\(\int (\pi -c^2 \pi x^2)^{3/2} (a+b \arccos (c x)) \, dx\) [38]

Optimal result

Integrand size = 24, antiderivative size = 122 \[ \int \left (\pi -c^2 \pi x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=\frac {3}{16} b c \pi ^{3/2} x^2-\frac {b \pi ^{3/2} \left (1-c^2 x^2\right )^2}{16 c}+\frac {3}{8} \pi x \sqrt {\pi -c^2 \pi x^2} (a+b \arccos (c x))+\frac {1}{4} x \left (\pi -c^2 \pi x^2\right )^{3/2} (a+b \arccos (c x))-\frac {3 \pi ^{3/2} (a+b \arccos (c x))^2}{16 b c} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.93 \[ \int \left (\pi -c^2 \pi x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=-\frac {\pi ^{3/2} \left (-80 a c x \sqrt {1-c^2 x^2}+32 a c^3 x^3 \sqrt {1-c^2 x^2}+24 b \arccos (c x)^2-48 a \arcsin (c x)-16 b \cos (2 \arccos (c x))+b \cos (4 \arccos (c x))+4 b \arccos (c x) (-8 \sin (2 \arccos (c x))+\sin (4 \arccos (c x)))\right )}{128 c} \] Input:

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

Output:

-1/128*(Pi^(3/2)*(-80*a*c*x*Sqrt[1 - c^2*x^2] + 32*a*c^3*x^3*Sqrt[1 - c^2* 
x^2] + 24*b*ArcCos[c*x]^2 - 48*a*ArcSin[c*x] - 16*b*Cos[2*ArcCos[c*x]] + b 
*Cos[4*ArcCos[c*x]] + 4*b*ArcCos[c*x]*(-8*Sin[2*ArcCos[c*x]] + Sin[4*ArcCo 
s[c*x]])))/c
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07, 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.

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

\(\Big \downarrow \) 5159

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5157

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5153

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

Input:

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

Output:

(b*c*Pi^(3/2)*(x^2/2 - (c^2*x^4)/4))/4 + (x*(Pi - c^2*Pi*x^2)^(3/2)*(a + b 
*ArcCos[c*x]))/4 + (3*Pi*((b*c*Sqrt[Pi]*x^2)/4 + (x*Sqrt[Pi - c^2*Pi*x^2]* 
(a + b*ArcCos[c*x]))/2 - (Sqrt[Pi]*(a + b*ArcCos[c*x])^2)/(4*b*c)))/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 [A] (verified)

Time = 0.41 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {a x \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4}+\frac {3 a \pi x \sqrt {-\pi \,c^{2} x^{2}+\pi }}{8}+\frac {3 a \,\pi ^{2} \arctan \left (\frac {\sqrt {\pi \,c^{2}}\, x}{\sqrt {-\pi \,c^{2} x^{2}+\pi }}\right )}{8 \sqrt {\pi \,c^{2}}}-\frac {b \,\pi ^{\frac {3}{2}} \left (16 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{3} c^{3}+4 c^{4} x^{4}-40 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x c -20 c^{2} x^{2}+12 \arccos \left (c x \right )^{2}+25\right )}{64 c}\)	\(152\)
parts	\(\frac {a x \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4}+\frac {3 a \pi x \sqrt {-\pi \,c^{2} x^{2}+\pi }}{8}+\frac {3 a \,\pi ^{2} \arctan \left (\frac {\sqrt {\pi \,c^{2}}\, x}{\sqrt {-\pi \,c^{2} x^{2}+\pi }}\right )}{8 \sqrt {\pi \,c^{2}}}-\frac {b \,\pi ^{\frac {3}{2}} \left (16 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{3} c^{3}+4 c^{4} x^{4}-40 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x c -20 c^{2} x^{2}+12 \arccos \left (c x \right )^{2}+25\right )}{64 c}\)	\(152\)

Input:

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

Output:

1/4*a*x*(-Pi*c^2*x^2+Pi)^(3/2)+3/8*a*Pi*x*(-Pi*c^2*x^2+Pi)^(1/2)+3/8*a*Pi^ 
2/(Pi*c^2)^(1/2)*arctan((Pi*c^2)^(1/2)*x/(-Pi*c^2*x^2+Pi)^(1/2))-1/64*b*Pi 
^(3/2)*(16*(-c^2*x^2+1)^(1/2)*arccos(c*x)*x^3*c^3+4*c^4*x^4-40*(-c^2*x^2+1 
)^(1/2)*arccos(c*x)*x*c-20*c^2*x^2+12*arccos(c*x)^2+25)/c
 

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.38 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.56 \[ \int \left (\pi -c^2 \pi x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=\begin {cases} - \frac {\pi ^{\frac {3}{2}} a c^{2} x^{3} \sqrt {- c^{2} x^{2} + 1}}{4} + \frac {5 \pi ^{\frac {3}{2}} a x \sqrt {- c^{2} x^{2} + 1}}{8} - \frac {3 \pi ^{\frac {3}{2}} a \operatorname {acos}{\left (c x \right )}}{8 c} - \frac {\pi ^{\frac {3}{2}} b c^{3} x^{4}}{16} - \frac {\pi ^{\frac {3}{2}} b c^{2} x^{3} \sqrt {- c^{2} x^{2} + 1} \operatorname {acos}{\left (c x \right )}}{4} + \frac {5 \pi ^{\frac {3}{2}} b c x^{2}}{16} + \frac {5 \pi ^{\frac {3}{2}} b x \sqrt {- c^{2} x^{2} + 1} \operatorname {acos}{\left (c x \right )}}{8} - \frac {3 \pi ^{\frac {3}{2}} b \operatorname {acos}^{2}{\left (c x \right )}}{16 c} & \text {for}\: c \neq 0 \\\pi ^{\frac {3}{2}} x \left (a + \frac {\pi b}{2}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-pi**(3/2)*a*c**2*x**3*sqrt(-c**2*x**2 + 1)/4 + 5*pi**(3/2)*a*x 
*sqrt(-c**2*x**2 + 1)/8 - 3*pi**(3/2)*a*acos(c*x)/(8*c) - pi**(3/2)*b*c**3 
*x**4/16 - pi**(3/2)*b*c**2*x**3*sqrt(-c**2*x**2 + 1)*acos(c*x)/4 + 5*pi** 
(3/2)*b*c*x**2/16 + 5*pi**(3/2)*b*x*sqrt(-c**2*x**2 + 1)*acos(c*x)/8 - 3*p 
i**(3/2)*b*acos(c*x)**2/(16*c), Ne(c, 0)), (pi**(3/2)*x*(a + pi*b/2), True 
))
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(pi)*pi*(3*asin(c*x)*a - 2*sqrt( - c**2*x**2 + 1)*a*c**3*x**3 + 5*sqr 
t( - 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)