\(\int \frac {\sqrt {d-c^2 d x^2}}{a+b \arccos (c x)} \, dx\) [70]

Optimal result

Integrand size = 26, antiderivative size = 169 \[ \int \frac {\sqrt {d-c^2 d x^2}}{a+b \arccos (c x)} \, dx=\frac {\sqrt {d-c^2 d x^2} \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{2 b c \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \log (a+b \arccos (c x))}{2 b c \sqrt {1-c^2 x^2}}+\frac {\sqrt {d-c^2 d x^2} \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{2 b c \sqrt {1-c^2 x^2}} \] Output:

1/2*(-c^2*d*x^2+d)^(1/2)*cos(2*a/b)*Ci(2*(a+b*arccos(c*x))/b)/b/c/(-c^2*x^ 
2+1)^(1/2)-1/2*(-c^2*d*x^2+d)^(1/2)*ln(a+b*arccos(c*x))/b/c/(-c^2*x^2+1)^( 
1/2)+1/2*(-c^2*d*x^2+d)^(1/2)*sin(2*a/b)*Si(2*(a+b*arccos(c*x))/b)/b/c/(-c 
^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {d-c^2 d x^2}}{a+b \arccos (c x)} \, dx=\frac {\sqrt {d \left (1-c^2 x^2\right )} \left (\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\log (a+b \arccos (c x))+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{2 b c \sqrt {1-c^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5169, 3042, 3793, 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.

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

\(\Big \downarrow \) 5169

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 5169

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[(-(b*c)^(-1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Subst[ 
Int[x^n*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, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (2 i \sqrt {-c^{2} x^{2}+1}\, \ln \left (a +b \arccos \left (c x \right )\right )+2 \ln \left (a +b \arccos \left (c x \right )\right ) c x +\operatorname {expIntegral}_{1}\left (2 i \arccos \left (c x \right )+\frac {2 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+2 a \right )}{b}}+\operatorname {expIntegral}_{1}\left (-2 i \arccos \left (c x \right )-\frac {2 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+2 a \right )}{b}}\right )}{4 c \left (c^{2} x^{2}-1\right ) b}\)	\(170\)

Input:

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

Output:

1/4*(-d*(c^2*x^2-1))^(1/2)*(I*c^2*x^2+c*x*(-c^2*x^2+1)^(1/2)-I)*(2*I*(-c^2 
*x^2+1)^(1/2)*ln(a+b*arccos(c*x))+2*ln(a+b*arccos(c*x))*c*x+Ei(1,2*I*arcco 
s(c*x)+2*I*a/b)*exp(I*(b*arccos(c*x)+2*a)/b)+Ei(1,-2*I*arccos(c*x)-2*I*a/b 
)*exp(-I*(-b*arccos(c*x)+2*a)/b))/c/(c^2*x^2-1)/b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {d-c^2 d x^2}}{a+b \arccos (c x)} \, dx=\frac {1}{2} \, c \sqrt {d} {\left (\frac {2 \, \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac {2 \, \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} - \frac {\operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} - \frac {\log \left (b \arccos \left (c x\right ) + a\right )}{b c^{2}}\right )} \] Input:

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

Output:

1/2*c*sqrt(d)*(2*cos(a/b)^2*cos_integral(2*a/b + 2*arccos(c*x))/(b*c^2) + 
2*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arccos(c*x))/(b*c^2) - cos_inte 
gral(2*a/b + 2*arccos(c*x))/(b*c^2) - log(b*arccos(c*x) + a)/(b*c^2))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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