\(\int (a-b \arcsin (1-d x^2))^3 \, dx\) [284]

Optimal result

Integrand size = 16, antiderivative size = 115 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^3 \, dx=-24 a b^2 x-\frac {48 b^3 \sqrt {2 d x^2-d^2 x^4}}{d x}+24 b^3 x \arcsin \left (1-d x^2\right )+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^3 \] Output:

-24*a*b^2*x-48*b^3*(-d^2*x^4+2*d*x^2)^(1/2)/d/x-24*b^3*x*arcsin(d*x^2-1)+6 
*b*(-d^2*x^4+2*d*x^2)^(1/2)*(a+b*arcsin(d*x^2-1))^2/d/x+x*(a+b*arcsin(d*x^ 
2-1))^3
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.44 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^3 \, dx=\frac {a \left (a^2-24 b^2\right ) d x^2+6 b \left (a^2-8 b^2\right ) \sqrt {d x^2 \left (2-d x^2\right )}-3 b \left (a^2 d x^2-8 b^2 d x^2+4 a b \sqrt {-d x^2 \left (-2+d x^2\right )}\right ) \arcsin \left (1-d x^2\right )+3 b^2 \left (a d x^2+2 b \sqrt {-d x^2 \left (-2+d x^2\right )}\right ) \arcsin \left (1-d x^2\right )^2-b^3 d x^2 \arcsin \left (1-d x^2\right )^3}{d x} \] Input:

Integrate[(a - b*ArcSin[1 - d*x^2])^3,x]
 

Output:

(a*(a^2 - 24*b^2)*d*x^2 + 6*b*(a^2 - 8*b^2)*Sqrt[d*x^2*(2 - d*x^2)] - 3*b* 
(a^2*d*x^2 - 8*b^2*d*x^2 + 4*a*b*Sqrt[-(d*x^2*(-2 + d*x^2))])*ArcSin[1 - d 
*x^2] + 3*b^2*(a*d*x^2 + 2*b*Sqrt[-(d*x^2*(-2 + d*x^2))])*ArcSin[1 - d*x^2 
]^2 - b^3*d*x^2*ArcSin[1 - d*x^2]^3)/(d*x)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5313, 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.

\(\displaystyle \int \left (a-b \arcsin \left (1-d x^2\right )\right )^3 \, dx\)

\(\Big \downarrow \) 5313

\(\displaystyle -24 b^2 \int \left (a-b \arcsin \left (1-d x^2\right )\right )dx+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^3\)

\(\Big \downarrow \) 2009

\(\displaystyle -24 b^2 \left (a x+b (-x) \arcsin \left (1-d x^2\right )+\frac {2 b \sqrt {2 d x^2-d^2 x^4}}{d x}\right )+\frac {6 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^2}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^3\)

Input:

Int[(a - b*ArcSin[1 - d*x^2])^3,x]
 

Output:

(6*b*Sqrt[2*d*x^2 - d^2*x^4]*(a - b*ArcSin[1 - d*x^2])^2)/(d*x) + x*(a - b 
*ArcSin[1 - d*x^2])^3 - 24*b^2*(a*x + (2*b*Sqrt[2*d*x^2 - d^2*x^4])/(d*x) 
- b*x*ArcSin[1 - d*x^2])
 

Defintions of rubi rules used

rule 2009

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

rule 5313

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*( 
a + b*ArcSin[c + d*x^2])^n, x] + (Simp[2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*((a 
 + b*ArcSin[c + d*x^2])^(n - 1)/(d*x)), x] - Simp[4*b^2*n*(n - 1)   Int[(a 
+ b*ArcSin[c + d*x^2])^(n - 2), x], x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^ 
2, 1] && GtQ[n, 1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(437\) vs. \(2(106)=212\).

Time = 0.13 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.81

method	result	size
orering	\(x {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{3}-\frac {6 \left (d \,x^{2}-4\right ) {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{2} b x}{\sqrt {-\left (d \,x^{2}-1\right )^{2}+1}}-\frac {2 \left (d \,x^{2}-2\right ) x \left (\frac {24 \left (a +b \arcsin \left (d \,x^{2}-1\right )\right ) b^{2} d^{2} x^{2}}{-\left (d \,x^{2}-1\right )^{2}+1}+\frac {6 {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{2} b d}{\sqrt {-\left (d \,x^{2}-1\right )^{2}+1}}+\frac {12 {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{2} b \,d^{2} x^{2} \left (d \,x^{2}-1\right )}{{\left (-\left (d \,x^{2}-1\right )^{2}+1\right )}^{\frac {3}{2}}}\right )}{d}-\frac {\left (d \,x^{2}-2\right )^{2} \left (\frac {48 b^{3} d^{3} x^{3}}{{\left (-\left (d \,x^{2}-1\right )^{2}+1\right )}^{\frac {3}{2}}}+\frac {72 \left (a +b \arcsin \left (d \,x^{2}-1\right )\right ) b^{2} d^{2} x}{-\left (d \,x^{2}-1\right )^{2}+1}+\frac {144 \left (a +b \arcsin \left (d \,x^{2}-1\right )\right ) b^{2} d^{3} x^{3} \left (d \,x^{2}-1\right )}{{\left (-\left (d \,x^{2}-1\right )^{2}+1\right )}^{2}}+\frac {36 {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{2} b \,d^{2} \left (d \,x^{2}-1\right ) x}{{\left (-\left (d \,x^{2}-1\right )^{2}+1\right )}^{\frac {3}{2}}}+\frac {72 {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{2} b \,d^{3} x^{3} \left (d \,x^{2}-1\right )^{2}}{{\left (-\left (d \,x^{2}-1\right )^{2}+1\right )}^{\frac {5}{2}}}+\frac {24 {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{2} b \,d^{3} x^{3}}{{\left (-\left (d \,x^{2}-1\right )^{2}+1\right )}^{\frac {3}{2}}}\right )}{d^{2}}\)	\(438\)

Input:

int((a+b*arcsin(d*x^2-1))^3,x,method=_RETURNVERBOSE)
 

Output:

x*(a+b*arcsin(d*x^2-1))^3-6*(d*x^2-4)*(a+b*arcsin(d*x^2-1))^2*b*x/(-(d*x^2 
-1)^2+1)^(1/2)-2/d*(d*x^2-2)*x*(24*(a+b*arcsin(d*x^2-1))*b^2*d^2*x^2/(-(d* 
x^2-1)^2+1)+6*(a+b*arcsin(d*x^2-1))^2*b*d/(-(d*x^2-1)^2+1)^(1/2)+12*(a+b*a 
rcsin(d*x^2-1))^2*b*d^2*x^2/(-(d*x^2-1)^2+1)^(3/2)*(d*x^2-1))-1/d^2*(d*x^2 
-2)^2*(48*b^3*d^3*x^3/(-(d*x^2-1)^2+1)^(3/2)+72*(a+b*arcsin(d*x^2-1))*b^2* 
d^2*x/(-(d*x^2-1)^2+1)+144*(a+b*arcsin(d*x^2-1))*b^2*d^3*x^3/(-(d*x^2-1)^2 
+1)^2*(d*x^2-1)+36*(a+b*arcsin(d*x^2-1))^2*b*d^2/(-(d*x^2-1)^2+1)^(3/2)*(d 
*x^2-1)*x+72*(a+b*arcsin(d*x^2-1))^2*b*d^3*x^3/(-(d*x^2-1)^2+1)^(5/2)*(d*x 
^2-1)^2+24*(a+b*arcsin(d*x^2-1))^2*b*d^3*x^3/(-(d*x^2-1)^2+1)^(3/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.25 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^3 \, dx=\frac {b^{3} d x^{2} \arcsin \left (d x^{2} - 1\right )^{3} + 3 \, a b^{2} d x^{2} \arcsin \left (d x^{2} - 1\right )^{2} + 3 \, {\left (a^{2} b - 8 \, b^{3}\right )} d x^{2} \arcsin \left (d x^{2} - 1\right ) + {\left (a^{3} - 24 \, a b^{2}\right )} d x^{2} + 6 \, \sqrt {-d^{2} x^{4} + 2 \, d x^{2}} {\left (b^{3} \arcsin \left (d x^{2} - 1\right )^{2} + 2 \, a b^{2} \arcsin \left (d x^{2} - 1\right ) + a^{2} b - 8 \, b^{3}\right )}}{d x} \] Input:

integrate((a+b*arcsin(d*x^2-1))^3,x, algorithm="fricas")
 

Output:

(b^3*d*x^2*arcsin(d*x^2 - 1)^3 + 3*a*b^2*d*x^2*arcsin(d*x^2 - 1)^2 + 3*(a^ 
2*b - 8*b^3)*d*x^2*arcsin(d*x^2 - 1) + (a^3 - 24*a*b^2)*d*x^2 + 6*sqrt(-d^ 
2*x^4 + 2*d*x^2)*(b^3*arcsin(d*x^2 - 1)^2 + 2*a*b^2*arcsin(d*x^2 - 1) + a^ 
2*b - 8*b^3))/(d*x)
 

Sympy [F]

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

integrate((a+b*asin(d*x**2-1))**3,x)
 

Output:

Integral((a + b*asin(d*x**2 - 1))**3, x)
 

Maxima [F]

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

integrate((a+b*arcsin(d*x^2-1))^3,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (106) = 212\).

Time = 0.52 (sec) , antiderivative size = 346, normalized size of antiderivative = 3.01 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^3 \, dx=3 \, {\left (x \arcsin \left (d x^{2} - 1\right ) - \frac {2 \, \sqrt {2} \mathrm {sgn}\left (x\right )}{\sqrt {d}} + \frac {2 \, \sqrt {-d^{2} x^{2} + 2 \, d}}{d \mathrm {sgn}\left (x\right )}\right )} a^{2} b + 3 \, {\left (x \arcsin \left (d x^{2} - 1\right )^{2} + \frac {2 \, {\left (\sqrt {2} \pi \sqrt {d} {\left | d \right |} - 4 \, \sqrt {2} d^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{d {\left | d \right |}} + \frac {4 \, {\left (\sqrt {-d^{2} x^{2} + 2 \, d} \arcsin \left (d x^{2} - 1\right ) + \frac {2 \, {\left (\sqrt {2} \sqrt {d} - \sqrt {d^{2} x^{2}}\right )} d}{{\left | d \right |}}\right )}}{d \mathrm {sgn}\left (x\right )}\right )} a b^{2} + \frac {1}{2} \, {\left (2 \, x \arcsin \left (d x^{2} - 1\right )^{3} - \frac {3 \, {\left (\sqrt {2} \pi ^{2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \pi \sqrt {d} {\left | d \right |} - 32 \, \sqrt {2} d^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{d^{2}} + \frac {12 \, {\left (\sqrt {-d^{2} x^{2} + 2 \, d} \arcsin \left (d x^{2} - 1\right )^{2} - \frac {2 \, {\left (2 \, \sqrt {d^{2} x^{2}} \arcsin \left (\frac {d^{2} x^{2} - d}{d}\right ) - \frac {4 \, {\left (\sqrt {2} \sqrt {d} - \sqrt {-d^{2} x^{2} + 2 \, d}\right )} d}{{\left | d \right |}} - \frac {\sqrt {2} \pi \sqrt {d} {\left | d \right |} - 4 \, \sqrt {2} d^{\frac {3}{2}}}{{\left | d \right |}}\right )} d}{{\left | d \right |}}\right )}}{d \mathrm {sgn}\left (x\right )}\right )} b^{3} + a^{3} x \] Input:

integrate((a+b*arcsin(d*x^2-1))^3,x, algorithm="giac")
 

Output:

3*(x*arcsin(d*x^2 - 1) - 2*sqrt(2)*sgn(x)/sqrt(d) + 2*sqrt(-d^2*x^2 + 2*d) 
/(d*sgn(x)))*a^2*b + 3*(x*arcsin(d*x^2 - 1)^2 + 2*(sqrt(2)*pi*sqrt(d)*abs( 
d) - 4*sqrt(2)*d^(3/2))*sgn(x)/(d*abs(d)) + 4*(sqrt(-d^2*x^2 + 2*d)*arcsin 
(d*x^2 - 1) + 2*(sqrt(2)*sqrt(d) - sqrt(d^2*x^2))*d/abs(d))/(d*sgn(x)))*a* 
b^2 + 1/2*(2*x*arcsin(d*x^2 - 1)^3 - 3*(sqrt(2)*pi^2*d^(3/2) + 8*sqrt(2)*p 
i*sqrt(d)*abs(d) - 32*sqrt(2)*d^(3/2))*sgn(x)/d^2 + 12*(sqrt(-d^2*x^2 + 2* 
d)*arcsin(d*x^2 - 1)^2 - 2*(2*sqrt(d^2*x^2)*arcsin((d^2*x^2 - d)/d) - 4*(s 
qrt(2)*sqrt(d) - sqrt(-d^2*x^2 + 2*d))*d/abs(d) - (sqrt(2)*pi*sqrt(d)*abs( 
d) - 4*sqrt(2)*d^(3/2))/abs(d))*d/abs(d))/(d*sgn(x)))*b^3 + a^3*x
 

Mupad [F(-1)]

Timed out. \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^3 \, dx=\int {\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^3 \,d x \] Input:

int((a + b*asin(d*x^2 - 1))^3,x)
 

Output:

int((a + b*asin(d*x^2 - 1))^3, x)
 

Reduce [F]

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

int((a+b*asin(d*x^2-1))^3,x)
 

Output:

3*int(asin(d*x**2 - 1),x)*a**2*b + int(asin(d*x**2 - 1)**3,x)*b**3 + 3*int 
(asin(d*x**2 - 1)**2,x)*a*b**2 + a**3*x