\(\int \frac {1}{(a-i b \arcsin (1+i d x^2))^2} \, dx\) [261]

Optimal result

Integrand size = 20, antiderivative size = 244 \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^2} \, dx=-\frac {\sqrt {-2 i d x^2+d^2 x^4}}{2 b d x \left (a-i b \arcsin \left (1+i d x^2\right )\right )}+\frac {x \operatorname {CosIntegral}\left (\frac {i \left (a-i b \arcsin \left (1+i d x^2\right )\right )}{2 b}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {x \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right ) \text {Shi}\left (\frac {a-i b \arcsin \left (1+i d x^2\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \] Output:

-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))+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*(I*cosh(1/2*a/b) 
+sinh(1/2*a/b))*Shi(1/2*(a-I*b*arcsin(1+I*d*x^2))/b)/b^2/(cos(1/2*arcsin(1 
+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^2} \, dx=\frac {-\frac {2 b \sqrt {d x^2 \left (-2 i+d x^2\right )}}{d \left (a-i b \arcsin \left (1+i d x^2\right )\right )}+\frac {x^2 \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (\frac {i a}{b}+\arcsin \left (1+i d x^2\right )\right )\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )-\left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {1}{2} \left (\frac {i a}{b}+\arcsin \left (1+i d x^2\right )\right )\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )}}{4 b^2 x} \] Input:

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

Output:

((-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*Sinh[ 
a/(2*b)]) - (Cosh[a/(2*b)] - I*Sinh[a/(2*b)])*SinIntegral[((I*a)/b + ArcSi 
n[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)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 244, 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

\(\displaystyle \frac {x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (\frac {i \left (a-i b \arcsin \left (i d x^2+1\right )\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {x \left (\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {Shi}\left (\frac {a-i b \arcsin \left (i d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {\sqrt {d^2 x^4-2 i d x^2}}{2 b d x \left (a-i b \arcsin \left (1+i d x^2\right )\right )}\)

Input:

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

Output:

-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[((I/2)*(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*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)])*SinhIntegral[(a - I*b*ArcS 
in[1 + I*d*x^2])/(2*b)])/(4*b^2*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 
 + I*d*x^2]/2]))
 

Defintions of rubi rules used

rule 5324

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

((b^2*d*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*b*d)*integral(sqrt(d^2*x^2 - 2*I*d)*x/(2*a*b*d*x^2 - 4*I*a*b + 
(b^2*d*x^2 - 2*I*b^2)*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)), x) - sqrt(d^2*x^2 - 2*I*d))/(b^2*d*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*b*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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