Integrand size = 21, antiderivative size = 1082 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\frac {b c \sqrt {-d} \sqrt {1-c^2 x^2}}{16 e^2 \left (c^2 d+e\right ) \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {b c \sqrt {-d} \sqrt {1-c^2 x^2}}{16 e^2 \left (c^2 d+e\right ) \left (\sqrt {-d}+\sqrt {e} x\right )}-\frac {\sqrt {-d} (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {-d}-\sqrt {e} x\right )^2}+\frac {5 (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\sqrt {-d} (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {-d}+\sqrt {e} x\right )^2}-\frac {5 (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {-d}+\sqrt {e} x\right )}+\frac {b c^3 d \text {arctanh}\left (\frac {\sqrt {e}-c^2 \sqrt {-d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{16 e^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {5 b c \text {arctanh}\left (\frac {\sqrt {e}-c^2 \sqrt {-d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{16 e^{5/2} \sqrt {c^2 d+e}}+\frac {b c^3 d \text {arctanh}\left (\frac {\sqrt {e}+c^2 \sqrt {-d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{16 e^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {5 b c \text {arctanh}\left (\frac {\sqrt {e}+c^2 \sqrt {-d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{16 e^{5/2} \sqrt {c^2 d+e}}+\frac {3 (a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 (a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}}+\frac {3 (a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 (a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}}+\frac {3 i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}}+\frac {3 i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{16 \sqrt {-d} e^{5/2}} \]
1/16*b*c^3*d*arctanh((-c^2*x*(-d)^(1/2)+e^(1/2))/(c^2*d+e)^(1/2)/(-c^2*x^2 +1)^(1/2))/e^(5/2)/(c^2*d+e)^(3/2)+1/16*b*c^3*d*arctanh((c^2*x*(-d)^(1/2)+ e^(1/2))/(c^2*d+e)^(1/2)/(-c^2*x^2+1)^(1/2))/e^(5/2)/(c^2*d+e)^(3/2)+3/16* (a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)- (c^2*d+e)^(1/2)))/e^(5/2)/(-d)^(1/2)-3/16*(a+b*arcsin(c*x))*ln(1+(I*c*x+(- c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/e^(5/2)/(-d)^( 1/2)+3/16*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*( -d)^(1/2)+(c^2*d+e)^(1/2)))/e^(5/2)/(-d)^(1/2)-3/16*(a+b*arcsin(c*x))*ln(1 +(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/e^(5 /2)/(-d)^(1/2)+3/16*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c *(-d)^(1/2)+(c^2*d+e)^(1/2)))/e^(5/2)/(-d)^(1/2)-3/16*I*b*polylog(2,(I*c*x +(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/e^(5/2)/(-d )^(1/2)+3/16*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^( 1/2)-(c^2*d+e)^(1/2)))/e^(5/2)/(-d)^(1/2)-3/16*I*b*polylog(2,(I*c*x+(-c^2* x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/e^(5/2)/(-d)^(1/2) -1/16*(a+b*arcsin(c*x))*(-d)^(1/2)/e^(5/2)/((-d)^(1/2)-x*e^(1/2))^2+5/16*( a+b*arcsin(c*x))/e^(5/2)/((-d)^(1/2)-x*e^(1/2))+1/16*(a+b*arcsin(c*x))*(-d )^(1/2)/e^(5/2)/((-d)^(1/2)+x*e^(1/2))^2-5/16*(a+b*arcsin(c*x))/e^(5/2)/(( -d)^(1/2)+x*e^(1/2))-5/16*b*c*arctanh((-c^2*x*(-d)^(1/2)+e^(1/2))/(c^2*d+e )^(1/2)/(-c^2*x^2+1)^(1/2))/e^(5/2)/(c^2*d+e)^(1/2)-5/16*b*c*arctanh((c...
Time = 4.47 (sec) , antiderivative size = 1014, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\frac {-\frac {i b c \sqrt {d} \sqrt {e} \sqrt {1-c^2 x^2}}{\left (c^2 d+e\right ) \left (-i \sqrt {d}+\sqrt {e} x\right )}+\frac {i b c \sqrt {d} \sqrt {e} \sqrt {1-c^2 x^2}}{\left (c^2 d+e\right ) \left (i \sqrt {d}+\sqrt {e} x\right )}+\frac {4 a d \sqrt {e} x}{\left (d+e x^2\right )^2}-\frac {10 a \sqrt {e} x}{d+e x^2}+\frac {i b \sqrt {d} \arcsin (c x)}{\left (\sqrt {d}+i \sqrt {e} x\right )^2}+\frac {i b \sqrt {d} \arcsin (c x)}{\left (i \sqrt {d}+\sqrt {e} x\right )^2}-\frac {5 b \arcsin (c x)}{i \sqrt {d}+\sqrt {e} x}+\frac {6 a \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}-5 i b \left (\frac {\arcsin (c x)}{\sqrt {d}+i \sqrt {e} x}-\frac {c \arctan \left (\frac {i \sqrt {e}+c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}\right )-\frac {5 b c \text {arctanh}\left (\frac {\sqrt {e}+i c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}+\frac {3 i b \arcsin (c x) \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+\log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )}{\sqrt {d}}-\frac {3 i b \arcsin (c x) \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )}{\sqrt {d}}+\frac {b c^3 d \left (\log (4)+\log \left (\frac {e \sqrt {c^2 d+e} \left (\sqrt {e}-i c^2 \sqrt {d} x+\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}\right )}{c^3 \left (d+i \sqrt {d} \sqrt {e} x\right )}\right )\right )}{\left (c^2 d+e\right )^{3/2}}+\frac {b c^3 d \left (\log (4)+\log \left (\frac {e \sqrt {c^2 d+e} \left (\sqrt {e}+i c^2 \sqrt {d} x+\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}\right )}{c^3 \left (d-i \sqrt {d} \sqrt {e} x\right )}\right )\right )}{\left (c^2 d+e\right )^{3/2}}+\frac {3 b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )}{\sqrt {d}}-\frac {3 b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )}{\sqrt {d}}-\frac {3 b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )}{\sqrt {d}}+\frac {3 b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )}{\sqrt {d}}}{16 e^{5/2}} \]
(((-I)*b*c*Sqrt[d]*Sqrt[e]*Sqrt[1 - c^2*x^2])/((c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]*x)) + (I*b*c*Sqrt[d]*Sqrt[e]*Sqrt[1 - c^2*x^2])/((c^2*d + e)*(I*S qrt[d] + Sqrt[e]*x)) + (4*a*d*Sqrt[e]*x)/(d + e*x^2)^2 - (10*a*Sqrt[e]*x)/ (d + e*x^2) + (I*b*Sqrt[d]*ArcSin[c*x])/(Sqrt[d] + I*Sqrt[e]*x)^2 + (I*b*S qrt[d]*ArcSin[c*x])/(I*Sqrt[d] + Sqrt[e]*x)^2 - (5*b*ArcSin[c*x])/(I*Sqrt[ d] + Sqrt[e]*x) + (6*a*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] - (5*I)*b*(Arc Sin[c*x]/(Sqrt[d] + I*Sqrt[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]) - (5*b*c*ArcTanh[(Sq rt[e] + I*c^2*Sqrt[d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/Sqrt[c^2*d + e] + ((3*I)*b*ArcSin[c*x]*(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])]))/Sqrt[d] - ((3*I)*b*ArcSin[c*x]*(Log[1 + (Sqrt[e]*E^( I*ArcSin[c*x]))/(c*Sqrt[d] - Sqrt[c^2*d + e])] + Log[1 + (Sqrt[e]*E^(I*Arc Sin[c*x]))/(c*Sqrt[d] + Sqrt[c^2*d + e])]))/Sqrt[d] + (b*c^3*d*(Log[4] + L og[(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*c^3 *d*(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) + (3*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] - Sqrt[c^2* d + e])])/Sqrt[d] - (3*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(-(c*Sq...
Time = 3.70 (sec) , antiderivative size = 1082, 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.
| \(\displaystyle \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5232 |
| \(\displaystyle \int \left (\frac {d^2 (a+b \arcsin (c x))}{e^2 \left (d+e x^2\right )^3}-\frac {2 d (a+b \arcsin (c x))}{e^2 \left (d+e x^2\right )^2}+\frac {a+b \arcsin (c x)}{e^2 \left (d+e x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b d \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^{5/2} \left (d c^2+e\right )^{3/2}}+\frac {b d \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^{5/2} \left (d c^2+e\right )^{3/2}}-\frac {5 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 e^{5/2} \sqrt {d c^2+e}}-\frac {5 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 e^{5/2} \sqrt {d c^2+e}}+\frac {b \sqrt {-d} \sqrt {1-c^2 x^2} c}{16 e^2 \left (d c^2+e\right ) \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {b \sqrt {-d} \sqrt {1-c^2 x^2} c}{16 e^2 \left (d c^2+e\right ) \left (\sqrt {e} x+\sqrt {-d}\right )}+\frac {5 (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {-d}-\sqrt {e} x\right )}-\frac {5 (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {e} x+\sqrt {-d}\right )}-\frac {\sqrt {-d} (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {-d}-\sqrt {e} x\right )^2}+\frac {\sqrt {-d} (a+b \arcsin (c x))}{16 e^{5/2} \left (\sqrt {e} x+\sqrt {-d}\right )^2}+\frac {3 (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 \sqrt {-d} e^{5/2}}-\frac {3 (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 \sqrt {-d} e^{5/2}}+\frac {3 (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 \sqrt {-d} e^{5/2}}-\frac {3 (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 \sqrt {-d} e^{5/2}}+\frac {3 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 \sqrt {-d} e^{5/2}}-\frac {3 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 \sqrt {-d} e^{5/2}}+\frac {3 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 \sqrt {-d} e^{5/2}}-\frac {3 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 \sqrt {-d} e^{5/2}}\) |
(b*c*Sqrt[-d]*Sqrt[1 - c^2*x^2])/(16*e^2*(c^2*d + e)*(Sqrt[-d] - Sqrt[e]*x )) + (b*c*Sqrt[-d]*Sqrt[1 - c^2*x^2])/(16*e^2*(c^2*d + e)*(Sqrt[-d] + Sqrt [e]*x)) - (Sqrt[-d]*(a + b*ArcSin[c*x]))/(16*e^(5/2)*(Sqrt[-d] - Sqrt[e]*x )^2) + (5*(a + b*ArcSin[c*x]))/(16*e^(5/2)*(Sqrt[-d] - Sqrt[e]*x)) + (Sqrt [-d]*(a + b*ArcSin[c*x]))/(16*e^(5/2)*(Sqrt[-d] + Sqrt[e]*x)^2) - (5*(a + b*ArcSin[c*x]))/(16*e^(5/2)*(Sqrt[-d] + Sqrt[e]*x)) + (b*c^3*d*ArcTanh[(Sq rt[e] - c^2*Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(5/2)* (c^2*d + e)^(3/2)) - (5*b*c*ArcTanh[(Sqrt[e] - c^2*Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(5/2)*Sqrt[c^2*d + e]) + (b*c^3*d*ArcTanh [(Sqrt[e] + c^2*Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(5 /2)*(c^2*d + e)^(3/2)) - (5*b*c*ArcTanh[(Sqrt[e] + c^2*Sqrt[-d]*x)/(Sqrt[c ^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(5/2)*Sqrt[c^2*d + e]) + (3*(a + b*Ar cSin[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(16*Sqrt[-d]*e^(5/2)) - (3*(a + b*ArcSin[c*x])*Log[1 + (Sqrt[e]*E^ (I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(16*Sqrt[-d]*e^(5/2)) + (3*(a + b*ArcSin[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(16*Sqrt[-d]*e^(5/2)) - (3*(a + b*ArcSin[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(16*Sqrt [-d]*e^(5/2)) + (((3*I)/16)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I* c*Sqrt[-d] - Sqrt[c^2*d + e]))])/(Sqrt[-d]*e^(5/2)) - (((3*I)/16)*b*Pol...
3.7.46.3.1 Defintions of rubi rules used
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 = 9.73 (sec) , antiderivative size = 1748, normalized size of antiderivative = 1.62
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1748\) |
| default | \(\text {Expression too large to display}\) | \(1748\) |
| parts | \(\text {Expression too large to display}\) | \(1750\) |
1/c^5*(a*c^6*((-5/8/e*c^3*x^3-3/8/e^2*d*c^3*x)/(c^2*e*x^2+c^2*d)^2+3/8/e^2 /c/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2)))+b*c^6*(-1/8*(3*arcsin(c*x)*d^2*c^5 *x+5*arcsin(c*x)*d*c^5*e*x^3-d^2*c^4*(-c^2*x^2+1)^(1/2)-(-c^2*x^2+1)^(1/2) *c^4*d*e*x^2+3*arcsin(c*x)*c^3*d*e*x+5*arcsin(c*x)*e^2*c^3*x^3)/e^2/(c^2*d +e)/(c^2*e*x^2+c^2*d)^2+1/2*((2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e)*e)^(1/2 )*(-2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2*d^2*c^4+2*c^2*e*d-(d*c^2*(c^2*d+e))^ (1/2)*e)*d*c^2*arctanh(e*(I*c*x+(-c^2*x^2+1)^(1/2))/((2*c^2*d+2*(d*c^2*(c^ 2*d+e))^(1/2)+e)*e)^(1/2))/(c^2*d+e)^2/e^5-3/16/(c^2*d+e)/e^2*c^2*d*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))/ _R1)+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))-3/16/(c^2*d+e)/e^2*c^2*d*sum(_R1/(-_R1^2*e+2*c^2*d+e)*(I*a rcsin(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)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(-4*c^2*d-2*e)*_Z^2+e))+5/8*((2*c^2*d +2*(d*c^2*(c^2*d+e))^(1/2)+e)*e)^(1/2)*(-2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2 *d^2*c^4+2*c^2*e*d-(d*c^2*(c^2*d+e))^(1/2)*e)*arctanh(e*(I*c*x+(-c^2*x^2+1 )^(1/2))/((2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e)*e)^(1/2))/(c^2*d+e)^2/e^4- 5/8*(-e*(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e))^(1/2)*(2*c^2*d+2*(d*c^2*(c^ 2*d+e))^(1/2)+e)*arctan(e*(I*c*x+(-c^2*x^2+1)^(1/2))/((-2*c^2*d+2*(d*c^2*( c^2*d+e))^(1/2)-e)*e)^(1/2))/(c^2*d+e)/e^4-1/2*(-e*(2*c^2*d-2*(d*c^2*(c^2* d+e))^(1/2)+e))^(1/2)*(2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e)*c^2*d*arcta...
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^{4} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]