2*sin_integral(2*a/b + 2*arccos 
(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + b*d^(3/2)*arccos(c*x)*sin_integ 
ral(4*a/b + 4*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + 2*b*d^(3...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(d)*d*(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**2)/(acos(c*x)**2*b**2 + 2*aco 
s(c*x)*a*b + a**2),x)*c**2)