3.4.19 \(\int \frac {1}{(a+i b \text {ArcSin}(1-i d x^2))^2} \, dx\) [319]

Optimal. Leaf size=245 \[ -\frac {\sqrt {2 i d x^2+d^2 x^4}}{2 b d x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )}+\frac {x \text {CosIntegral}\left (-\frac {i \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )}{2 b}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )}+\frac {x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )} \]

[Out]

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/(
cos(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))

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Rubi [A]
time = 0.03, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4909} \begin {gather*} \frac {x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \text {CosIntegral}\left (-\frac {i \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {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} \text {ArcSin}\left (1-i d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {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 \text {ArcSin}\left (1-i d x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-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*Arc
Sin[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]))

Rule 4909

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> Simp[-Sqrt[-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*S
in[a/(2*b)])*(SinIntegral[(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]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+i b \sin ^{-1}\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 \sin ^{-1}\left (1-i d x^2\right )\right )}+\frac {x \text {Ci}\left (-\frac {i \left (a+i b \sin ^{-1}\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} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\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} \sin ^{-1}\left (1-i d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}\\ \end {align*}

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Mathematica [A]
time = 0.96, size = 197, normalized size = 0.80 \begin {gather*} \frac {-\frac {2 b \sqrt {d x^2 \left (2 i+d x^2\right )}}{d \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )}+\frac {x^2 \left (\text {CosIntegral}\left (\frac {1}{2} \left (-\frac {i a}{b}+\text {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} \text {ArcSin}\left (1-i d x^2\right )\right )\right )}{\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )}}{4 b^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-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/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)

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Maple [F]
time = 0.92, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsinh \left (d \,x^{2}+i\right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)) + integrate(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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*log(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)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real I

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(d*x^2 + 1i))^2,x)

[Out]

int(1/(a + b*asinh(d*x^2 + 1i))^2, x)

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