3.4.38 \(\int \frac {1}{\sqrt {a-i b \text {ArcSin}(1+i d x^2)}} \, dx\) [338]

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

[Out]

-x*FresnelC((a-I*b*arcsin(1+I*d*x^2))^(1/2)/(-I*b)^(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)-x*FresnelS((a-I*b*arcsin(1+I*d*x^2))^(1/2)/
(-I*b)^(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)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 231, 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 = {4903} \begin {gather*} -\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 \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi } \sqrt {-i b}}\right )}{\sqrt {-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 {\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 \text {ArcSin}\left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )}{\sqrt {-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]

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

[Out]

-((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]))) - (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[Arc
Sin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2]))

Rule 4903

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[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] - 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 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 )}-\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 )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.00, size = 180, normalized size = 0.78 \begin {gather*} \frac {\sqrt {\pi } x \left (-\text {FresnelC}\left (\frac {\sqrt {a-i b \text {ArcSin}\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 )-S\left (\frac {\sqrt {a-i b \text {ArcSin}\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 )\right )}{\sqrt {-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[1/Sqrt[a - I*b*ArcSin[1 + I*d*x^2]],x]

[Out]

(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)])) - 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]))

________________________________________________________________________________________

Maple [F]
time = 0.90, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {a +b \arcsinh \left (d \,x^{2}-i\right )}}\, dx\]

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)

________________________________________________________________________________________

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))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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))^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real -I

________________________________________________________________________________________

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))^(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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{\sqrt {a+b\,\mathrm {asinh}\left (d\,x^2-\mathrm {i}\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________