\(\int \frac {x^5 (a+b \arcsin (c x))}{(d+e x^2)^3} \, dx\) [467]

Optimal result

Integrand size = 21, antiderivative size = 705 \[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\frac {b c d x \sqrt {1-c^2 x^2}}{8 e^2 \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac {d^2 (a+b \arcsin (c x))}{4 e^3 \left (d+e x^2\right )^2}+\frac {d (a+b \arcsin (c x))}{e^3 \left (d+e x^2\right )}-\frac {i (a+b \arcsin (c x))^2}{2 b e^3}-\frac {b c \sqrt {d} \arctan \left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{e^3 \sqrt {c^2 d+e}}+\frac {b c \sqrt {d} \left (2 c^2 d+e\right ) \arctan \left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{8 e^3 \left (c^2 d+e\right )^{3/2}}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^3} \] Output:

1/8*b*c*d*x*(-c^2*x^2+1)^(1/2)/e^2/(c^2*d+e)/(e*x^2+d)-1/4*d^2*(a+b*arcsin 
(c*x))/e^3/(e*x^2+d)^2+d*(a+b*arcsin(c*x))/e^3/(e*x^2+d)-1/2*I*(a+b*arcsin 
(c*x))^2/b/e^3-b*c*d^(1/2)*arctan((c^2*d+e)^(1/2)*x/d^(1/2)/(-c^2*x^2+1)^( 
1/2))/e^3/(c^2*d+e)^(1/2)+1/8*b*c*d^(1/2)*(2*c^2*d+e)*arctan((c^2*d+e)^(1/ 
2)*x/d^(1/2)/(-c^2*x^2+1)^(1/2))/e^3/(c^2*d+e)^(3/2)+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^3+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^3+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^3+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^3-1/2*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^3-1/2*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^3-1/2*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^3-1/2*I*b*poly 
log(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^3
 

Mathematica [A] (warning: unable to verify)

Time = 4.72 (sec) , antiderivative size = 973, normalized size of antiderivative = 1.38 \[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[(x^5*(a + b*ArcSin[c*x]))/(d + e*x^2)^3,x]
 

Output:

((-4*a*d^2)/(d + e*x^2)^2 + (16*a*d)/(d + e*x^2) + 8*a*Log[d + e*x^2] + b* 
((c*d*Sqrt[e]*Sqrt[1 - c^2*x^2])/((c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]*x)) 
+ (c*d*Sqrt[e]*Sqrt[1 - c^2*x^2])/((c^2*d + e)*(I*Sqrt[d] + Sqrt[e]*x)) + 
(7*Sqrt[d]*ArcSin[c*x])/(Sqrt[d] - I*Sqrt[e]*x) - (d*ArcSin[c*x])/(Sqrt[d] 
 + I*Sqrt[e]*x)^2 + (7*Sqrt[d]*ArcSin[c*x])/(Sqrt[d] + I*Sqrt[e]*x) + (d*A 
rcSin[c*x])/(I*Sqrt[d] + Sqrt[e]*x)^2 - (8*I)*ArcSin[c*x]^2 - (7*c*Sqrt[d] 
*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] + ((7*I)*c*Sqrt[d]*ArcTanh[(Sqrt[e] + I*c^2*Sqrt[d]*x)/(Sq 
rt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/Sqrt[c^2*d + e] + 8*ArcSin[c*x]*Log[1 + 
 (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] - Sqrt[c^2*d + e])] + 8*ArcSin[c*x 
]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(-(c*Sqrt[d]) + Sqrt[c^2*d + e])] + 
8*ArcSin[c*x]*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sqrt[c^2*d 
+ e])] + 8*ArcSin[c*x]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sq 
rt[c^2*d + e])] + (I*c^3*d^(3/2)*Log[(e*Sqrt[c^2*d + e]*(Sqrt[e] - I*c^2*S 
qrt[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) - (I*c^3*d^(3/2)*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) - (8*I)*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[ 
c*x]))/(c*Sqrt[d] - Sqrt[c^2*d + e])] - (8*I)*PolyLog[2, (Sqrt[e]*E^(I*Arc 
Sin[c*x]))/(-(c*Sqrt[d]) + Sqrt[c^2*d + e])] - (8*I)*PolyLog[2, -((Sqrt...
 

Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 705, 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.

\(\displaystyle \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx\)

\(\Big \downarrow \) 5232

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^3}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^3}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^3}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e^3}-\frac {d^2 (a+b \arcsin (c x))}{4 e^3 \left (d+e x^2\right )^2}+\frac {d (a+b \arcsin (c x))}{e^3 \left (d+e x^2\right )}-\frac {i (a+b \arcsin (c x))^2}{2 b e^3}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e^3}+\frac {b c \sqrt {d} \left (2 c^2 d+e\right ) \arctan \left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{8 e^3 \left (c^2 d+e\right )^{3/2}}-\frac {b c \sqrt {d} \arctan \left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{e^3 \sqrt {c^2 d+e}}+\frac {b c d x \sqrt {1-c^2 x^2}}{8 e^2 \left (c^2 d+e\right ) \left (d+e x^2\right )}\)

Input:

Int[(x^5*(a + b*ArcSin[c*x]))/(d + e*x^2)^3,x]
 

Output:

(b*c*d*x*Sqrt[1 - c^2*x^2])/(8*e^2*(c^2*d + e)*(d + e*x^2)) - (d^2*(a + b* 
ArcSin[c*x]))/(4*e^3*(d + e*x^2)^2) + (d*(a + b*ArcSin[c*x]))/(e^3*(d + e* 
x^2)) - ((I/2)*(a + b*ArcSin[c*x])^2)/(b*e^3) - (b*c*Sqrt[d]*ArcTan[(Sqrt[ 
c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(e^3*Sqrt[c^2*d + e]) + (b*c*S 
qrt[d]*(2*c^2*d + e)*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2] 
)])/(8*e^3*(c^2*d + 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])])/(2*e^3) + ((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^3) + ((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^3) + ((a + b*ArcSin[c*x])*Log[1 + (Sq 
rt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*e^3) - ((I/ 
2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + 
 e]))])/e^3 - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d 
] - Sqrt[c^2*d + e])])/e^3 - ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c* 
x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/e^3 - ((I/2)*b*PolyLog[2, (Sqrt[e 
]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/e^3
 

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 = 23.83 (sec) , antiderivative size = 3508, normalized size of antiderivative = 4.98

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3508\)
default	\(\text {Expression too large to display}\)	\(3508\)
parts	\(\text {Expression too large to display}\)	\(3513\)

Input:

int(x^5*(a+b*arcsin(c*x))/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
 

Output:

1/c^6*(a*c^6*(1/2/e^3*ln(c^2*e*x^2+c^2*d)+c^2*d/e^3/(c^2*e*x^2+c^2*d)-1/4* 
c^4*d^2/e^3/(c^2*e*x^2+c^2*d)^2)+b*c^6*(5/8*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^4*d^2+2*c^2*d 
*e+e^2)/e^3-I/e^3/(c^2*d+e)*c^2*d*arcsin(c*x)^2-1/2*I/e^3/(c^2*d+e)*c^2*d* 
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))+1/8*c^2*d*(6*c^4*d^2*arcsin(c*x)+8* 
arcsin(c*x)*c^4*d*e*x^2-I*c^4*d^2-2*I*c^4*d*e*x^2-I*e^2*c^4*x^4+(-c^2*x^2+ 
1)^(1/2)*c^3*d*e*x+(-c^2*x^2+1)^(1/2)*e^2*c^3*x^3+6*c^2*d*e*arcsin(c*x)+8* 
arcsin(c*x)*e^2*c^2*x^2)/e^3/(c^2*d+e)/(c^2*e*x^2+c^2*d)^2-I/e^2/(c^2*d+e) 
*arcsin(c*x)^2-1/2*I/e^2/(c^2*d+e)*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))+ 
(2*c^2*d+2*(c^2*d*(c^2*d+e))^(1/2)+e)*d^2*c^4*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)/e^5/(c^2*d+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^4*d^2*arcsin(c*x)^2/(c^4*d^2+2*c^2*d*e+e^2)/e^5-3/4*I*(c^2*d*(c^2 
*d+e))^(1/2)/e^3/(c^2*d+e)^2*c^2*d*arctanh(1/4*(4*c^2*d-2*e*(I*c*x+(-c^2*x 
^2+1)^(1/2))^2+2*e)/(c^4*d^2+c^2*d*e)^(1/2))-1/8*I*(c^2*d*(c^2*d+e))^(1...
 

Fricas [F]

\[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \] Input:

integrate(x^5*(a+b*arcsin(c*x))/(e*x^2+d)^3,x, algorithm="fricas")
 

Output:

integral((b*x^5*arcsin(c*x) + a*x^5)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 
+ d^3), x)
 

Sympy [F]

\[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^{5} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{3}}\, dx \] Input:

integrate(x**5*(a+b*asin(c*x))/(e*x**2+d)**3,x)
 

Output:

Integral(x**5*(a + b*asin(c*x))/(d + e*x**2)**3, x)
 

Maxima [F]

\[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \] Input:

integrate(x^5*(a+b*arcsin(c*x))/(e*x^2+d)^3,x, algorithm="maxima")
 

Output:

1/4*a*((4*d*e*x^2 + 3*d^2)/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3) + 2*log(e*x^2 
 + d)/e^3) + b*integrate(x^5*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(e 
^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(x^5*(a+b*arcsin(c*x))/(e*x^2+d)^3,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. \[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^5\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \] Input:

int((x^5*(a + b*asin(c*x)))/(d + e*x^2)^3,x)
 

Output:

int((x^5*(a + b*asin(c*x)))/(d + e*x^2)^3, x)
 

Reduce [F]

\[ \int \frac {x^5 (a+b \arcsin (c x))}{\left (d+e x^2\right )^3} \, dx=\frac {4 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{5}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{2} e^{3}+8 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{5}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b d \,e^{4} x^{2}+4 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{5}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,e^{5} x^{4}+2 \,\mathrm {log}\left (e \,x^{2}+d \right ) a \,d^{2}+4 \,\mathrm {log}\left (e \,x^{2}+d \right ) a d e \,x^{2}+2 \,\mathrm {log}\left (e \,x^{2}+d \right ) a \,e^{2} x^{4}+a \,d^{2}-2 a \,e^{2} x^{4}}{4 e^{3} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:

int(x^5*(a+b*asin(c*x))/(e*x^2+d)^3,x)
 

Output:

(4*int((asin(c*x)*x**5)/(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6) 
,x)*b*d**2*e**3 + 8*int((asin(c*x)*x**5)/(d**3 + 3*d**2*e*x**2 + 3*d*e**2* 
x**4 + e**3*x**6),x)*b*d*e**4*x**2 + 4*int((asin(c*x)*x**5)/(d**3 + 3*d**2 
*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x)*b*e**5*x**4 + 2*log(d + e*x**2)*a* 
d**2 + 4*log(d + e*x**2)*a*d*e*x**2 + 2*log(d + e*x**2)*a*e**2*x**4 + a*d* 
*2 - 2*a*e**2*x**4)/(4*e**3*(d**2 + 2*d*e*x**2 + e**2*x**4))