\(\int \frac {\sqrt {d-c^2 d x^2}}{(a+b \arccos (c x))^2} \, dx\) [76]

Optimal result

Integrand size = 26, antiderivative size = 162 \[ \int \frac {\sqrt {d-c^2 d x^2}}{(a+b \arccos (c x))^2} \, dx=\frac {\sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}{b c (a+b \arccos (c x))}+\frac {\sqrt {d-c^2 d x^2} \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{b^2 c \sqrt {1-c^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {d-c^2 d x^2}}{(a+b \arccos (c x))^2} \, dx=-\frac {\sqrt {d-c^2 d x^2} \left (b \left (-1+c^2 x^2\right )-(a+b \arccos (c x)) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+(a+b \arccos (c x)) \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{b^2 c \sqrt {1-c^2 x^2} (a+b \arccos (c x))} \] Input:

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

Output:

-((Sqrt[d - c^2*d*x^2]*(b*(-1 + c^2*x^2) - (a + b*ArcCos[c*x])*CosIntegral 
[2*(a/b + ArcCos[c*x])]*Sin[(2*a)/b] + (a + b*ArcCos[c*x])*Cos[(2*a)/b]*Si 
nIntegral[2*(a/b + ArcCos[c*x])]))/(b^2*c*Sqrt[1 - c^2*x^2]*(a + b*ArcCos[ 
c*x])))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.80, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {5167, 5147, 25, 4906, 27, 3042, 3784, 25, 3042, 3780, 3783}

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.

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

\(\Big \downarrow \) 5167

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

\(\Big \downarrow \) 5147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

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

Output:

(Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2])/(b*c*(a + b*ArcCos[c*x])) - (Sqrt[ 
d - c^2*d*x^2]*(-(CosIntegral[(2*(a + b*ArcCos[c*x]))/b]*Sin[(2*a)/b]) + C 
os[(2*a)/b]*SinIntegral[(2*(a + b*ArcCos[c*x]))/b]))/(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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* 
e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* 
f)/d]   Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] 
&& NeQ[d*e - c*f, 0]
 

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 5147

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.72

method	result	size
default	\(-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (-2 i b \,c^{3} x^{3}+2 x^{2} c^{2} \sqrt {-c^{2} x^{2}+1}\, b +\operatorname {expIntegral}_{1}\left (-2 i \arccos \left (c x \right )-\frac {2 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+2 a \right )}{b}} b \arccos \left (c x \right )-\operatorname {expIntegral}_{1}\left (2 i \arccos \left (c x \right )+\frac {2 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+2 a \right )}{b}} b \arccos \left (c x \right )+2 i c x b +\operatorname {expIntegral}_{1}\left (-2 i \arccos \left (c x \right )-\frac {2 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+2 a \right )}{b}} a -\operatorname {expIntegral}_{1}\left (2 i \arccos \left (c x \right )+\frac {2 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+2 a \right )}{b}} a -2 \sqrt {-c^{2} x^{2}+1}\, b \right )}{2 c \left (c^{2} x^{2}-1\right ) b^{2} \left (a +b \arccos \left (c x \right )\right )}\)	\(279\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-(c^2*x^2 - 2*(b^2*c^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c^ 
2)*integrate(x/(b^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b), x) 
- 1)*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. 348 vs. \(2 (150) = 300\).

Time = 0.68 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {d-c^2 d x^2}}{(a+b \arccos (c x))^2} \, dx=-{\left (\frac {b c^{2} x^{2}}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {2 \, b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac {2 \, b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {2 \, a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac {2 \, a \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {b \arccos \left (c x\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {a \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac {b}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}}\right )} c \sqrt {d} \] Input:

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

Output:

-(b*c^2*x^2/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 2*b*arccos(c*x)*cos(a/b)*c 
os_integral(2*a/b + 2*arccos(c*x))*sin(a/b)/(b^3*c^2*arccos(c*x) + a*b^2*c 
^2) + 2*b*arccos(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) - 2*a*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) + 2*a*cos(a/b)^2*sin_inte 
gral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - b*arccos(c 
*x)*sin_integral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) 
- a*sin_integral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) 
- b/(b^3*c^2*arccos(c*x) + a*b^2*c^2))*c*sqrt(d)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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