\(\int \frac {x^3 (a+b \arcsin (c x))}{(d+e x^2)^2} \, dx\) [459]

Optimal result

Integrand size = 21, antiderivative size = 574 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\frac {d (a+b \arcsin (c x))}{2 e^2 \left (d+e x^2\right )}-\frac {i (a+b \arcsin (c x))^2}{2 b e^2}-\frac {b c \sqrt {d} \arctan \left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 e^2 \sqrt {c^2 d+e}}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2} \] Output:

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* 
b*c*d^(1/2)*arctan((c^2*d+e)^(1/2)*x/d^(1/2)/(-c^2*x^2+1)^(1/2))/e^2/(c^2* 
d+e)^(1/2)+1/2*(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)))/e^2+1/2*(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)))/e^2+1/2*(a+b*ar 
csin(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)))/e^2+1/2*(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)))/e^2-1/2*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)))/e^2-1/2*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)))/e^ 
2-1/2*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)))/e^2-1/2*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)))/e^2
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 593, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d+e x^2\right )^2} \, dx=\frac {\frac {2 a d}{d+e x^2}+2 a \log \left (d+e x^2\right )+b \left (\sqrt {d} \left (\frac {\arcsin (c x)}{\sqrt {d}+i \sqrt {e} x}-\frac {c \arctan \left (\frac {i \sqrt {e}+c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}\right )-i \sqrt {d} \left (-\frac {\arcsin (c x)}{i \sqrt {d}+\sqrt {e} x}-\frac {c \text {arctanh}\left (\frac {\sqrt {e}+i c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}\right )-i \left (\arcsin (c x) \left (\arcsin (c x)+2 i \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )-i \left (\arcsin (c x) \left (\arcsin (c x)+2 i \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+\log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )}{4 e^2} \] Input:

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

Output:

((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)
 

Rubi [A] (verified)

Time = 1.43 (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

\(\displaystyle \frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^2}+\frac {d (a+b \arcsin (c x))}{2 e^2 \left (d+e x^2\right )}-\frac {i (a+b \arcsin (c x))^2}{2 b e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e^2}-\frac {b c \sqrt {d} \arctan \left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 e^2 \sqrt {c^2 d+e}}\)

Input:

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

Output:

(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
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5232

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 5.78 (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	\(\text {Expression too large to display}\)	\(2113\)

Input:

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

Output:

1/c^4*(1/2*a*c^6*d/e^2/(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*(-1/8*I*(c^2*d*(c^2*d+e))^(1/2)/c^2/d/(c^2*d+e)/e*polylog(2,e*(I*c* 
x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d+2*(c^2*d*(c^2*d+e))^(1/2)+e))+I*(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^2 
*d*arcsin(c*x)^2/(c^2*d+e)/e^4-1/4*(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)*ln(1-e*(I*c*x+(-c^2*x^2+1)^(1/2)) 
^2/(2*c^2*d-2*(c^2*d*(c^2*d+e))^(1/2)+e))*arcsin(c*x)/c^2/d/(c^2*d+e)/e^2+ 
1/2*I*(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^2*d*polylog(2,e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d-2*(c^2 
*d*(c^2*d+e))^(1/2)+e))/(c^2*d+e)/e^4+1/8*I*(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)*polylog(2,e*(I*c*x+(-c^2 
*x^2+1)^(1/2))^2/(2*c^2*d-2*(c^2*d*(c^2*d+e))^(1/2)+e))/c^2/d/(c^2*d+e)/e^ 
2-1/4*I*(c^2*d*(c^2*d+e))^(1/2)/c^2/d/(c^2*d+e)/e*arcsin(c*x)^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^2* 
d*ln(1-e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d-2*(c^2*d*(c^2*d+e))^(1/2)+e 
))*arcsin(c*x)/(c^2*d+e)/e^4+1/4*(c^2*d*(c^2*d+e))^(1/2)/c^2/d/(c^2*d+e)/e 
*arcsin(c*x)*ln(1-e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d+2*(c^2*d*(c^2*d+ 
e))^(1/2)+e))+1/4*I*(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)*arcsin(c*x)^2/c^2/d/(c^2*d+e)/e^2+(2*c^2*d+2*(c^ 
2*d*(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*...
 

Fricas [F]

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

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

Output:

integral((b*x^3*arcsin(c*x) + a*x^3)/(e^2*x^4 + 2*d*e*x^2 + d^2), x)
 

Sympy [F]

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

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

Output:

Integral(x**3*(a + b*asin(c*x))/(d + e*x**2)**2, x)
 

Maxima [F]

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

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

Output:

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)
 

Giac [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*int((asin(c*x)*x**3)/(d**2 + 2*d*e*x**2 + e**2*x**4),x)*b*d*e**2 + 2*in 
t((asin(c*x)*x**3)/(d**2 + 2*d*e*x**2 + e**2*x**4),x)*b*e**3*x**2 + log(d 
+ e*x**2)*a*d + log(d + e*x**2)*a*e*x**2 - a*e*x**2)/(2*e**2*(d + e*x**2))