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

Optimal. Leaf size=194 \[ \frac{x \left (-\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (-\frac{i \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )} \]

[Out]

(x*CosIntegral[((-I/2)*(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]))

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Rubi [A]  time = 0.0478033, antiderivative size = 194, 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 = {4816} \[ \frac{x \left (-\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (-\frac{i \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*CosIntegral[((-I/2)*(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]))

Rule 4816

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> -Simp[(x*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*Co
sIntegral[(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*ArcSin[c + d*x^2])])/(2*b*(Cos[ArcS
in[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}{a+i b \sin ^{-1}\left (1-i d x^2\right )} \, dx &=\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 (i \cosh \left (\frac{a}{2 b}\right )-\sinh \left (\frac{a}{2 b}\right )\right )}{2 b \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 (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} \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b \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 = 0.711214, size = 150, normalized size = 0.77 \[ \frac{x \left (\left (-\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\sin ^{-1}\left (1-i d x^2\right )-\frac{i a}{b}\right )\right )+\left (-\sinh \left (\frac{a}{2 b}\right )-i \cosh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(CosIntegral[(((-I)*a)/b + ArcSin[1 - I*d*x^2])/2]*(I*Cosh[a/(2*b)] - Sinh[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[Arc
Sin[1 - I*d*x^2]/2]))

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( i+d{x}^{2} \right ) \right ) ^{-1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(b*log(d*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I) + a), 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)),x)

[Out]

Exception raised: TypeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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