\(\int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx\) [453]

Optimal result

Integrand size = 19, antiderivative size = 491 \[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=-\frac {i (a+b \arcsin (c x))^2}{2 b e}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e} \] Output:

-1/2*I*(a+b*arcsin(c*x))^2/b/e+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+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+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+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-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-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-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-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
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 399, normalized size of antiderivative = 0.81 \[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=-\frac {i \left (b \arcsin (c x)^2+i b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+i b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+i b \arcsin (c x) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+i b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+i a \log \left (d+e x^2\right )+b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )}{2 e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, 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.

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

\(\Big \downarrow \) 5232

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((-1/2*I)*(a + b*ArcSin[c*x])^2)/(b*e) + ((a + b*ArcSin[c*x])*Log[1 - (Sqr 
t[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(2*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) + ((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) + ((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) - ((I 
/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d 
+ e]))])/e - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] 
 - Sqrt[c^2*d + e])])/e - ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]) 
)/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/e - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^( 
I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/e
 

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 = 2.43 (sec) , antiderivative size = 1965, normalized size of antiderivative = 4.00

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1965\)
default	\(\text {Expression too large to display}\)	\(1965\)
parts	\(\text {Expression too large to display}\)	\(1966\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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