3.7.25 \(\int \frac {x^3 (a+b \arcsin (c x))}{d+e x^2} \, dx\) [625]

3.7.25.1 Optimal result

Integrand size = 21, antiderivative size = 559 \[ \int \frac {x^3 (a+b \arcsin (c x))}{d+e x^2} \, dx=\frac {b x \sqrt {1-c^2 x^2}}{4 c e}-\frac {b \arcsin (c x)}{4 c^2 e}+\frac {x^2 (a+b \arcsin (c x))}{2 e}+\frac {i d (a+b \arcsin (c x))^2}{2 b e^2}-\frac {d (a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {d (a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {d (a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}-\frac {d (a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {i b d \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {i b d \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {i b d \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2}+\frac {i b d \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^2} \]

output

-1/4*b*arcsin(c*x)/c^2/e+1/2*x^2*(a+b*arcsin(c*x))/e+1/2*I*d*(a+b*arcsin(c 
*x))^2/b/e^2-1/2*d*(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*d*(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*d 
*(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*d*(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*d*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*d*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*d*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*d*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+1/4*b*x*(-c^2*x^2+1 
)^(1/2)/c/e
 

3.7.25.2 Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 475, normalized size of antiderivative = 0.85 \[ \int \frac {x^3 (a+b \arcsin (c x))}{d+e x^2} \, dx=\frac {2 a c^2 e x^2-2 a c^2 d \log \left (d+e x^2\right )+b \left (e \left (c x \sqrt {1-c^2 x^2}+2 c^2 x^2 \arcsin (c x)-2 \arctan \left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )\right )+i c^2 d \left (\arcsin (c x) \left (\arcsin (c x)+2 i \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )+i c^2 d \left (\arcsin (c x) \left (\arcsin (c x)+2 i \left (\log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+\log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )\right )}{4 c^2 e^2} \]

input

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

output

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

3.7.25.3 Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 559, 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.

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

\(\Big \downarrow \) 5232

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

\(\Big \downarrow \) 2009

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

input

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

output

(b*x*Sqrt[1 - c^2*x^2])/(4*c*e) - (b*ArcSin[c*x])/(4*c^2*e) + (x^2*(a + b* 
ArcSin[c*x]))/(2*e) + ((I/2)*d*(a + b*ArcSin[c*x])^2)/(b*e^2) - (d*(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) - (d*(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) - (d*(a + b*ArcSin[c*x])*L 
og[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*e 
^2) - (d*(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*d*PolyLog[2, -((Sqrt[e]*E^(I* 
ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e]))])/e^2 + ((I/2)*b*d*PolyLog 
[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/e^2 + ( 
(I/2)*b*d*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^ 
2*d + e]))])/e^2 + ((I/2)*b*d*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c* 
Sqrt[-d] + Sqrt[c^2*d + e])])/e^2
 

3.7.25.3.1 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]
 

3.7.25.4 Maple [C] (warning: unable to verify)

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

Time = 2.30 (sec) , antiderivative size = 2088, normalized size of antiderivative = 3.74

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2088\)
default	\(\text {Expression too large to display}\)	\(2088\)
parts	\(\text {Expression too large to display}\)	\(2095\)

input

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

output

1/c^4*(1/2*a*c^4/e*x^2-1/2*a*c^4*d/e^2*ln(c^2*e*x^2+c^2*d)+b*c^2*(1/4*I*(2 
*c^2*d-2*(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))*d*c^2/e^3+1/16*(I+2*arcsin(c*x)) 
/e*(2*c^2*x^2-2*I*c*x*(-c^2*x^2+1)^(1/2)-1)+I*(2*c^2*d-2*(d*c^2*(c^2*d+e)) 
^(1/2)+e)*arcsin(c*x)^2*c^4*d^2/e^4+1/2*I*(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/ 
2)+e)*c^2*d*arcsin(c*x)^2/e^3-(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e)/e^4*d^ 
2*c^4*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)+1/2*I*(2*c^2*d-2*(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))*c^4*d 
^2/e^4-1/2*(2*c^2*d-2*(d*c^2*(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*(d*c^2*(c^2*d+e))^(1/2)+e))*c^2*d*arcsin(c*x)+1/ 
16*(2*I*c*x*(-c^2*x^2+1)^(1/2)+2*c^2*x^2-1)*(-I+2*arcsin(c*x))/e-1/2*I*(-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*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^3/(c^2*d+e)-I*(-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*arcsin(c*x)^2/e^3/(c^2*d+e)-1 
/2*I*(-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^2*c^4*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)+(-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)/e^4*d^2*c^4/(c^2*d+e)*ln...
 

3.7.25.5 Fricas [F]

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

input

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

output

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

3.7.25.6 Sympy [F]

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

input

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

output

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

3.7.25.7 Maxima [F]

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

input

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

output

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

3.7.25.8 Giac [F]

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

input

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

output

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

3.7.25.9 Mupad [F(-1)]

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

input

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

output

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