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

Optimal. Leaf size=231 \[ -\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi } \sqrt{i b}}\right )}{\sqrt{i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right )}{\sqrt{i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )} \]

[Out]

-((Sqrt[Pi]*x*FresnelS[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] - I*Sinh[a/(2*b)
]))/(Sqrt[I*b]*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))) - (Sqrt[Pi]*x*FresnelC[Sqrt[a + I*b
*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(Sqrt[I*b]*(Cos[ArcSin[1 - I*d*
x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))

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Rubi [A]  time = 0.0299369, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {4819} \[ -\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi } \sqrt{i b}}\right )}{\sqrt{i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right )}{\sqrt{i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[Pi]*x*FresnelS[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] - I*Sinh[a/(2*b)
]))/(Sqrt[I*b]*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))) - (Sqrt[Pi]*x*FresnelC[Sqrt[a + I*b
*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(Sqrt[I*b]*(Cos[ArcSin[1 - I*d*
x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))

Rule 4819

Int[1/Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> -Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a/
(2*b)])*FresnelC[(1*Sqrt[a + b*ArcSin[c + d*x^2]])/(Sqrt[b*c]*Sqrt[Pi])])/(Sqrt[b*c]*(Cos[ArcSin[c + d*x^2]/2]
- c*Sin[ArcSin[c + d*x^2]/2])), x] - Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelS[(1/(Sqrt[b*c]*
Sqrt[Pi]))*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[b*c]*(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}{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}} \, dx &=-\frac{\sqrt{\pi } x S\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right ) \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right )}{\sqrt{i 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{\sqrt{\pi } x C\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right ) \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right )}{\sqrt{i 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 = 0.0037899, size = 180, normalized size = 0.78 \[ \frac{\sqrt{\pi } x \left (\left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) \left (-S\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right )\right )-\left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi } \sqrt{i b}}\right )\right )}{\sqrt{i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi]*x*(-(FresnelS[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] - I*Sinh[a/(2*b
)])) - FresnelC[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)])))/(S
qrt[I*b]*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{a+b{\it Arcsinh} \left ( i+d{x}^{2} \right ) }}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{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))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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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))**(1/2),x)

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{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))^(1/2),x, algorithm="giac")

[Out]

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