Integrand size = 21, antiderivative size = 1092 \[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Output:
1/16*b*c*(-c^2*x^2+1)^(1/2)/(-d)^(1/2)/e/(c^2*d+e)/((-d)^(1/2)-e^(1/2)*x)+ 1/16*b*c*(-c^2*x^2+1)^(1/2)/(-d)^(1/2)/e/(c^2*d+e)/((-d)^(1/2)+e^(1/2)*x)- 1/16*(a+b*arcsin(c*x))/(-d)^(1/2)/e^(3/2)/((-d)^(1/2)-e^(1/2)*x)^2-1/16*(a +b*arcsin(c*x))/d/e^(3/2)/((-d)^(1/2)-e^(1/2)*x)+1/16*(a+b*arcsin(c*x))/(- d)^(1/2)/e^(3/2)/((-d)^(1/2)+e^(1/2)*x)^2+1/16*(a+b*arcsin(c*x))/d/e^(3/2) /((-d)^(1/2)+e^(1/2)*x)-1/16*b*c^3*arctanh((e^(1/2)-c^2*(-d)^(1/2)*x)/(c^2 *d+e)^(1/2)/(-c^2*x^2+1)^(1/2))/e^(3/2)/(c^2*d+e)^(3/2)+1/16*b*c*arctanh(( e^(1/2)-c^2*(-d)^(1/2)*x)/(c^2*d+e)^(1/2)/(-c^2*x^2+1)^(1/2))/d/e^(3/2)/(c ^2*d+e)^(1/2)-1/16*b*c^3*arctanh((e^(1/2)+c^2*(-d)^(1/2)*x)/(c^2*d+e)^(1/2 )/(-c^2*x^2+1)^(1/2))/e^(3/2)/(c^2*d+e)^(3/2)+1/16*b*c*arctanh((e^(1/2)+c^ 2*(-d)^(1/2)*x)/(c^2*d+e)^(1/2)/(-c^2*x^2+1)^(1/2))/d/e^(3/2)/(c^2*d+e)^(1 /2)-1/16*(a+b*arcsin(c*x))*ln(1-e^(1/2)*(I*c*x+(-c^2*x^2+1)^(1/2))/(I*c*(- d)^(1/2)-(c^2*d+e)^(1/2)))/(-d)^(3/2)/e^(3/2)+1/16*(a+b*arcsin(c*x))*ln(1+ e^(1/2)*(I*c*x+(-c^2*x^2+1)^(1/2))/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/(-d)^ (3/2)/e^(3/2)-1/16*(a+b*arcsin(c*x))*ln(1-e^(1/2)*(I*c*x+(-c^2*x^2+1)^(1/2 ))/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/(-d)^(3/2)/e^(3/2)+1/16*(a+b*arcsin(c *x))*ln(1+e^(1/2)*(I*c*x+(-c^2*x^2+1)^(1/2))/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/ 2)))/(-d)^(3/2)/e^(3/2)+1/16*I*b*polylog(2,e^(1/2)*(I*c*x+(-c^2*x^2+1)^(1/ 2))/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/(-d)^(3/2)/e^(3/2)-1/16*I*b*polylog( 2,-e^(1/2)*(I*c*x+(-c^2*x^2+1)^(1/2))/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))...
Time = 4.02 (sec) , antiderivative size = 1014, normalized size of antiderivative = 0.93 \[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Input:
Integrate[(x^2*(a + b*ArcSin[c*x]))/(d + e*x^2)^3,x]
Output:
(((2*I)*b*c*Sqrt[e]*Sqrt[1 - c^2*x^2])/(Sqrt[d]*(c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]*x)) - ((2*I)*b*c*Sqrt[e]*Sqrt[1 - c^2*x^2])/(Sqrt[d]*(c^2*d + e) *(I*Sqrt[d] + Sqrt[e]*x)) - (8*a*Sqrt[e]*x)/(d + e*x^2)^2 + (4*a*Sqrt[e]*x )/(d^2 + d*e*x^2) + ((2*I)*b*ArcSin[c*x])/(Sqrt[d]*(Sqrt[d] - I*Sqrt[e]*x) ^2) - ((2*I)*b*ArcSin[c*x])/(Sqrt[d]*(Sqrt[d] + I*Sqrt[e]*x)^2) + (4*a*Arc Tan[(Sqrt[e]*x)/Sqrt[d]])/d^(3/2) + ((2*I)*b*(ArcSin[c*x]/(Sqrt[d] + I*Sqr t[e]*x) - (c*ArcTan[(I*Sqrt[e] + c^2*Sqrt[d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/Sqrt[c^2*d + e]))/d + (2*b*(ArcSin[c*x]/(I*Sqrt[d] + Sqrt[e]*x ) + (c*ArcTanh[(Sqrt[e] + I*c^2*Sqrt[d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x ^2])])/Sqrt[c^2*d + e]))/d - (2*b*c^3*(Log[4] + Log[(e*Sqrt[c^2*d + e]*(Sq rt[e] - I*c^2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2]))/(c^3*(d + I* Sqrt[d]*Sqrt[e]*x))]))/(c^2*d + e)^(3/2) - (2*b*c^3*(Log[4] + Log[(e*Sqrt[ c^2*d + e]*(Sqrt[e] + I*c^2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2]) )/(c^3*(d - I*Sqrt[d]*Sqrt[e]*x))]))/(c^2*d + e)^(3/2) - (b*(ArcSin[c*x]*( ArcSin[c*x] + (2*I)*(Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] - Sqrt [c^2*d + e])] + Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sqrt[c^2* d + e])])) + 2*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(-(c*Sqrt[d]) + Sqrt [c^2*d + e])] + 2*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sq rt[c^2*d + e]))]))/d^(3/2) + (b*(ArcSin[c*x]*(ArcSin[c*x] + (2*I)*(Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(-(c*Sqrt[d]) + Sqrt[c^2*d + e])] + Log[1 ...
Time = 3.16 (sec) , antiderivative size = 1092, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5232, 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.
| \(\displaystyle \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5232 |
| \(\displaystyle \int \left (\frac {a+b \arcsin (c x)}{e \left (d+e x^2\right )^2}-\frac {d (a+b \arcsin (c x))}{e \left (d+e x^2\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b \text {arctanh}\left (\frac {\sqrt {e}-c^2 \sqrt {-d} x}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c^3}{16 e^{3/2} \left (d c^2+e\right )^{3/2}}-\frac {b \text {arctanh}\left (\frac {\sqrt {-d} x c^2+\sqrt {e}}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c^3}{16 e^{3/2} \left (d c^2+e\right )^{3/2}}+\frac {b \text {arctanh}\left (\frac {\sqrt {e}-c^2 \sqrt {-d} x}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c}{16 d e^{3/2} \sqrt {d c^2+e}}+\frac {b \text {arctanh}\left (\frac {\sqrt {-d} x c^2+\sqrt {e}}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c}{16 d e^{3/2} \sqrt {d c^2+e}}+\frac {b \sqrt {1-c^2 x^2} c}{16 \sqrt {-d} e \left (d c^2+e\right ) \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {b \sqrt {1-c^2 x^2} c}{16 \sqrt {-d} e \left (d c^2+e\right ) \left (\sqrt {e} x+\sqrt {-d}\right )}-\frac {a+b \arcsin (c x)}{16 d e^{3/2} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {a+b \arcsin (c x)}{16 d e^{3/2} \left (\sqrt {e} x+\sqrt {-d}\right )}-\frac {a+b \arcsin (c x)}{16 \sqrt {-d} e^{3/2} \left (\sqrt {-d}-\sqrt {e} x\right )^2}+\frac {a+b \arcsin (c x)}{16 \sqrt {-d} e^{3/2} \left (\sqrt {e} x+\sqrt {-d}\right )^2}-\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{16 (-d)^{3/2} e^{3/2}}+\frac {(a+b \arcsin (c x)) \log \left (\frac {e^{i \arcsin (c x)} \sqrt {e}}{i c \sqrt {-d}-\sqrt {d c^2+e}}+1\right )}{16 (-d)^{3/2} e^{3/2}}-\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{16 (-d)^{3/2} e^{3/2}}+\frac {(a+b \arcsin (c x)) \log \left (\frac {e^{i \arcsin (c x)} \sqrt {e}}{i \sqrt {-d} c+\sqrt {d c^2+e}}+1\right )}{16 (-d)^{3/2} e^{3/2}}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{16 (-d)^{3/2} e^{3/2}}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{16 (-d)^{3/2} e^{3/2}}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{16 (-d)^{3/2} e^{3/2}}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{16 (-d)^{3/2} e^{3/2}}\) |
Input:
Int[(x^2*(a + b*ArcSin[c*x]))/(d + e*x^2)^3,x]
Output:
(b*c*Sqrt[1 - c^2*x^2])/(16*Sqrt[-d]*e*(c^2*d + e)*(Sqrt[-d] - Sqrt[e]*x)) + (b*c*Sqrt[1 - c^2*x^2])/(16*Sqrt[-d]*e*(c^2*d + e)*(Sqrt[-d] + Sqrt[e]* x)) - (a + b*ArcSin[c*x])/(16*Sqrt[-d]*e^(3/2)*(Sqrt[-d] - Sqrt[e]*x)^2) - (a + b*ArcSin[c*x])/(16*d*e^(3/2)*(Sqrt[-d] - Sqrt[e]*x)) + (a + b*ArcSin [c*x])/(16*Sqrt[-d]*e^(3/2)*(Sqrt[-d] + Sqrt[e]*x)^2) + (a + b*ArcSin[c*x] )/(16*d*e^(3/2)*(Sqrt[-d] + Sqrt[e]*x)) - (b*c^3*ArcTanh[(Sqrt[e] - c^2*Sq rt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(3/2)*(c^2*d + e)^(3 /2)) + (b*c*ArcTanh[(Sqrt[e] - c^2*Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c ^2*x^2])])/(16*d*e^(3/2)*Sqrt[c^2*d + e]) - (b*c^3*ArcTanh[(Sqrt[e] + c^2* Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(3/2)*(c^2*d + e)^ (3/2)) + (b*c*ArcTanh[(Sqrt[e] + c^2*Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*d*e^(3/2)*Sqrt[c^2*d + e]) - ((a + b*ArcSin[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(16*(-d)^( 3/2)*e^(3/2)) + ((a + b*ArcSin[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/( I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(16*(-d)^(3/2)*e^(3/2)) - ((a + b*ArcSin [c*x])*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e] )])/(16*(-d)^(3/2)*e^(3/2)) + ((a + b*ArcSin[c*x])*Log[1 + (Sqrt[e]*E^(I*A rcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(16*(-d)^(3/2)*e^(3/2)) - ((I/16)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^ 2*d + e]))])/((-d)^(3/2)*e^(3/2)) + ((I/16)*b*PolyLog[2, (Sqrt[e]*E^(I*...
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSin[c*x])^n, ( f*x)^m*(d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[p] && IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 32.56 (sec) , antiderivative size = 1224, normalized size of antiderivative = 1.12
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1224\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1231\) |
| default | \(\text {Expression too large to display}\) | \(1231\) |
Input:
int(x^2*(a+b*arcsin(c*x))/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
Output:
a*((1/8/d*x^3-1/8/e*x)/(e*x^2+d)^2+1/8/e/d/(d*e)^(1/2)*arctan(e*x/(d*e)^(1 /2)))+b/c^3*(1/8*c^4*(arcsin(c*x)*d*c^5*e*x^3-arcsin(c*x)*d^2*c^5*x-c^4*d* e*x^2*(-c^2*x^2+1)^(1/2)-d^2*c^4*(-c^2*x^2+1)^(1/2)+arcsin(c*x)*c^3*x^3*e^ 2-arcsin(c*x)*c^3*x*d*e)/e/d/(c^2*d+e)/(c^2*e*x^2+c^2*d)^2-1/8*((2*c^2*d+2 *(c^2*d*(c^2*d+e))^(1/2)+e)*e)^(1/2)*(-2*(c^2*d*(c^2*d+e))^(1/2)*c^2*d+2*c ^4*d^2+2*c^2*d*e-(c^2*d*(c^2*d+e))^(1/2)*e)*c^4*arctanh(e*(I*c*x+(-c^2*x^2 +1)^(1/2))/((2*c^2*d+2*(c^2*d*(c^2*d+e))^(1/2)+e)*e)^(1/2))/(c^2*d+e)^2/d/ e^3+1/8*((2*c^2*d+2*(c^2*d*(c^2*d+e))^(1/2)+e)*e)^(1/2)*(2*c^2*d-2*(c^2*d* (c^2*d+e))^(1/2)+e)*arctanh(e*(I*c*x+(-c^2*x^2+1)^(1/2))/((2*c^2*d+2*(c^2* d*(c^2*d+e))^(1/2)+e)*e)^(1/2))*c^4/(c^2*d+e)/d/e^3-1/8*(-e*(2*c^2*d-2*(c^ 2*d*(c^2*d+e))^(1/2)+e))^(1/2)*(2*c^4*d^2+2*(c^2*d*(c^2*d+e))^(1/2)*c^2*d+ 2*c^2*d*e+(c^2*d*(c^2*d+e))^(1/2)*e)*c^4*arctan(e*(I*c*x+(-c^2*x^2+1)^(1/2 ))/((-2*c^2*d+2*(c^2*d*(c^2*d+e))^(1/2)-e)*e)^(1/2))/(c^2*d+e)^2/d/e^3+1/8 *(-e*(2*c^2*d-2*(c^2*d*(c^2*d+e))^(1/2)+e))^(1/2)*(2*c^2*d+2*(c^2*d*(c^2*d +e))^(1/2)+e)*arctan(e*(I*c*x+(-c^2*x^2+1)^(1/2))/((-2*c^2*d+2*(c^2*d*(c^2 *d+e))^(1/2)-e)*e)^(1/2))*c^4/(c^2*d+e)/d/e^3+1/16/(c^2*d+e)/d*c^4*sum(1/_ R1/(_R1^2*e-2*c^2*d-e)*(I*arcsin(c*x)*ln((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R 1)+dilog((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(-4*c^2*d- 2*e)*_Z^2+e))+1/16/(c^2*d+e)/d*c^4*sum(_R1/(_R1^2*e-2*c^2*d-e)*(I*arcsin(c *x)*ln((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-I*c*x-(-c^2*x^2+1...
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \] Input:
integrate(x^2*(a+b*arcsin(c*x))/(e*x^2+d)^3,x, algorithm="fricas")
Output:
integral((b*x^2*arcsin(c*x) + a*x^2)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^{2} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{3}}\, dx \] Input:
integrate(x**2*(a+b*asin(c*x))/(e*x**2+d)**3,x)
Output:
Integral(x**2*(a + b*asin(c*x))/(d + e*x**2)**3, x)
Exception generated. \[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^2*(a+b*arcsin(c*x))/(e*x^2+d)^3,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \] Input:
integrate(x^2*(a+b*arcsin(c*x))/(e*x^2+d)^3,x, algorithm="giac")
Output:
integrate((b*arcsin(c*x) + a)*x^2/(e*x^2 + d)^3, x)
Timed out. \[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \] Input:
int((x^2*(a + b*asin(c*x)))/(d + e*x^2)^3,x)
Output:
int((x^2*(a + b*asin(c*x)))/(d + e*x^2)^3, x)
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\frac {\sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,d^{2}+2 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d e \,x^{2}+\sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,e^{2} x^{4}+8 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{2}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{4} e^{2}+16 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{2}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{3} e^{3} x^{2}+8 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{2}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{2} e^{4} x^{4}-a \,d^{2} e x +a d \,e^{2} x^{3}}{8 d^{2} e^{2} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:
int(x^2*(a+b*asin(c*x))/(e*x^2+d)^3,x)
Output:
(sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d**2 + 2*sqrt(e)*sqrt(d)* atan((e*x)/(sqrt(e)*sqrt(d)))*a*d*e*x**2 + sqrt(e)*sqrt(d)*atan((e*x)/(sqr t(e)*sqrt(d)))*a*e**2*x**4 + 8*int((asin(c*x)*x**2)/(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x)*b*d**4*e**2 + 16*int((asin(c*x)*x**2)/(d** 3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x)*b*d**3*e**3*x**2 + 8*int ((asin(c*x)*x**2)/(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x)*b* d**2*e**4*x**4 - a*d**2*e*x + a*d*e**2*x**3)/(8*d**2*e**2*(d**2 + 2*d*e*x* *2 + e**2*x**4))