\(\int (d-c^2 d x^2)^{3/2} (a+b \arccos (c x)) \, dx\) [52]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 185 \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=\frac {3 b c d x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {b d \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2}}{16 c}+\frac {3}{8} d x \sqrt {d-c^2 d x^2} (a+b \arccos (c x))+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x))-\frac {3 d \sqrt {d-c^2 d x^2} (a+b \arccos (c x))^2}{16 b c \sqrt {1-c^2 x^2}} \] Output:

3/16*b*c*d*x^2*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1/16*b*d*(-c^2*x^2+ 
1)^(3/2)*(-c^2*d*x^2+d)^(1/2)/c+3/8*d*x*(-c^2*d*x^2+d)^(1/2)*(a+b*arccos(c 
*x))+1/4*x*(-c^2*d*x^2+d)^(3/2)*(a+b*arccos(c*x))-3/16*d*(-c^2*d*x^2+d)^(1 
/2)*(a+b*arccos(c*x))^2/b/c/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.14 \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=\frac {-24 b d \sqrt {d-c^2 d x^2} \arccos (c x)^2-48 a d^{3/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+d \sqrt {d-c^2 d x^2} \left (16 a c x \left (5-2 c^2 x^2\right ) \sqrt {1-c^2 x^2}+16 b \cos (2 \arccos (c x))-b \cos (4 \arccos (c x))\right )-4 b d \sqrt {d-c^2 d x^2} \arccos (c x) (-8 \sin (2 \arccos (c x))+\sin (4 \arccos (c x)))}{128 c \sqrt {1-c^2 x^2}} \] Input:

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

Output:

(-24*b*d*Sqrt[d - c^2*d*x^2]*ArcCos[c*x]^2 - 48*a*d^(3/2)*Sqrt[1 - c^2*x^2 
]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + d*Sqrt[d - 
c^2*d*x^2]*(16*a*c*x*(5 - 2*c^2*x^2)*Sqrt[1 - c^2*x^2] + 16*b*Cos[2*ArcCos 
[c*x]] - b*Cos[4*ArcCos[c*x]]) - 4*b*d*Sqrt[d - c^2*d*x^2]*ArcCos[c*x]*(-8 
*Sin[2*ArcCos[c*x]] + Sin[4*ArcCos[c*x]]))/(128*c*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5159, 244, 2009, 5157, 15, 5153}

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.

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

\(\Big \downarrow \) 5159

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5157

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5153

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

Input:

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

Output:

(b*c*d*Sqrt[d - c^2*d*x^2]*(x^2/2 - (c^2*x^4)/4))/(4*Sqrt[1 - c^2*x^2]) + 
(x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcCos[c*x]))/4 + (3*d*((b*c*x^2*Sqrt[d - 
c^2*d*x^2])/(4*Sqrt[1 - c^2*x^2]) + (x*Sqrt[d - c^2*d*x^2]*(a + b*ArcCos[c 
*x]))/2 - (Sqrt[d - c^2*d*x^2]*(a + b*ArcCos[c*x])^2)/(4*b*c*Sqrt[1 - c^2* 
x^2])))/4
 

Defintions of rubi rules used

rule 15
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 244
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5153
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[(-(b*c*(n + 1))^(-1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2] 
]*(a + b*ArcCos[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^ 
2*d + e, 0] && NeQ[n, -1]
 

rule 5157
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcCos[c*x])^n/2), x] + (Simp[(1/2 
)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]]   Int[(a + b*ArcCos[c*x])^n/Sqrt[ 
1 - c^2*x^2], x], x] + Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2 
]]   Int[x*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x 
] && EqQ[c^2*d + e, 0] && GtQ[n, 0]
 

rule 5159
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[x*(d + e*x^2)^p*((a + b*ArcCos[c*x])^n/(2*p + 1)), x] + (S 
imp[2*d*(p/(2*p + 1))   Int[(d + e*x^2)^(p - 1)*(a + b*ArcCos[c*x])^n, x], 
x] + Simp[b*c*(n/(2*p + 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}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 479, normalized size of antiderivative = 2.59

method result size
default \(\frac {a x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{4}+\frac {3 a d x \sqrt {-c^{2} d \,x^{2}+d}}{8}+\frac {3 a \,d^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 \sqrt {c^{2} d}}+b \left (\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2} d}{16 \left (c^{2} x^{2}-1\right ) c}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (8 c^{5} x^{5}-12 c^{3} x^{3}+8 i \sqrt {-c^{2} x^{2}+1}\, x^{4} c^{4}+4 c x -8 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+i \sqrt {-c^{2} x^{2}+1}\right ) \left (i+4 \arccos \left (c x \right )\right ) d}{256 \left (c^{2} x^{2}-1\right ) c}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+2 c^{3} x^{3}+i \sqrt {-c^{2} x^{2}+1}-2 c x \right ) \left (-i+2 \arccos \left (c x \right )\right ) d}{16 \left (c^{2} x^{2}-1\right ) c}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (5 i+12 \arccos \left (c x \right )\right ) \cos \left (3 \arccos \left (c x \right )\right ) d}{256 \left (c^{2} x^{2}-1\right ) c}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (17 i+28 \arccos \left (c x \right )\right ) \sin \left (3 \arccos \left (c x \right )\right ) d}{256 \left (c^{2} x^{2}-1\right ) c}\right )\) \(479\)
parts \(\frac {a x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{4}+\frac {3 a d x \sqrt {-c^{2} d \,x^{2}+d}}{8}+\frac {3 a \,d^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 \sqrt {c^{2} d}}+b \left (\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2} d}{16 \left (c^{2} x^{2}-1\right ) c}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (8 c^{5} x^{5}-12 c^{3} x^{3}+8 i \sqrt {-c^{2} x^{2}+1}\, x^{4} c^{4}+4 c x -8 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+i \sqrt {-c^{2} x^{2}+1}\right ) \left (i+4 \arccos \left (c x \right )\right ) d}{256 \left (c^{2} x^{2}-1\right ) c}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+2 c^{3} x^{3}+i \sqrt {-c^{2} x^{2}+1}-2 c x \right ) \left (-i+2 \arccos \left (c x \right )\right ) d}{16 \left (c^{2} x^{2}-1\right ) c}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (5 i+12 \arccos \left (c x \right )\right ) \cos \left (3 \arccos \left (c x \right )\right ) d}{256 \left (c^{2} x^{2}-1\right ) c}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (17 i+28 \arccos \left (c x \right )\right ) \sin \left (3 \arccos \left (c x \right )\right ) d}{256 \left (c^{2} x^{2}-1\right ) c}\right )\) \(479\)

Input:

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

Output:

1/4*a*x*(-c^2*d*x^2+d)^(3/2)+3/8*a*d*x*(-c^2*d*x^2+d)^(1/2)+3/8*a*d^2/(c^2 
*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+b*(3/16*(-d*(c^2*x^ 
2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)/c*arccos(c*x)^2*d-1/256*(-d*(c^ 
2*x^2-1))^(1/2)*(8*c^5*x^5-12*c^3*x^3+8*I*(-c^2*x^2+1)^(1/2)*x^4*c^4+4*c*x 
-8*I*(-c^2*x^2+1)^(1/2)*x^2*c^2+I*(-c^2*x^2+1)^(1/2))*(I+4*arccos(c*x))*d/ 
(c^2*x^2-1)/c+1/16*(-d*(c^2*x^2-1))^(1/2)*(-2*I*(-c^2*x^2+1)^(1/2)*x^2*c^2 
+2*c^3*x^3+I*(-c^2*x^2+1)^(1/2)-2*c*x)*(-I+2*arccos(c*x))*d/(c^2*x^2-1)/c- 
3/256*(-d*(c^2*x^2-1))^(1/2)*(-I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(5*I+12 
*arccos(c*x))*cos(3*arccos(c*x))*d/(c^2*x^2-1)/c-1/256*(-d*(c^2*x^2-1))^(1 
/2)*(I*c^2*x^2+c*x*(-c^2*x^2+1)^(1/2)-I)*(17*I+28*arccos(c*x))*sin(3*arcco 
s(c*x))*d/(c^2*x^2-1)/c)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

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

Output:

b*sqrt(d)*integrate(-(c^2*d*x^2 - d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2( 
sqrt(c*x + 1)*sqrt(-c*x + 1), c*x), x) + 1/8*(2*(-c^2*d*x^2 + d)^(3/2)*x + 
 3*sqrt(-c^2*d*x^2 + d)*d*x + 3*d^(3/2)*arcsin(c*x)/c)*a
 

Giac [F(-2)]

Exception generated. \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=\frac {\sqrt {d}\, d \left (3 \mathit {asin} \left (c x \right ) a -2 \sqrt {-c^{2} x^{2}+1}\, a \,c^{3} x^{3}+5 \sqrt {-c^{2} x^{2}+1}\, a c x -8 \left (\int \sqrt {-c^{2} x^{2}+1}\, \mathit {acos} \left (c x \right ) x^{2}d x \right ) b \,c^{3}+8 \left (\int \sqrt {-c^{2} x^{2}+1}\, \mathit {acos} \left (c x \right )d x \right ) b c \right )}{8 c} \] Input:

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

Output:

(sqrt(d)*d*(3*asin(c*x)*a - 2*sqrt( - c**2*x**2 + 1)*a*c**3*x**3 + 5*sqrt( 
 - c**2*x**2 + 1)*a*c*x - 8*int(sqrt( - c**2*x**2 + 1)*acos(c*x)*x**2,x)*b 
*c**3 + 8*int(sqrt( - c**2*x**2 + 1)*acos(c*x),x)*b*c))/(8*c)