\(\int \frac {x^4 (a+b \arcsin (c x))}{(d-c^2 d x^2)^{3/2}} \, dx\) [118]

Optimal result

Integrand size = 27, antiderivative size = 214 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=-\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \arcsin (c x))}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {3 x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 c^4 d^2}-\frac {3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{4 b c^5 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^5 d \sqrt {d-c^2 d x^2}} \] Output:

-1/4*b*x^2*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^(1/2)+x^3*(a+b*arcsin(c 
*x))/c^2/d/(-c^2*d*x^2+d)^(1/2)+3/2*x*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x 
))/c^4/d^2-3/4*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))^2/b/c^5/d/(-c^2*d*x^2+ 
d)^(1/2)+1/2*b*(-c^2*x^2+1)^(1/2)*ln(-c^2*x^2+1)/c^5/d/(-c^2*d*x^2+d)^(1/2 
)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.81 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {-4 a c \sqrt {d} x \left (-3+c^2 x^2\right )+12 a \sqrt {d-c^2 d x^2} \arctan \left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+b \sqrt {d} \left (8 c x \arcsin (c x)+\sqrt {1-c^2 x^2} \left (-6 \arcsin (c x)^2+\cos (2 \arcsin (c x))+4 \log \left (1-c^2 x^2\right )+2 \arcsin (c x) \sin (2 \arcsin (c x))\right )\right )}{8 c^5 d^{3/2} \sqrt {d-c^2 d x^2}} \] Input:

Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]
 

Output:

(-4*a*c*Sqrt[d]*x*(-3 + c^2*x^2) + 12*a*Sqrt[d - c^2*d*x^2]*ArcTan[(c*x*Sq 
rt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + b*Sqrt[d]*(8*c*x*ArcSin[c*x 
] + Sqrt[1 - c^2*x^2]*(-6*ArcSin[c*x]^2 + Cos[2*ArcSin[c*x]] + 4*Log[1 - c 
^2*x^2] + 2*ArcSin[c*x]*Sin[2*ArcSin[c*x]])))/(8*c^5*d^(3/2)*Sqrt[d - c^2* 
d*x^2])
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5206, 243, 49, 2009, 5210, 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.

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

\(\Big \downarrow \) 5206

\(\displaystyle -\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}}dx}{c^2 d}-\frac {b \sqrt {1-c^2 x^2} \int \frac {x^3}{1-c^2 x^2}dx}{c d \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \arcsin (c x))}{c^2 d \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 243

\(\displaystyle -\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}}dx}{c^2 d}-\frac {b \sqrt {1-c^2 x^2} \int \frac {x^2}{1-c^2 x^2}dx^2}{2 c d \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \arcsin (c x))}{c^2 d \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5210

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5152

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

Input:

Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]
 

Output:

(x^3*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) - (3*((b*x^2*Sqrt[1 
- c^2*x^2])/(4*c*Sqrt[d - c^2*d*x^2]) - (x*Sqrt[d - c^2*d*x^2]*(a + b*ArcS 
in[c*x]))/(2*c^2*d) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(4*b*c^3*S 
qrt[d - c^2*d*x^2])))/(c^2*d) - (b*Sqrt[1 - c^2*x^2]*(-(x^2/c^2) - Log[1 - 
 c^2*x^2]/c^4))/(2*c*d*Sqrt[d - c^2*d*x^2])
 

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 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 5206

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + 
 b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + (-Simp[f^2*((m - 1)/(2*e*(p + 1))) 
 Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Simp 
[b*f*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f*x)^(m - 
 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{ 
a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && IG 
tQ[m, 1]
 

rule 5210

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + 
 b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Simp[f^2*((m - 1)/(c^2*(m + 2*p 
 + 1)))   Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + S 
imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f* 
x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; 
FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m 
, 1] && NeQ[m + 2*p + 1, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.02

method	result	size
default	\(-\frac {a \,x^{3}}{2 c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {3 a x}{2 c^{4} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {3 a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{4} d \sqrt {c^{2} d}}+\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{4 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}}{16 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {9 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 d^{2} c^{4} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{16 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}\)	\(432\)
parts	\(-\frac {a \,x^{3}}{2 c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {3 a x}{2 c^{4} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {3 a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{4} d \sqrt {c^{2} d}}+\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{4 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}}{16 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {9 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 d^{2} c^{4} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{16 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}\)	\(432\)

Input:

int(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*a*x^3/c^2/d/(-c^2*d*x^2+d)^(1/2)+3/2*a/c^4*x/d/(-c^2*d*x^2+d)^(1/2)-3 
/2*a/c^4/d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+3/4* 
b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^5/(c^2*x^2-1)*arcsin(c*x 
)^2+I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^5/(c^2*x^2-1)*arcs 
in(c*x)-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^5/(c^2*x^2-1)*ln 
(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/16*b*(-d*(c^2*x^2-1))^(1/2)/d^2/c^5/(c^ 
2*x^2-1)*(-c^2*x^2+1)^(1/2)-9/8*b*(-d*(c^2*x^2-1))^(1/2)/d^2/c^4/(c^2*x^2- 
1)*arcsin(c*x)*x-1/16*b*(-d*(c^2*x^2-1))^(1/2)/d^2/c^5/(c^2*x^2-1)*cos(3*a 
rcsin(c*x))-1/8*b*(-d*(c^2*x^2-1))^(1/2)/d^2/c^5/(c^2*x^2-1)*arcsin(c*x)*s 
in(3*arcsin(c*x))
 

Fricas [F]

\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="fricas" 
)
 

Output:

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

Sympy [F]

\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {x^{4} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**4*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(3/2),x)
 

Output:

Integral(x**4*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima" 
)
 

Output:

-1/2*a*(x^3/(sqrt(-c^2*d*x^2 + d)*c^2*d) - 3*x/(sqrt(-c^2*d*x^2 + d)*c^4*d 
) + 3*arcsin(c*x)/(c^5*d^(3/2))) - b*integrate(x^4*arctan2(c*x, sqrt(c*x + 
 1)*sqrt(-c*x + 1))/((c^2*d*x^2 - d)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqr 
t(d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(3/2),x)
 

Output:

int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {-3 \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right )^{2} b -6 \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right ) a -2 \mathit {asin} \left (c x \right ) b \,c^{3} x^{3}+2 \mathit {asin} \left (c x \right ) b c x -4 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {asin} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}}d x \right ) b c -\sqrt {-c^{2} x^{2}+1}\, b \,c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}\, b -2 a \,c^{3} x^{3}+6 a c x}{4 \sqrt {d}\, \sqrt {-c^{2} x^{2}+1}\, c^{5} d} \] Input:

int(x^4*(a+b*asin(c*x))/(-c^2*d*x^2+d)^(3/2),x)
 

Output:

( - 3*sqrt( - c**2*x**2 + 1)*asin(c*x)**2*b - 6*sqrt( - c**2*x**2 + 1)*asi 
n(c*x)*a - 2*asin(c*x)*b*c**3*x**3 + 2*asin(c*x)*b*c*x - 4*sqrt( - c**2*x* 
*2 + 1)*int(asin(c*x)/(sqrt( - c**2*x**2 + 1)*c**2*x**2 - sqrt( - c**2*x** 
2 + 1)),x)*b*c - sqrt( - c**2*x**2 + 1)*b*c**2*x**2 + sqrt( - c**2*x**2 + 
1)*b - 2*a*c**3*x**3 + 6*a*c*x)/(4*sqrt(d)*sqrt( - c**2*x**2 + 1)*c**5*d)