3.4.30 \(\int \sqrt {a+i b \text {ArcSin}(1-i d x^2)} \, dx\) [330]

Optimal. Leaf size=263 \[ x \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}+\frac {\sqrt {\pi } x S\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {-\frac {i}{b}} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )}-\frac {\sqrt {-\frac {i}{b}} b \sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )} \]

[Out]

x*FresnelS((-I/b)^(1/2)*(a-I*b*arcsin(-1+I*d*x^2))^(1/2)/Pi^(1/2))*(cosh(1/2*a/b)+I*sinh(1/2*a/b))*Pi^(1/2)/(c
os(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))/(-I/b)^(1/2)-b*x*FresnelC((-I/b)^(1/2)*(a-I*b*arcsin(-
1+I*d*x^2))^(1/2)/Pi^(1/2))*(I*cosh(1/2*a/b)+sinh(1/2*a/b))*(-I/b)^(1/2)*Pi^(1/2)/(cos(1/2*arcsin(-1+I*d*x^2))
+sin(1/2*arcsin(-1+I*d*x^2)))+x*(a-I*b*arcsin(-1+I*d*x^2))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 263, normalized size of antiderivative = 1.00, 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 = {4895} \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-\frac {i}{b}} b x \left (\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\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 )}+\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 \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {i}{b}} \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 )}+x \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )} \end {gather*}

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

Rule 4895

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]]]/(Sqrt
[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[A
rcSin[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 {-\frac {i}{b}} b \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 (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\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 )}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 259, normalized size = 0.98 \begin {gather*} \frac {x \left (\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )} \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 )-\sqrt {\pi } \text {FresnelC}\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\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 {\pi } S\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\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 )\right )}{\sqrt {-\frac {i}{b}} \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 )} \end {gather*}

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]) -
Sqrt[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)]
) + Sqrt[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)])))/(Sqrt[(-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 = 0.92, size = 0, normalized size = 0.00 \[\int \sqrt {a +b \arcsinh \left (d \,x^{2}+i\right )}\, dx\]

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 = 0.00, 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((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 = 0.00, 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((a+b*arcsinh(I+d*x^2))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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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((a+b*asinh(I+d*x**2))**(1/2),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((a+b*arcsinh(I+d*x^2))^(1/2),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 \sqrt {a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(d*x^2 + 1i))^(1/2),x)

[Out]

int((a + b*asinh(d*x^2 + 1i))^(1/2), x)

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