3.319 \(\int \frac{1}{(a+i b \sin ^{-1}(1-i d x^2))^2} \, dx\)

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

[Out]

-Sqrt[(2*I)*d*x^2 + d^2*x^4]/(2*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*(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]))

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Rubi [A]  time = 0.0426255, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {4825} \[ \frac{x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (-\frac{i \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+\frac{x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{\sqrt{d^2 x^4+2 i d x^2}}{2 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[(2*I)*d*x^2 + d^2*x^4]/(2*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*(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 4825

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*
Sin[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[ArcSi
n[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*}

Mathematica [A]  time = 1.3219, size = 197, normalized size = 0.8 \[ \frac{\frac{x^2 \left (\left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\sin ^{-1}\left (1-i d x^2\right )-\frac{i a}{b}\right )\right )+\left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}-\frac{2 b \sqrt{d x^2 \left (d x^2+2 i\right )}}{d \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}}{4 b^2 x} \]

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 = 0.064, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( i+d{x}^{2} \right ) \right ) ^{-2}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} -\frac{d^{2} x^{4} + 3 i \, d x^{2} +{\left (d^{\frac{3}{2}} x^{3} + 2 i \, \sqrt{d} x\right )} \sqrt{d x^{2} + 2 i} - 2}{2 \, a b d^{2} x^{3} + 4 i \, a b d x +{\left (2 \, b^{2} d^{2} x^{3} + 4 i \, b^{2} d x +{\left (2 \, b^{2} d^{\frac{3}{2}} x^{2} + 2 i \, b^{2} \sqrt{d}\right )} \sqrt{d x^{2} + 2 i}\right )} \log \left (d x^{2} + \sqrt{d x^{2} + 2 i} \sqrt{d} x + i\right ) +{\left (2 \, a b d^{\frac{3}{2}} x^{2} + 2 i \, a b \sqrt{d}\right )} \sqrt{d x^{2} + 2 i}} + \int \frac{2 \, d^{3} x^{6} + 6 i \, d^{2} x^{4} +{\left (2 \, d^{2} x^{4} + 2 i \, d x^{2} - 4\right )}{\left (d x^{2} + 2 i\right )} + 2 \,{\left (2 \, d^{\frac{5}{2}} x^{5} + 4 i \, d^{\frac{3}{2}} x^{3} - \sqrt{d} x\right )} \sqrt{d x^{2} + 2 i} + 8 i}{4 \, a b d^{3} x^{6} + 16 i \, a b d^{2} x^{4} - 16 \, a b d x^{2} +{\left (4 \, a b d^{2} x^{4} + 8 i \, a b d x^{2} - 4 \, a b\right )}{\left (d x^{2} + 2 i\right )} +{\left (4 \, b^{2} d^{3} x^{6} + 16 i \, b^{2} d^{2} x^{4} - 16 \, b^{2} d x^{2} + 4 \,{\left (b^{2} d^{2} x^{4} + 2 i \, b^{2} d x^{2} - b^{2}\right )}{\left (d x^{2} + 2 i\right )} +{\left (8 \, b^{2} d^{\frac{5}{2}} x^{5} + 24 i \, b^{2} d^{\frac{3}{2}} x^{3} - 16 \, b^{2} \sqrt{d} x\right )} \sqrt{d x^{2} + 2 i}\right )} \log \left (d x^{2} + \sqrt{d x^{2} + 2 i} \sqrt{d} x + i\right ) +{\left (8 \, a b d^{\frac{5}{2}} x^{5} + 24 i \, a b d^{\frac{3}{2}} x^{3} - 16 \, a b \sqrt{d} x\right )} \sqrt{d x^{2} + 2 i}}\,{d x} \end{align*}

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]

-(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)/(2*a*b*d^2*x^3 + 4*I*a*b*d*x + (2
*b^2*d^2*x^3 + 4*I*b^2*d*x + (2*b^2*d^(3/2)*x^2 + 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) + (2*a*b*d^(3/2)*x^2 + 2*I*a*b*sqrt(d))*sqrt(d*x^2 + 2*I)) + integrate((2*d^3*x^6 + 6*I*d
^2*x^4 + (2*d^2*x^4 + 2*I*d*x^2 - 4)*(d*x^2 + 2*I) + 2*(2*d^(5/2)*x^5 + 4*I*d^(3/2)*x^3 - sqrt(d)*x)*sqrt(d*x^
2 + 2*I) + 8*I)/(4*a*b*d^3*x^6 + 16*I*a*b*d^2*x^4 - 16*a*b*d*x^2 + (4*a*b*d^2*x^4 + 8*I*a*b*d*x^2 - 4*a*b)*(d*
x^2 + 2*I) + (4*b^2*d^3*x^6 + 16*I*b^2*d^2*x^4 - 16*b^2*d*x^2 + 4*(b^2*d^2*x^4 + 2*I*b^2*d*x^2 - b^2)*(d*x^2 +
2*I) + (8*b^2*d^(5/2)*x^5 + 24*I*b^2*d^(3/2)*x^3 - 16*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) + (8*a*b*d^(5/2)*x^5 + 24*I*a*b*d^(3/2)*x^3 - 16*a*b*sqrt(d)*x)*sqrt(d*x^2 + 2*I)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (b^{2} d x \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) + a b d x\right )}{\rm integral}\left (\frac{\sqrt{d^{2} x^{4} + 2 i \, d x^{2}}}{2 \, a b d x^{2} + 4 i \, a b +{\left (2 \, b^{2} d x^{2} + 4 i \, b^{2}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right )}, x\right ) - \sqrt{d^{2} x^{4} + 2 i \, d x^{2}}}{2 \,{\left (b^{2} d x \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) + a b d x\right )}} \end{align*}

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

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

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsinh}\left (d x^{2} + i\right ) + a\right )}^{2}}\,{d x} \end{align*}

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]

integrate((b*arcsinh(d*x^2 + I) + a)^(-2), x)