\(\int \frac {x^3 (a+b \arcsin (c x))}{(d-c^2 d x^2)^3} \, dx\) [47]

Optimal result

Integrand size = 25, antiderivative size = 100 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=-\frac {b x^3}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {b x}{4 c^3 d^3 \sqrt {1-c^2 x^2}}-\frac {b \arcsin (c x)}{4 c^4 d^3}+\frac {x^4 (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2} \] Output:

-1/12*b*x^3/c/d^3/(-c^2*x^2+1)^(3/2)+1/4*b*x/c^3/d^3/(-c^2*x^2+1)^(1/2)-1/ 
4*b*arcsin(c*x)/c^4/d^3+1/4*x^4*(a+b*arcsin(c*x))/d^3/(-c^2*x^2+1)^2
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\frac {b c x \left (3-4 c^2 x^2\right ) \sqrt {1-c^2 x^2}+a \left (-3+6 c^2 x^2\right )+3 b \left (-1+2 c^2 x^2\right ) \arcsin (c x)}{12 c^4 d^3 \left (-1+c^2 x^2\right )^2} \] Input:

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

Output:

(b*c*x*(3 - 4*c^2*x^2)*Sqrt[1 - c^2*x^2] + a*(-3 + 6*c^2*x^2) + 3*b*(-1 + 
2*c^2*x^2)*ArcSin[c*x])/(12*c^4*d^3*(-1 + c^2*x^2)^2)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5186, 252, 252, 223}

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.

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

\(\Big \downarrow \) 5186

\(\displaystyle \frac {x^4 (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {b c \int \frac {x^4}{\left (1-c^2 x^2\right )^{5/2}}dx}{4 d^3}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {x^4 (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {b c \left (\frac {x^3}{3 c^2 \left (1-c^2 x^2\right )^{3/2}}-\frac {\int \frac {x^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{c^2}\right )}{4 d^3}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {x^4 (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {b c \left (\frac {x^3}{3 c^2 \left (1-c^2 x^2\right )^{3/2}}-\frac {\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}dx}{c^2}}{c^2}\right )}{4 d^3}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {x^4 (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {b c \left (\frac {x^3}{3 c^2 \left (1-c^2 x^2\right )^{3/2}}-\frac {\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}}{c^2}\right )}{4 d^3}\)

Input:

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

Output:

(x^4*(a + b*ArcSin[c*x]))/(4*d^3*(1 - c^2*x^2)^2) - (b*c*(x^3/(3*c^2*(1 - 
c^2*x^2)^(3/2)) - (x/(c^2*Sqrt[1 - c^2*x^2]) - ArcSin[c*x]/c^3)/c^2))/(4*d 
^3)
 

Defintions of rubi rules used

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 5186

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b 
*ArcSin[c*x])^n/(d*f*(m + 1))), x] - Simp[b*c*(n/(f*(m + 1)))*Simp[(d + e*x 
^2)^p/(1 - c^2*x^2)^p]   Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*A 
rcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^ 
2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.76

method	result	size
orering	\(-\frac {\left (c x -1\right ) \left (c x +1\right ) \left (4 c^{4} x^{4}+3 c^{2} x^{2}-4\right ) \left (a +b \arcsin \left (c x \right )\right )}{4 c^{4} \left (-c^{2} d \,x^{2}+d \right )^{3}}-\frac {\left (4 c^{2} x^{2}-3\right ) \left (c x -1\right )^{2} \left (c x +1\right )^{2} \left (\frac {3 x^{2} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c^{2} d \,x^{2}+d \right )^{3}}+\frac {x^{3} b c}{\sqrt {-c^{2} x^{2}+1}\, \left (-c^{2} d \,x^{2}+d \right )^{3}}+\frac {6 x^{4} \left (a +b \arcsin \left (c x \right )\right ) c^{2} d}{\left (-c^{2} d \,x^{2}+d \right )^{4}}\right )}{12 x^{2} c^{4}}\)	\(176\)
derivativedivides	\(\frac {-\frac {a \left (-\frac {1}{16 \left (c x +1\right )^{2}}+\frac {3}{16 \left (c x +1\right )}-\frac {1}{16 \left (c x -1\right )^{2}}-\frac {3}{16 \left (c x -1\right )}\right )}{d^{3}}-\frac {b \left (-\frac {\arcsin \left (c x \right )}{16 \left (c x +1\right )^{2}}+\frac {3 \arcsin \left (c x \right )}{16 \left (c x +1\right )}-\frac {\arcsin \left (c x \right )}{16 \left (c x -1\right )^{2}}-\frac {3 \arcsin \left (c x \right )}{16 \left (c x -1\right )}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{48 \left (c x -1\right )^{2}}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{6 c x -6}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{48 \left (c x +1\right )^{2}}+\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{6 c x +6}\right )}{d^{3}}}{c^{4}}\)	\(212\)
default	\(\frac {-\frac {a \left (-\frac {1}{16 \left (c x +1\right )^{2}}+\frac {3}{16 \left (c x +1\right )}-\frac {1}{16 \left (c x -1\right )^{2}}-\frac {3}{16 \left (c x -1\right )}\right )}{d^{3}}-\frac {b \left (-\frac {\arcsin \left (c x \right )}{16 \left (c x +1\right )^{2}}+\frac {3 \arcsin \left (c x \right )}{16 \left (c x +1\right )}-\frac {\arcsin \left (c x \right )}{16 \left (c x -1\right )^{2}}-\frac {3 \arcsin \left (c x \right )}{16 \left (c x -1\right )}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{48 \left (c x -1\right )^{2}}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{6 c x -6}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{48 \left (c x +1\right )^{2}}+\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{6 c x +6}\right )}{d^{3}}}{c^{4}}\)	\(212\)
parts	\(-\frac {a \left (-\frac {1}{16 c^{4} \left (c x -1\right )^{2}}-\frac {3}{16 c^{4} \left (c x -1\right )}-\frac {1}{16 c^{4} \left (c x +1\right )^{2}}+\frac {3}{16 c^{4} \left (c x +1\right )}\right )}{d^{3}}-\frac {b \left (-\frac {\arcsin \left (c x \right )}{16 \left (c x +1\right )^{2}}+\frac {3 \arcsin \left (c x \right )}{16 \left (c x +1\right )}-\frac {\arcsin \left (c x \right )}{16 \left (c x -1\right )^{2}}-\frac {3 \arcsin \left (c x \right )}{16 \left (c x -1\right )}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{48 \left (c x -1\right )^{2}}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{6 c x -6}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{48 \left (c x +1\right )^{2}}+\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{6 c x +6}\right )}{d^{3} c^{4}}\)	\(223\)

Input:

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

Output:

-1/4*(c*x-1)*(c*x+1)*(4*c^4*x^4+3*c^2*x^2-4)/c^4*(a+b*arcsin(c*x))/(-c^2*d 
*x^2+d)^3-1/12/x^2*(4*c^2*x^2-3)/c^4*(c*x-1)^2*(c*x+1)^2*(3*x^2*(a+b*arcsi 
n(c*x))/(-c^2*d*x^2+d)^3+x^3*b*c/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^3+6*x^4 
*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^4*c^2*d)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.91 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\frac {3 \, a c^{4} x^{4} + 3 \, {\left (2 \, b c^{2} x^{2} - b\right )} \arcsin \left (c x\right ) - {\left (4 \, b c^{3} x^{3} - 3 \, b c x\right )} \sqrt {-c^{2} x^{2} + 1}}{12 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \] Input:

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

Output:

1/12*(3*a*c^4*x^4 + 3*(2*b*c^2*x^2 - b)*arcsin(c*x) - (4*b*c^3*x^3 - 3*b*c 
*x)*sqrt(-c^2*x^2 + 1))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3)
 

Sympy [F]

\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=- \frac {\int \frac {a x^{3}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b x^{3} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \] Input:

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

Output:

-(Integral(a*x**3/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integr 
al(b*x**3*asin(c*x)/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x))/d**3
 

Maxima [F]

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

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

Output:

1/4*(2*c^2*x^2 - 1)*a/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3) + 1/4*((2*c^ 
2*x^2 - 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 4*(c^8*d^3*x^4 - 2 
*c^6*d^3*x^2 + c^4*d^3)*integrate(1/4*(2*c^2*x^2 - 1)*e^(1/2*log(c*x + 1) 
+ 1/2*log(-c*x + 1))/(c^11*d^3*x^8 - 3*c^9*d^3*x^6 + 3*c^7*d^3*x^4 - c^5*d 
^3*x^2 + (c^9*d^3*x^6 - 3*c^7*d^3*x^4 + 3*c^5*d^3*x^2 - c^3*d^3)*e^(log(c* 
x + 1) + log(-c*x + 1))), x))*b/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.24 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\frac {b x^{4} \arcsin \left (c x\right )}{4 \, {\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac {a x^{4}}{4 \, {\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac {b x^{3}}{12 \, {\left (c^{2} x^{2} - 1\right )} \sqrt {-c^{2} x^{2} + 1} c d^{3}} + \frac {b x}{4 \, \sqrt {-c^{2} x^{2} + 1} c^{3} d^{3}} - \frac {b \arcsin \left (c x\right )}{4 \, c^{4} d^{3}} - \frac {a}{4 \, c^{4} d^{3}} \] Input:

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

Output:

1/4*b*x^4*arcsin(c*x)/((c^2*x^2 - 1)^2*d^3) + 1/4*a*x^4/((c^2*x^2 - 1)^2*d 
^3) + 1/12*b*x^3/((c^2*x^2 - 1)*sqrt(-c^2*x^2 + 1)*c*d^3) + 1/4*b*x/(sqrt( 
-c^2*x^2 + 1)*c^3*d^3) - 1/4*b*arcsin(c*x)/(c^4*d^3) - 1/4*a/(c^4*d^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\frac {-4 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{3}}{c^{6} x^{6}-3 c^{4} x^{4}+3 c^{2} x^{2}-1}d x \right ) b \,c^{4} x^{4}+8 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{3}}{c^{6} x^{6}-3 c^{4} x^{4}+3 c^{2} x^{2}-1}d x \right ) b \,c^{2} x^{2}-4 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{3}}{c^{6} x^{6}-3 c^{4} x^{4}+3 c^{2} x^{2}-1}d x \right ) b +a \,x^{4}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )} \] Input:

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

Output:

( - 4*int((asin(c*x)*x**3)/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1),x)* 
b*c**4*x**4 + 8*int((asin(c*x)*x**3)/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x** 
2 - 1),x)*b*c**2*x**2 - 4*int((asin(c*x)*x**3)/(c**6*x**6 - 3*c**4*x**4 + 
3*c**2*x**2 - 1),x)*b + a*x**4)/(4*d**3*(c**4*x**4 - 2*c**2*x**2 + 1))