\(\int \frac {1}{a-b \arcsin (1-d x^2)} \, dx\) [287]

Optimal result

Integrand size = 16, antiderivative size = 168 \[ \int \frac {1}{a-b \arcsin \left (1-d x^2\right )} \, dx=\frac {x \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \] Output:

1/2*x*Ci(-1/2*(a+b*arcsin(d*x^2-1))/b)*(cos(1/2*a/b)+sin(1/2*a/b))/b/(cos( 
1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))-1/2*x*(cos(1/2*a/b)-sin(1/2 
*a/b))*Si(1/2*a/b+1/2*arcsin(d*x^2-1))/b/(cos(1/2*arcsin(d*x^2-1))+sin(1/2 
*arcsin(d*x^2-1)))
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.77 \[ \int \frac {1}{a-b \arcsin \left (1-d x^2\right )} \, dx=\frac {\left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right ) \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (-\frac {a}{b}+\arcsin \left (1-d x^2\right )\right )\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )+\left (-\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )\right )}{2 b d x} \] Input:

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

Output:

((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)]))/(2*b*d*x)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 168, 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 = {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.

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

\(\Big \downarrow \) 5315

\(\displaystyle \frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (-\frac {a-b \arcsin \left (1-d x^2\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \arcsin \left (1-d x^2\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}\)

Input:

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

Output:

(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*(Co 
s[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]))
 

Defintions of rubi rules used

rule 5315

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]
 

Maple [F]

\[\int \frac {1}{a +b \arcsin \left (d \,x^{2}-1\right )}d x\]

Input:

int(1/(a+b*arcsin(d*x^2-1)),x)
 

Output:

int(1/(a+b*arcsin(d*x^2-1)),x)
 

Fricas [F]

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

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

Output:

integral(1/(b*arcsin(d*x^2 - 1) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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