\(\int (\pi -c^2 \pi x^2)^{5/2} (a+b \arcsin (c x)) \, dx\) [37]

Optimal result

Integrand size = 24, antiderivative size = 178 \[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=-\frac {5}{32} b c \pi ^{5/2} x^2+\frac {5 b \pi ^{5/2} \left (1-c^2 x^2\right )^2}{96 c}+\frac {b \pi ^{5/2} \left (1-c^2 x^2\right )^3}{36 c}+\frac {5}{16} \pi ^2 x \sqrt {\pi -c^2 \pi x^2} (a+b \arcsin (c x))+\frac {5}{24} \pi x \left (\pi -c^2 \pi x^2\right )^{3/2} (a+b \arcsin (c x))+\frac {1}{6} x \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {5 \pi ^{5/2} (a+b \arcsin (c x))^2}{32 b c} \] Output:

-5/32*b*c*Pi^(5/2)*x^2+5/96*b*Pi^(5/2)*(-c^2*x^2+1)^2/c+1/36*b*Pi^(5/2)*(- 
c^2*x^2+1)^3/c+5/16*Pi^2*x*(-Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsin(c*x))+5/24*P 
i*x*(-Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsin(c*x))+1/6*x*(-Pi*c^2*x^2+Pi)^(5/2)* 
(a+b*arcsin(c*x))+5/32*Pi^(5/2)*(a+b*arcsin(c*x))^2/b/c
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.88 \[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\frac {\pi ^{5/2} \left (1584 a c x \sqrt {1-c^2 x^2}-1248 a c^3 x^3 \sqrt {1-c^2 x^2}+384 a c^5 x^5 \sqrt {1-c^2 x^2}+360 b \arcsin (c x)^2+270 b \cos (2 \arcsin (c x))+27 b \cos (4 \arcsin (c x))+2 b \cos (6 \arcsin (c x))+12 \arcsin (c x) (60 a+45 b \sin (2 \arcsin (c x))+9 b \sin (4 \arcsin (c x))+b \sin (6 \arcsin (c x)))\right )}{2304 c} \] Input:

Integrate[(Pi - c^2*Pi*x^2)^(5/2)*(a + b*ArcSin[c*x]),x]
 

Output:

(Pi^(5/2)*(1584*a*c*x*Sqrt[1 - c^2*x^2] - 1248*a*c^3*x^3*Sqrt[1 - c^2*x^2] 
 + 384*a*c^5*x^5*Sqrt[1 - c^2*x^2] + 360*b*ArcSin[c*x]^2 + 270*b*Cos[2*Arc 
Sin[c*x]] + 27*b*Cos[4*ArcSin[c*x]] + 2*b*Cos[6*ArcSin[c*x]] + 12*ArcSin[c 
*x]*(60*a + 45*b*Sin[2*ArcSin[c*x]] + 9*b*Sin[4*ArcSin[c*x]] + b*Sin[6*Arc 
Sin[c*x]])))/(2304*c)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5158, 241, 5158, 244, 2009, 5156, 15, 5152}

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.

\(\displaystyle \int \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx\)

\(\Big \downarrow \) 5158

\(\displaystyle \frac {5}{6} \pi \int \left (\pi -c^2 \pi x^2\right )^{3/2} (a+b \arcsin (c x))dx-\frac {1}{6} \pi ^{5/2} b c \int x \left (1-c^2 x^2\right )^2dx+\frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))\)

\(\Big \downarrow \) 241

\(\displaystyle \frac {5}{6} \pi \int \left (\pi -c^2 \pi x^2\right )^{3/2} (a+b \arcsin (c x))dx+\frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {\pi ^{5/2} b \left (1-c^2 x^2\right )^3}{36 c}\)

\(\Big \downarrow \) 5158

\(\displaystyle \frac {5}{6} \pi \left (\frac {3}{4} \pi \int \sqrt {\pi -c^2 \pi x^2} (a+b \arcsin (c x))dx-\frac {1}{4} \pi ^{3/2} b c \int x \left (1-c^2 x^2\right )dx+\frac {1}{4} x \left (\pi -\pi c^2 x^2\right )^{3/2} (a+b \arcsin (c x))\right )+\frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {\pi ^{5/2} b \left (1-c^2 x^2\right )^3}{36 c}\)

\(\Big \downarrow \) 244

\(\displaystyle \frac {5}{6} \pi \left (\frac {3}{4} \pi \int \sqrt {\pi -c^2 \pi x^2} (a+b \arcsin (c x))dx-\frac {1}{4} \pi ^{3/2} b c \int \left (x-c^2 x^3\right )dx+\frac {1}{4} x \left (\pi -\pi c^2 x^2\right )^{3/2} (a+b \arcsin (c x))\right )+\frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {\pi ^{5/2} b \left (1-c^2 x^2\right )^3}{36 c}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5}{6} \pi \left (\frac {3}{4} \pi \int \sqrt {\pi -c^2 \pi x^2} (a+b \arcsin (c x))dx+\frac {1}{4} x \left (\pi -\pi c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} \pi ^{3/2} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {\pi ^{5/2} b \left (1-c^2 x^2\right )^3}{36 c}\)

\(\Big \downarrow \) 5156

\(\displaystyle \frac {5}{6} \pi \left (\frac {3}{4} \pi \left (\frac {1}{2} \sqrt {\pi } \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx-\frac {1}{2} \sqrt {\pi } b c \int xdx+\frac {1}{2} x \sqrt {\pi -\pi c^2 x^2} (a+b \arcsin (c x))\right )+\frac {1}{4} x \left (\pi -\pi c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} \pi ^{3/2} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {\pi ^{5/2} b \left (1-c^2 x^2\right )^3}{36 c}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {5}{6} \pi \left (\frac {3}{4} \pi \left (\frac {1}{2} \sqrt {\pi } \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} x \sqrt {\pi -\pi c^2 x^2} (a+b \arcsin (c x))-\frac {1}{4} \sqrt {\pi } b c x^2\right )+\frac {1}{4} x \left (\pi -\pi c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} \pi ^{3/2} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {\pi ^{5/2} b \left (1-c^2 x^2\right )^3}{36 c}\)

\(\Big \downarrow \) 5152

\(\displaystyle \frac {1}{6} x \left (\pi -\pi c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {5}{6} \pi \left (\frac {1}{4} x \left (\pi -\pi c^2 x^2\right )^{3/2} (a+b \arcsin (c x))+\frac {3}{4} \pi \left (\frac {1}{2} x \sqrt {\pi -\pi c^2 x^2} (a+b \arcsin (c x))+\frac {\sqrt {\pi } (a+b \arcsin (c x))^2}{4 b c}-\frac {1}{4} \sqrt {\pi } b c x^2\right )-\frac {1}{4} \pi ^{3/2} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {\pi ^{5/2} b \left (1-c^2 x^2\right )^3}{36 c}\)

Input:

Int[(Pi - c^2*Pi*x^2)^(5/2)*(a + b*ArcSin[c*x]),x]
 

Output:

(b*Pi^(5/2)*(1 - c^2*x^2)^3)/(36*c) + (x*(Pi - c^2*Pi*x^2)^(5/2)*(a + b*Ar 
cSin[c*x]))/6 + (5*Pi*(-1/4*(b*c*Pi^(3/2)*(x^2/2 - (c^2*x^4)/4)) + (x*(Pi 
- c^2*Pi*x^2)^(3/2)*(a + b*ArcSin[c*x]))/4 + (3*Pi*(-1/4*(b*c*Sqrt[Pi]*x^2 
) + (x*Sqrt[Pi - c^2*Pi*x^2]*(a + b*ArcSin[c*x]))/2 + (Sqrt[Pi]*(a + b*Arc 
Sin[c*x])^2)/(4*b*c)))/4))/6
 

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 241

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ 
(2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -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 5152

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

rule 5156

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

rule 5158

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[x*(d + e*x^2)^p*((a + b*ArcSin[c*x])^n/(2*p + 1)), x] + (S 
imp[2*d*(p/(2*p + 1))   Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[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*ArcSin[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 = 0.33 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {a x \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{6}+\frac {5 a \pi x \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{24}+\frac {5 a \,\pi ^{2} x \sqrt {-\pi \,c^{2} x^{2}+\pi }}{16}+\frac {5 a \,\pi ^{3} \arctan \left (\frac {\sqrt {\pi \,c^{2}}\, x}{\sqrt {-\pi \,c^{2} x^{2}+\pi }}\right )}{16 \sqrt {\pi \,c^{2}}}+\frac {b \,\pi ^{\frac {5}{2}} \left (48 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x^{5} c^{5}-8 c^{6} x^{6}-156 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x^{3} c^{3}+39 c^{4} x^{4}+198 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -99 c^{2} x^{2}+45 \arcsin \left (c x \right )^{2}+68\right )}{288 c}\)	\(204\)
parts	\(\frac {a x \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{6}+\frac {5 a \pi x \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{24}+\frac {5 a \,\pi ^{2} x \sqrt {-\pi \,c^{2} x^{2}+\pi }}{16}+\frac {5 a \,\pi ^{3} \arctan \left (\frac {\sqrt {\pi \,c^{2}}\, x}{\sqrt {-\pi \,c^{2} x^{2}+\pi }}\right )}{16 \sqrt {\pi \,c^{2}}}+\frac {b \,\pi ^{\frac {5}{2}} \left (48 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x^{5} c^{5}-8 c^{6} x^{6}-156 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x^{3} c^{3}+39 c^{4} x^{4}+198 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -99 c^{2} x^{2}+45 \arcsin \left (c x \right )^{2}+68\right )}{288 c}\)	\(204\)

Input:

int((-Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)
 

Output:

1/6*a*x*(-Pi*c^2*x^2+Pi)^(5/2)+5/24*a*Pi*x*(-Pi*c^2*x^2+Pi)^(3/2)+5/16*a*P 
i^2*x*(-Pi*c^2*x^2+Pi)^(1/2)+5/16*a*Pi^3/(Pi*c^2)^(1/2)*arctan((Pi*c^2)^(1 
/2)*x/(-Pi*c^2*x^2+Pi)^(1/2))+1/288*b*Pi^(5/2)*(48*(-c^2*x^2+1)^(1/2)*arcs 
in(c*x)*x^5*c^5-8*c^6*x^6-156*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*x^3*c^3+39*c^ 
4*x^4+198*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c*x-99*c^2*x^2+45*arcsin(c*x)^2+6 
8)/c
 

Fricas [F]

\[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\int { {\left (\pi - \pi c^{2} x^{2}\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \] Input:

integrate((-pi*c^2*x^2+pi)^(5/2)*(a+b*arcsin(c*x)),x, algorithm="fricas")
 

Output:

integral(sqrt(pi - pi*c^2*x^2)*(pi^2*a*c^4*x^4 - 2*pi^2*a*c^2*x^2 + pi^2*a 
 + (pi^2*b*c^4*x^4 - 2*pi^2*b*c^2*x^2 + pi^2*b)*arcsin(c*x)), x)
 

Sympy [A] (verification not implemented)

Time = 17.82 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.49 \[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {\pi ^{\frac {5}{2}} a c^{4} x^{5} \sqrt {- c^{2} x^{2} + 1}}{6} - \frac {13 \pi ^{\frac {5}{2}} a c^{2} x^{3} \sqrt {- c^{2} x^{2} + 1}}{24} + \frac {11 \pi ^{\frac {5}{2}} a x \sqrt {- c^{2} x^{2} + 1}}{16} + \frac {5 \pi ^{\frac {5}{2}} a \operatorname {asin}{\left (c x \right )}}{16 c} - \frac {\pi ^{\frac {5}{2}} b c^{5} x^{6}}{36} + \frac {\pi ^{\frac {5}{2}} b c^{4} x^{5} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{6} + \frac {13 \pi ^{\frac {5}{2}} b c^{3} x^{4}}{96} - \frac {13 \pi ^{\frac {5}{2}} b c^{2} x^{3} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{24} - \frac {11 \pi ^{\frac {5}{2}} b c x^{2}}{32} + \frac {11 \pi ^{\frac {5}{2}} b x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{16} + \frac {5 \pi ^{\frac {5}{2}} b \operatorname {asin}^{2}{\left (c x \right )}}{32 c} & \text {for}\: c \neq 0 \\\pi ^{\frac {5}{2}} a x & \text {otherwise} \end {cases} \] Input:

integrate((-pi*c**2*x**2+pi)**(5/2)*(a+b*asin(c*x)),x)
 

Output:

Piecewise((pi**(5/2)*a*c**4*x**5*sqrt(-c**2*x**2 + 1)/6 - 13*pi**(5/2)*a*c 
**2*x**3*sqrt(-c**2*x**2 + 1)/24 + 11*pi**(5/2)*a*x*sqrt(-c**2*x**2 + 1)/1 
6 + 5*pi**(5/2)*a*asin(c*x)/(16*c) - pi**(5/2)*b*c**5*x**6/36 + pi**(5/2)* 
b*c**4*x**5*sqrt(-c**2*x**2 + 1)*asin(c*x)/6 + 13*pi**(5/2)*b*c**3*x**4/96 
 - 13*pi**(5/2)*b*c**2*x**3*sqrt(-c**2*x**2 + 1)*asin(c*x)/24 - 11*pi**(5/ 
2)*b*c*x**2/32 + 11*pi**(5/2)*b*x*sqrt(-c**2*x**2 + 1)*asin(c*x)/16 + 5*pi 
**(5/2)*b*asin(c*x)**2/(32*c), Ne(c, 0)), (pi**(5/2)*a*x, True))
 

Maxima [F]

\[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\int { {\left (\pi - \pi c^{2} x^{2}\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \] Input:

integrate((-pi*c^2*x^2+pi)^(5/2)*(a+b*arcsin(c*x)),x, algorithm="maxima")
 

Output:

sqrt(pi)*b*integrate((pi^2*c^4*x^4 - 2*pi^2*c^2*x^2 + pi^2)*sqrt(c*x + 1)* 
sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/48*(15*p 
i^2*sqrt(pi - pi*c^2*x^2)*x + 10*pi*(pi - pi*c^2*x^2)^(3/2)*x + 8*(pi - pi 
*c^2*x^2)^(5/2)*x + 15*pi^(5/2)*arcsin(c*x)/c)*a
 

Giac [F(-2)]

Exception generated. \[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\text {Exception raised: TypeError} \] Input:

integrate((-pi*c^2*x^2+pi)^(5/2)*(a+b*arcsin(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. \[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (\Pi -\Pi \,c^2\,x^2\right )}^{5/2} \,d x \] Input:

int((a + b*asin(c*x))*(Pi - Pi*c^2*x^2)^(5/2),x)
 

Output:

int((a + b*asin(c*x))*(Pi - Pi*c^2*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \left (\pi -c^2 \pi x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\frac {\sqrt {\pi }\, \pi ^{2} \left (15 \mathit {asin} \left (c x \right ) a +8 \sqrt {-c^{2} x^{2}+1}\, a \,c^{5} x^{5}-26 \sqrt {-c^{2} x^{2}+1}\, a \,c^{3} x^{3}+33 \sqrt {-c^{2} x^{2}+1}\, a c x +48 \left (\int \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right ) x^{4}d x \right ) b \,c^{5}-96 \left (\int \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right ) x^{2}d x \right ) b \,c^{3}+48 \left (\int \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right )d x \right ) b c \right )}{48 c} \] Input:

int((-Pi*c^2*x^2+Pi)^(5/2)*(a+b*asin(c*x)),x)
 

Output:

(sqrt(pi)*pi**2*(15*asin(c*x)*a + 8*sqrt( - c**2*x**2 + 1)*a*c**5*x**5 - 2 
6*sqrt( - c**2*x**2 + 1)*a*c**3*x**3 + 33*sqrt( - c**2*x**2 + 1)*a*c*x + 4 
8*int(sqrt( - c**2*x**2 + 1)*asin(c*x)*x**4,x)*b*c**5 - 96*int(sqrt( - c** 
2*x**2 + 1)*asin(c*x)*x**2,x)*b*c**3 + 48*int(sqrt( - c**2*x**2 + 1)*asin( 
c*x),x)*b*c))/(48*c)