Integrand size = 16, antiderivative size = 240 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {x \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \] Output:
-1/4*(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2-1))^2+1/8*x/b^2/(a+b *arcsin(d*x^2-1))-1/16*x*Ci(-1/2*(a+b*arcsin(d*x^2-1))/b)*(cos(1/2*a/b)+si n(1/2*a/b))/b^3/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))+1/16*x *(cos(1/2*a/b)-sin(1/2*a/b))*Si(1/2*a/b+1/2*arcsin(d*x^2-1))/b^3/(cos(1/2* arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))
Time = 0.42 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=-\frac {\frac {4 b^2 \sqrt {-d x^2 \left (-2+d x^2\right )}}{d \left (a-b \arcsin \left (1-d x^2\right )\right )^2}-\frac {2 b x^2}{a-b \arcsin \left (1-d x^2\right )}+\frac {\left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right ) \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (-\frac {a}{b}+\arcsin \left (1-d x^2\right )\right )\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )+\left (-\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )\right )}{d}}{16 b^3 x} \] Input:
Integrate[(a - b*ArcSin[1 - d*x^2])^(-3),x]
Output:
-1/16*((4*b^2*Sqrt[-(d*x^2*(-2 + d*x^2))])/(d*(a - b*ArcSin[1 - d*x^2])^2) - (2*b*x^2)/(a - b*ArcSin[1 - d*x^2]) + ((Cos[ArcSin[1 - d*x^2]/2] - Sin[ ArcSin[1 - d*x^2]/2])*(CosIntegral[(-(a/b) + ArcSin[1 - d*x^2])/2]*(Cos[a/ (2*b)] + Sin[a/(2*b)]) + (-Cos[a/(2*b)] + Sin[a/(2*b)])*SinIntegral[(a - b *ArcSin[1 - d*x^2])/(2*b)]))/d)/(b^3*x)
Time = 0.36 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5327, 5315}
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.
\(\displaystyle \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx\) |
\(\Big \downarrow \) 5327 |
\(\displaystyle -\frac {\int \frac {1}{a-b \arcsin \left (1-d x^2\right )}dx}{8 b^2}+\frac {x}{8 b^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^2}\) |
\(\Big \downarrow \) 5315 |
\(\displaystyle -\frac {\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}}{8 b^2}+\frac {x}{8 b^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^2}\) |
Input:
Int[(a - b*ArcSin[1 - d*x^2])^(-3),x]
Output:
-1/4*Sqrt[2*d*x^2 - d^2*x^4]/(b*d*x*(a - b*ArcSin[1 - d*x^2])^2) + x/(8*b^ 2*(a - b*ArcSin[1 - d*x^2])) - ((x*CosIntegral[-1/2*(a - b*ArcSin[1 - d*x^ 2])/b]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(2*b*(Cos[ArcSin[1 - d*x^2]/2] - Sin [ArcSin[1 - d*x^2]/2])) - (x*(Cos[a/(2*b)] - Sin[a/(2*b)])*SinIntegral[a/( 2*b) - ArcSin[1 - d*x^2]/2])/(2*b*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])))/(8*b^2)
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(-x )*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*(CosIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))), x] - Simp[x*(c*Cos[a/(2*b)] + Sin[a/(2*b)])*(SinIntegral[(c/(2*b))*(a + b*Arc Sin[c + d*x^2])]/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2 ]))), x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*( (a + b*ArcSin[c + d*x^2])^(n + 2)/(4*b^2*(n + 1)*(n + 2))), x] + (Simp[Sqrt [-2*c*d*x^2 - d^2*x^4]*((a + b*ArcSin[c + d*x^2])^(n + 1)/(2*b*d*(n + 1)*x) ), x] - Simp[1/(4*b^2*(n + 1)*(n + 2)) Int[(a + b*ArcSin[c + d*x^2])^(n + 2), x], x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[ n, -2]
\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{3}}d x\]
Input:
int(1/(a+b*arcsin(d*x^2-1))^3,x)
Output:
int(1/(a+b*arcsin(d*x^2-1))^3,x)
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(1/(a+b*arcsin(d*x^2-1))^3,x, algorithm="fricas")
Output:
integral(1/(b^3*arcsin(d*x^2 - 1)^3 + 3*a*b^2*arcsin(d*x^2 - 1)^2 + 3*a^2* b*arcsin(d*x^2 - 1) + a^3), x)
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{3}}\, dx \] Input:
integrate(1/(a+b*asin(d*x**2-1))**3,x)
Output:
Integral((a + b*asin(d*x**2 - 1))**(-3), x)
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(1/(a+b*arcsin(d*x^2-1))^3,x, algorithm="maxima")
Output:
1/8*(b*d*x*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*d*x - 2*sqrt (-d*x^2 + 2)*b*sqrt(d) - 8*(b^4*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt (d)*x)^2 + 2*a*b^3*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a^2* b^2*d)*integrate(1/8/(b^3*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*b^2), x))/(b^4*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^2 + 2*a *b^3*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a^2*b^2*d)
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(1/(a+b*arcsin(d*x^2-1))^3,x, algorithm="giac")
Output:
integrate((b*arcsin(d*x^2 - 1) + a)^(-3), x)
Timed out. \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^3} \,d x \] Input:
int(1/(a + b*asin(d*x^2 - 1))^3,x)
Output:
int(1/(a + b*asin(d*x^2 - 1))^3, x)
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^3} \, dx=\int \frac {1}{\mathit {asin} \left (d \,x^{2}-1\right )^{3} b^{3}+3 \mathit {asin} \left (d \,x^{2}-1\right )^{2} a \,b^{2}+3 \mathit {asin} \left (d \,x^{2}-1\right ) a^{2} b +a^{3}}d x \] Input:
int(1/(a+b*asin(d*x^2-1))^3,x)
Output:
int(1/(asin(d*x**2 - 1)**3*b**3 + 3*asin(d*x**2 - 1)**2*a*b**2 + 3*asin(d* x**2 - 1)*a**2*b + a**3),x)