Integrand size = 16, antiderivative size = 216 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx=-\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )}-\frac {x \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {x \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \]
-1/4*x*Ci(1/2*(-a-b*arcsin(d*x^2-1))/b)*(cos(1/2*a/b)-sin(1/2*a/b))/b^2/(c os(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))-1/4*x*Si(1/2*a/b+1/2*arc sin(d*x^2-1))*(cos(1/2*a/b)+sin(1/2*a/b))/b^2/(cos(1/2*arcsin(d*x^2-1))+si n(1/2*arcsin(d*x^2-1)))-1/2*(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x ^2-1))
Time = 0.53 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx=\frac {2 b \sqrt {d x^2 \left (2-d x^2\right )}+\left (a-b \arcsin \left (1-d x^2\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right ) \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (-\frac {a}{b}+\arcsin \left (1-d x^2\right )\right )\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )+\left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )\right )}{4 b^2 d x \left (-a+b \arcsin \left (1-d x^2\right )\right )} \]
(2*b*Sqrt[d*x^2*(2 - d*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)])*SinI ntegral[(a - b*ArcSin[1 - d*x^2])/(2*b)]))/(4*b^2*d*x*(-a + b*ArcSin[1 - d *x^2]))
Time = 0.25 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {5324}
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.
\(\displaystyle \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx\) |
\(\Big \downarrow \) 5324 |
\(\displaystyle -\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )}\) |
-1/2*Sqrt[2*d*x^2 - d^2*x^4]/(b*d*x*(a - b*ArcSin[1 - d*x^2])) - (x*CosInt egral[-1/2*(a - b*ArcSin[1 - d*x^2])/b]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(4* b^2*(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])/(4*b^2*(Co s[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]))
3.5.13.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> Simp[-Sq rt[-2*c*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*ArcSin[c + d*x^2])), x] + (-Simp[x *(Cos[a/(2*b)] + c*Sin[a/(2*b)])*(CosIntegral[(c/(2*b))*(a + b*ArcSin[c + d *x^2])]/(4*b^2*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))), x ] + Simp[x*(Cos[a/(2*b)] - c*Sin[a/(2*b)])*(SinIntegral[(c/(2*b))*(a + b*Ar cSin[c + d*x^2])]/(4*b^2*(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]
\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{2}}d x\]
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{2}}\, dx \]
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{2}} \,d x } \]
1/2*(2*(b^2*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*b*d)*sqrt (d)*integrate(1/2*sqrt(-d*x^2 + 2)*x/(a*b*d*x^2 - 2*a*b + (b^2*d*x^2 - 2*b ^2)*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)), x) - sqrt(-d*x^2 + 2) *sqrt(d))/(b^2*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*b*d)
\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^2} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^2} \,d x \]