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

Optimal. Leaf size=262 \[ -\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+x \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )} \]

[Out]

x*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]] + (Sqrt[Pi]*x*FresnelS[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/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[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/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 = 0.0275163, antiderivative size = 262, 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 = {4811} \[ -\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+x \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]] + (Sqrt[Pi]*x*FresnelS[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/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[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/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 4811

Int[Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[x*Sqrt[a + b*ArcSin[c + d*x^2]], x] + (
-Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqr
t[c/b]*(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[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[c/b]*(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 \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )} \, dx &=x \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}+\frac{\sqrt{\pi } x S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right ) \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right )}{\sqrt{\frac{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{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right ) \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right )}{\sqrt{\frac{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 = 0.0518128, size = 259, normalized size = 0.99 \[ \frac{x \left (-\sqrt{\pi } \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right )+\sqrt{\pi } \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{\pi }}\right )+\sqrt{\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right ) \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}\right )}{\sqrt{\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2]) + Sq
rt[Pi]*FresnelS[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] - I*Sinh[a/(2*b)]) - Sq
rt[Pi]*FresnelC[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/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 = 0.062, size = 0, normalized size = 0. \begin{align*} \int \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((a+b*arcsinh(-I+d*x^2))^(1/2),x)

[Out]

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

[Out]

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

[Out]

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