Integrand size = 20, antiderivative size = 194 \[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=\frac {x \operatorname {CosIntegral}\left (-\frac {i \left (a+i b \arcsin \left (1-i d x^2\right )\right )}{2 b}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {x \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )} \] Output:
1/2*x*Ci(-1/2*I*(a-I*b*arcsin(-1+I*d*x^2))/b)*(I*cosh(1/2*a/b)-sinh(1/2*a/ b))/b/(cos(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))-1/2*x*(I*c osh(1/2*a/b)+sinh(1/2*a/b))*Si(1/2*I*a/b+1/2*arcsin(-1+I*d*x^2))/b/(cos(1/ 2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))
Time = 0.69 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.77 \[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=\frac {x \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (-\frac {i a}{b}+\arcsin \left (1-i d x^2\right )\right )\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )+\left (-i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )} \] Input:
Integrate[(a + I*b*ArcSin[1 - I*d*x^2])^(-1),x]
Output:
(x*(CosIntegral[(((-I)*a)/b + ArcSin[1 - I*d*x^2])/2]*(I*Cosh[a/(2*b)] - S inh[a/(2*b)]) + ((-I)*Cosh[a/(2*b)] - Sinh[a/(2*b)])*SinIntegral[((I/2)*a) /b - ArcSin[1 - I*d*x^2]/2]))/(2*b*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSi n[1 - I*d*x^2]/2]))
Time = 0.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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.
\(\displaystyle \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx\) |
\(\Big \downarrow \) 5315 |
\(\displaystyle \frac {x \left (-\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (-\frac {i \left (a+i b \arcsin \left (1-i d x^2\right )\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {x \left (\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}\) |
Input:
Int[(a + I*b*ArcSin[1 - I*d*x^2])^(-1),x]
Output:
(x*CosIntegral[((-1/2*I)*(a + I*b*ArcSin[1 - I*d*x^2]))/b]*(I*Cosh[a/(2*b) ] - Sinh[a/(2*b)]))/(2*b*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d* x^2]/2])) - (x*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b - ArcSin[1 - I*d*x^2]/2])/(2*b*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/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]
\[\int \frac {1}{a -i b \arcsin \left (i d \,x^{2}-1\right )}d x\]
Input:
int(1/(a-I*b*arcsin(-1+I*d*x^2)),x)
Output:
int(1/(a-I*b*arcsin(-1+I*d*x^2)),x)
\[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=\int { \frac {1}{-i \, b \arcsin \left (i \, d x^{2} - 1\right ) + a} \,d x } \] Input:
integrate(1/(a-I*b*arcsin(-1+I*d*x^2)),x, algorithm="fricas")
Output:
integral(2/(b*log(2*d^2*x^4 + 4*I*d*x^2 + 2*sqrt(d^2*x^2 + 2*I*d)*(d*x^3 + I*x) - 1) + 2*a), x)
\[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=\int \frac {1}{a - i b \operatorname {asin}{\left (i d x^{2} - 1 \right )}}\, dx \] Input:
integrate(1/(a-I*b*asin(-1+I*d*x**2)),x)
Output:
Integral(1/(a - I*b*asin(I*d*x**2 - 1)), x)
\[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=\int { \frac {1}{-i \, b \arcsin \left (i \, d x^{2} - 1\right ) + a} \,d x } \] Input:
integrate(1/(a-I*b*arcsin(-1+I*d*x^2)),x, algorithm="maxima")
Output:
integrate(1/(-I*b*arcsin(I*d*x^2 - 1) + a), x)
Exception generated. \[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a-I*b*arcsin(-1+I*d*x^2)),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=\int \frac {1}{a-b\,\mathrm {asin}\left (-1+d\,x^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}} \,d x \] Input:
int(1/(a - b*asin(d*x^2*1i - 1)*1i),x)
Output:
int(1/(a - b*asin(d*x^2*1i - 1)*1i), x)
\[ \int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )} \, dx=-\left (\int \frac {1}{\mathit {asin} \left (d i \,x^{2}-1\right ) b i -a}d x \right ) \] Input:
int(1/(a-I*b*asin(-1+I*d*x^2)),x)
Output:
- int(1/(asin(d*i*x**2 - 1)*b*i - a),x)