3.4.32 \(\int \frac {1}{(a+i b \text {ArcSin}(1-i d x^2))^{3/2}} \, dx\) [332]

Optimal. Leaf size=291 \[ -\frac {\sqrt {2 i d x^2+d^2 x^4}}{b d x \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}-\frac {\left (-\frac {i}{b}\right )^{3/2} \sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )}+\frac {\left (-\frac {i}{b}\right )^{3/2} \sqrt {\pi } x S\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )} \]

[Out]

-(-I/b)^(3/2)*x*FresnelC((-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)/(cos(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))+(-I/b)^(3/2)*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)/(cos(1/2*arcsin(-1+I*d*x^2))+
sin(1/2*arcsin(-1+I*d*x^2)))-(2*I*d*x^2+d^2*x^4)^(1/2)/b/d/x/(a-I*b*arcsin(-1+I*d*x^2))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 291, 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 = {4906} \begin {gather*} -\frac {\sqrt {d^2 x^4+2 i d x^2}}{b d x \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}-\frac {\sqrt {\pi } \left (-\frac {i}{b}\right )^{3/2} 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 \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 } \left (-\frac {i}{b}\right )^{3/2} 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 )}{\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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*b*ArcSin[1 - I*d*x^2])^(-3/2),x]

[Out]

-(Sqrt[(2*I)*d*x^2 + d^2*x^4]/(b*d*x*Sqrt[a + I*b*ArcSin[1 - I*d*x^2]])) - (((-I)/b)^(3/2)*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)]))/(Cos[ArcSin[1 -
I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]) + (((-I)/b)^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[(-I)/b]*Sqrt[a + I*b*Ar
cSin[1 - I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 -
I*d*x^2]/2])

Rule 4906

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[-Sqrt[-2*c*d*x^2 - d^2*x^4]/(b*d*x*S
qrt[a + b*ArcSin[c + d*x^2]]), x] + (-Simp[(c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*(FresnelC[Sq
rt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]]/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2])), x] +
Simp[(c/b)^(3/2)*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]]]/(Cos[(1/2)*ArcSin[c + d*x^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}{\left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}} \, dx &=-\frac {\sqrt {2 i d x^2+d^2 x^4}}{b d x \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}-\frac {\left (-\frac {i}{b}\right )^{3/2} \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 )}{\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 {\left (-\frac {i}{b}\right )^{3/2} \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 )}{\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 = 0.25, size = 291, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {2 i d x^2+d^2 x^4}}{b d x \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}-\frac {\left (-\frac {i}{b}\right )^{3/2} \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 (\cosh \left (\frac {a}{2 b}\right )-i \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 )}+\frac {\left (-\frac {i}{b}\right )^{3/2} \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 )}{\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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*b*ArcSin[1 - I*d*x^2])^(-3/2),x]

[Out]

-(Sqrt[(2*I)*d*x^2 + d^2*x^4]/(b*d*x*Sqrt[a + I*b*ArcSin[1 - I*d*x^2]])) - (((-I)/b)^(3/2)*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)]))/(Cos[ArcSin[1 -
I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]) + (((-I)/b)^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[(-I)/b]*Sqrt[a + I*b*Ar
cSin[1 - I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 -
I*d*x^2]/2])

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Maple [F]
time = 0.80, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsinh \left (d \,x^{2}+i\right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsinh(I+d*x^2))^(3/2),x)

[Out]

int(1/(a+b*arcsinh(I+d*x^2))^(3/2),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(1/(a+b*arcsinh(I+d*x^2))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(d*x^2 + I) + a)^(-3/2), 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(1/(a+b*arcsinh(I+d*x^2))^(3/2),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(1/(a+b*asinh(I+d*x**2))**(3/2),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(1/(a+b*arcsinh(I+d*x^2))^(3/2),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 \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(d*x^2 + 1i))^(3/2),x)

[Out]

int(1/(a + b*asinh(d*x^2 + 1i))^(3/2), x)

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