Integrand size = 20, antiderivative size = 245 \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx=-\frac {\sqrt {2 i d x^2+d^2 x^4}}{2 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )}+\frac {x \operatorname {CosIntegral}\left (-\frac {i \left (a+i b \arcsin \left (1-i d x^2\right )\right )}{2 b}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}+\frac {x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )} \]
1/4*x*Ci(-1/2*I*(a-I*b*arcsin(-1+I*d*x^2))/b)*(cosh(1/2*a/b)-I*sinh(1/2*a/ b))/b^2/(cos(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))+1/4*x*Si (1/2*I*a/b+1/2*arcsin(-1+I*d*x^2))*(cosh(1/2*a/b)+I*sinh(1/2*a/b))/b^2/(co s(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))-1/2*(2*I*d*x^2+d^2* x^4)^(1/2)/b/d/x/(a-I*b*arcsin(-1+I*d*x^2))
Time = 1.10 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx=\frac {-\frac {2 b \sqrt {d x^2 \left (2 i+d x^2\right )}}{d \left (a+i b \arcsin \left (1-i d x^2\right )\right )}+\frac {x^2 \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (-\frac {i a}{b}+\arcsin \left (1-i d x^2\right )\right )\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )+\left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}}{4 b^2 x} \]
((-2*b*Sqrt[d*x^2*(2*I + d*x^2)])/(d*(a + I*b*ArcSin[1 - I*d*x^2])) + (x^2 *(CosIntegral[(((-I)*a)/b + ArcSin[1 - I*d*x^2])/2]*(Cosh[a/(2*b)] - I*Sin h[a/(2*b)]) + (Cosh[a/(2*b)] + I*Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b - ArcSin[1 - I*d*x^2]/2]))/(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d* x^2]/2]))/(4*b^2*x)
Time = 0.29 (sec) , antiderivative size = 245, 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 = {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.
\(\displaystyle \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx\) |
\(\Big \downarrow \) 5324 |
\(\displaystyle \frac {x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (-\frac {i \left (a+i b \arcsin \left (1-i d x^2\right )\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}+\frac {x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {\sqrt {d^2 x^4+2 i d x^2}}{2 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )}\) |
-1/2*Sqrt[(2*I)*d*x^2 + d^2*x^4]/(b*d*x*(a + I*b*ArcSin[1 - I*d*x^2])) + ( x*CosIntegral[((-1/2*I)*(a + I*b*ArcSin[1 - I*d*x^2]))/b]*(Cosh[a/(2*b)] - I*Sinh[a/(2*b)]))/(4*b^2*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d *x^2]/2])) + (x*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b - ArcSin[1 - I*d*x^2]/2])/(4*b^2*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[ 1 - I*d*x^2]/2]))
3.4.19.3.1 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]
\[\int \frac {1}{{\left (a +b \,\operatorname {arcsinh}\left (d \,x^{2}+i\right )\right )}^{2}}d x\]
\[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x^{2} + i\right ) + a\right )}^{2}} \,d x } \]
1/2*(2*(b^2*d*log(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I) + a*b*d)*integral(1 /2*sqrt(d^2*x^2 + 2*I*d)*x/(a*b*d*x^2 + 2*I*a*b + (b^2*d*x^2 + 2*I*b^2)*lo g(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I)), x) - sqrt(d^2*x^2 + 2*I*d))/(b^2* d*log(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I) + a*b*d)
Exception generated. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx=\text {Exception raised: TypeError} \]
\[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x^{2} + i\right ) + a\right )}^{2}} \,d x } \]
-1/2*(d^2*x^4 + 3*I*d*x^2 + (d^(3/2)*x^3 + 2*I*sqrt(d)*x)*sqrt(d*x^2 + 2*I ) - 2)/(a*b*d^2*x^3 + 2*I*a*b*d*x + (b^2*d^2*x^3 + 2*I*b^2*d*x + (b^2*d^(3 /2)*x^2 + I*b^2*sqrt(d))*sqrt(d*x^2 + 2*I))*log(d*x^2 + sqrt(d*x^2 + 2*I)* sqrt(d)*x + I) + (a*b*d^(3/2)*x^2 + I*a*b*sqrt(d))*sqrt(d*x^2 + 2*I)) + in tegrate(1/2*(d^3*x^6 + 3*I*d^2*x^4 + (d^2*x^4 + I*d*x^2 - 2)*(d*x^2 + 2*I) + (2*d^(5/2)*x^5 + 4*I*d^(3/2)*x^3 - sqrt(d)*x)*sqrt(d*x^2 + 2*I) + 4*I)/ (a*b*d^3*x^6 + 4*I*a*b*d^2*x^4 - 4*a*b*d*x^2 + (a*b*d^2*x^4 + 2*I*a*b*d*x^ 2 - a*b)*(d*x^2 + 2*I) + (b^2*d^3*x^6 + 4*I*b^2*d^2*x^4 - 4*b^2*d*x^2 + (b ^2*d^2*x^4 + 2*I*b^2*d*x^2 - b^2)*(d*x^2 + 2*I) + 2*(b^2*d^(5/2)*x^5 + 3*I *b^2*d^(3/2)*x^3 - 2*b^2*sqrt(d)*x)*sqrt(d*x^2 + 2*I))*log(d*x^2 + sqrt(d* x^2 + 2*I)*sqrt(d)*x + I) + 2*(a*b*d^(5/2)*x^5 + 3*I*a*b*d^(3/2)*x^3 - 2*a *b*sqrt(d)*x)*sqrt(d*x^2 + 2*I)), x)
Exception generated. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^2} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^2} \,d x \]