Integrand size = 21, antiderivative size = 574 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\frac {d (a+b \arcsin (c x))}{2 e^2 \left (d+e x^2\right )}-\frac {i (a+b \arcsin (c x))^2}{2 b e^2}-\frac {b c \sqrt {d} \arctan \left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 e^2 \sqrt {c^2 d+e}}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2} \]
1/2*d*(a+b*arcsin(c*x))/e^2/(e*x^2+d)-1/2*I*(a+b*arcsin(c*x))^2/b/e^2+1/2* (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^2+1/2*(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^2+1/2*(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^2+ 1/2*(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^2-1/2*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^2-1/2*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^2-1/2*I*b*polyl og(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^2-1/2*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^2-1/2*b*c*arctan(x*(c^2*d+e)^(1/2)/d^(1/2)/(-c^2*x^2+1 )^(1/2))*d^(1/2)/e^2/(c^2*d+e)^(1/2)
Time = 0.87 (sec) , antiderivative size = 593, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\frac {\frac {2 a d}{d+e x^2}+2 a \log \left (d+e x^2\right )+b \left (\sqrt {d} \left (\frac {\arcsin (c x)}{\sqrt {d}+i \sqrt {e} x}-\frac {c \arctan \left (\frac {i \sqrt {e}+c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}\right )-i \sqrt {d} \left (-\frac {\arcsin (c x)}{i \sqrt {d}+\sqrt {e} x}-\frac {c \text {arctanh}\left (\frac {\sqrt {e}+i c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}\right )-i \left (\arcsin (c x) \left (\arcsin (c x)+2 i \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )-i \left (\arcsin (c x) \left (\arcsin (c x)+2 i \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+\log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )}{4 e^2} \]
((2*a*d)/(d + e*x^2) + 2*a*Log[d + e*x^2] + b*(Sqrt[d]*(ArcSin[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]) - I*Sqrt[d]*(-(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]) - I*(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])] + L og[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sqrt[c^2*d + e])])) + 2*Po lyLog[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] + Sqrt[c^2*d + e]))]) - I*(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] + Sqrt[c^2*d + e])])))/(4*e^2)
Time = 1.33 (sec) , antiderivative size = 574, 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.
| \(\displaystyle \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 5232 |
| \(\displaystyle \int \left (\frac {x (a+b \arcsin (c x))}{e \left (d+e x^2\right )}-\frac {d x (a+b \arcsin (c x))}{e \left (d+e x^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {d (a+b \arcsin (c x))}{2 e^2 \left (d+e x^2\right )}-\frac {i (a+b \arcsin (c x))^2}{2 b e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {b c \sqrt {d} \arctan \left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 e^2 \sqrt {c^2 d+e}}\) |
(d*(a + b*ArcSin[c*x]))/(2*e^2*(d + e*x^2)) - ((I/2)*(a + b*ArcSin[c*x])^2 )/(b*e^2) - (b*c*Sqrt[d]*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2* x^2])])/(2*e^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])])/(2*e^2) + ((a + b*ArcS in[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(2*e^2) + ((a + b*ArcSin[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/( I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*e^2) + ((a + b*ArcSin[c*x])*Log[1 + ( Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*e^2) - (( I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e]))])/e^2 - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[ -d] - Sqrt[c^2*d + e])])/e^2 - ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[ c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/e^2 - ((I/2)*b*PolyLog[2, (Sqrt [e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/e^2
3.7.33.3.1 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 = 1.85 (sec) , antiderivative size = 2101, normalized size of antiderivative = 3.66
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2101\) |
| default | \(\text {Expression too large to display}\) | \(2101\) |
| parts | \(\text {Expression too large to display}\) | \(2113\) |
1/c^4*(1/2*a*c^6/e^2*d/(c^2*e*x^2+c^2*d)+1/2*a*c^4/e^2*ln(c^2*e*x^2+c^2*d) +b*c^4*(-I*(2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e)*arcsin(c*x)^2*d*c^2/e^4-1 /2*I/e^2*sum((-_R1^2*e+4*c^2*d+2*e)/(-_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))-1/4*I*(d*c^2*(c^2*d+e))^(1 /2)/c^2/d/e/(c^2*d+e)*arcsin(c*x)^2-1/8*I*(d*c^2*(c^2*d+e))^(1/2)/c^2/d/e/ (c^2*d+e)*polylog(2,e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d+2*(d*c^2*(c^2* d+e))^(1/2)+e))+1/8*I*(2*d^2*c^4+2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2*c^2*e*d +(d*c^2*(c^2*d+e))^(1/2)*e)*polylog(2,e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^ 2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e))/e^2/d/c^2/(c^2*d+e)+1/4*I*(2*d^2*c^4+2*( d*c^2*(c^2*d+e))^(1/2)*d*c^2+2*c^2*e*d+(d*c^2*(c^2*d+e))^(1/2)*e)*arcsin(c *x)^2/e^2/d/c^2/(c^2*d+e)+I*(2*d^2*c^4+2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2*c ^2*e*d+(d*c^2*(c^2*d+e))^(1/2)*e)*d*c^2*arcsin(c*x)^2/e^4/(c^2*d+e)-(2*d^2 *c^4+2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2*c^2*e*d+(d*c^2*(c^2*d+e))^(1/2)*e)* d*c^2*ln(1-e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/ 2)+e))*arcsin(c*x)/e^4/(c^2*d+e)+1/2*I*(2*d^2*c^4+2*(d*c^2*(c^2*d+e))^(1/2 )*d*c^2+2*c^2*e*d+(d*c^2*(c^2*d+e))^(1/2)*e)*d*c^2*polylog(2,e*(I*c*x+(-c^ 2*x^2+1)^(1/2))^2/(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e))/e^4/(c^2*d+e)-1/4 *(2*d^2*c^4+2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2*c^2*e*d+(d*c^2*(c^2*d+e))^(1 /2)*e)*ln(1-e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d-2*(d*c^2*(c^2*d+e))...
\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^{3} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{2}}\, dx \]
\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
1/2*a*(d/(e^3*x^2 + d*e^2) + log(e*x^2 + d)/e^2) + b*integrate(x^3*arctan2 (c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(e^2*x^4 + 2*d*e*x^2 + d^2), x)
\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \]