\(\int \frac {(d-c^2 d x^2)^{3/2}}{a+b \arccos (c x)} \, dx\) [69]

Optimal result

Integrand size = 26, antiderivative size = 294 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2}}{a+b \arccos (c x)} \, dx=\frac {d \sqrt {d-c^2 d x^2} \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{2 b c \sqrt {1-c^2 x^2}}-\frac {d \sqrt {d-c^2 d x^2} \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{8 b c \sqrt {1-c^2 x^2}}-\frac {3 d \sqrt {d-c^2 d x^2} \log (a+b \arccos (c x))}{8 b c \sqrt {1-c^2 x^2}}+\frac {d \sqrt {d-c^2 d x^2} \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{2 b c \sqrt {1-c^2 x^2}}-\frac {d \sqrt {d-c^2 d x^2} \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{8 b c \sqrt {1-c^2 x^2}} \] Output:

1/2*d*(-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)-1/8*d*(-c^2*d*x^2+d)^(1/2)*cos(4*a/b)*Ci(4*(a+b*arccos(c*x))/ 
b)/b/c/(-c^2*x^2+1)^(1/2)-3/8*d*(-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*d*(-c^2*d*x^2+d)^(1/2)*sin(2*a/b)*Si(2*(a+b*arcc 
os(c*x))/b)/b/c/(-c^2*x^2+1)^(1/2)-1/8*d*(-c^2*d*x^2+d)^(1/2)*sin(4*a/b)*S 
i(4*(a+b*arccos(c*x))/b)/b/c/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.51 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2}}{a+b \arccos (c x)} \, dx=\frac {d \sqrt {d-c^2 d x^2} \left (4 \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )-4 \log (a+b \arccos (c x))+\log (8 (a+b \arccos (c x)))+4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{8 b c \sqrt {1-c^2 x^2}} \] Input:

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

Output:

(d*Sqrt[d - c^2*d*x^2]*(4*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcCos[c*x])] 
- Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcCos[c*x])] - 4*Log[a + b*ArcCos[c*x 
]] + Log[8*(a + b*ArcCos[c*x])] + 4*Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcC 
os[c*x])] - Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcCos[c*x])]))/(8*b*c*Sqrt[ 
1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.52, 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.

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

\(\Big \downarrow \) 5169

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((d*Sqrt[d - c^2*d*x^2]*(-1/2*(Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcCos[ 
c*x]))/b]) + (Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcCos[c*x]))/b])/8 + (3* 
Log[a + b*ArcCos[c*x]])/8 - (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcCos[c*x 
]))/b])/2 + (Sin[(4*a)/b]*SinIntegral[(4*(a + b*ArcCos[c*x]))/b])/8))/(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 = 0.22 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (-6 i \sqrt {-c^{2} x^{2}+1}\, \ln \left (a +b \arccos \left (c x \right )\right )-6 \ln \left (a +b \arccos \left (c x \right )\right ) c x +\operatorname {expIntegral}_{1}\left (4 i \arccos \left (c x \right )+\frac {4 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+4 a \right )}{b}}+\operatorname {expIntegral}_{1}\left (-4 i \arccos \left (c x \right )-\frac {4 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+4 a \right )}{b}}-4 \,\operatorname {expIntegral}_{1}\left (2 i \arccos \left (c x \right )+\frac {2 i a}{b}\right ) {\mathrm e}^{\frac {i \left (b \arccos \left (c x \right )+2 a \right )}{b}}-4 \,\operatorname {expIntegral}_{1}\left (-2 i \arccos \left (c x \right )-\frac {2 i a}{b}\right ) {\mathrm e}^{-\frac {i \left (-b \arccos \left (c x \right )+2 a \right )}{b}}\right ) d}{16 c \left (c^{2} x^{2}-1\right ) b}\)	\(244\)

Input:

int((-c^2*d*x^2+d)^(3/2)/(a+b*arccos(c*x)),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)*(-6*I*(- 
c^2*x^2+1)^(1/2)*ln(a+b*arccos(c*x))-6*ln(a+b*arccos(c*x))*c*x+Ei(1,4*I*ar 
ccos(c*x)+4*I*a/b)*exp(I*(b*arccos(c*x)+4*a)/b)+Ei(1,-4*I*arccos(c*x)-4*I* 
a/b)*exp(-I*(-b*arccos(c*x)+4*a)/b)-4*Ei(1,2*I*arccos(c*x)+2*I*a/b)*exp(I* 
(b*arccos(c*x)+2*a)/b)-4*Ei(1,-2*I*arccos(c*x)-2*I*a/b)*exp(-I*(-b*arccos( 
c*x)+2*a)/b))*d/c/(c^2*x^2-1)/b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.97 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2}}{a+b \arccos (c x)} \, dx=-\frac {1}{8} \, {\left (\frac {8 \, d^{\frac {3}{2}} \cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac {8 \, d^{\frac {3}{2}} \cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{2}} - \frac {8 \, d^{\frac {3}{2}} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{2}} - \frac {8 \, d^{\frac {3}{2}} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} - \frac {4 \, d^{\frac {3}{2}} \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{2}} - \frac {8 \, d^{\frac {3}{2}} \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac {d^{\frac {3}{2}} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac {4 \, d^{\frac {3}{2}} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac {3 \, d^{\frac {3}{2}} \log \left (b \arccos \left (c x\right ) + a\right )}{b c^{2}}\right )} c \] Input:

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

Output:

-1/8*(8*d^(3/2)*cos(a/b)^4*cos_integral(4*a/b + 4*arccos(c*x))/(b*c^2) + 8 
*d^(3/2)*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arccos(c*x))/(b*c^2) - 
 8*d^(3/2)*cos(a/b)^2*cos_integral(4*a/b + 4*arccos(c*x))/(b*c^2) - 8*d^(3 
/2)*cos(a/b)^2*cos_integral(2*a/b + 2*arccos(c*x))/(b*c^2) - 4*d^(3/2)*cos 
(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arccos(c*x))/(b*c^2) - 8*d^(3/2)*cos 
(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arccos(c*x))/(b*c^2) + d^(3/2)*cos_i 
ntegral(4*a/b + 4*arccos(c*x))/(b*c^2) + 4*d^(3/2)*cos_integral(2*a/b + 2* 
arccos(c*x))/(b*c^2) + 3*d^(3/2)*log(b*arccos(c*x) + a)/(b*c^2))*c
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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