3.4.18 \(\int \frac {1}{a+i b \text {ArcSin}(1-i d x^2)} \, dx\) [318]

Optimal. Leaf size=194 \[ \frac {x \text {CosIntegral}\left (-\frac {i \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )}{2 b}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )}-\frac {x \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )} \]

[Out]

1/2*x*Ci(-1/2*I*(a-I*b*arcsin(-1+I*d*x^2))/b)*(I*cosh(1/2*a/b)-sinh(1/2*a/b))/b/(cos(1/2*arcsin(-1+I*d*x^2))+s
in(1/2*arcsin(-1+I*d*x^2)))-1/2*x*Si(1/2*I*a/b+1/2*arcsin(-1+I*d*x^2))*(I*cosh(1/2*a/b)+sinh(1/2*a/b))/b/(cos(
1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))

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Rubi [A]
time = 0.04, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4900} \begin {gather*} \frac {x \left (-\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {CosIntegral}\left (-\frac {i \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )}{2 b}\right )}{2 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 {x \left (\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )}{2 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[(a + I*b*ArcSin[1 - I*d*x^2])^(-1),x]

[Out]

(x*CosIntegral[((-1/2*I)*(a + I*b*ArcSin[1 - I*d*x^2]))/b]*(I*Cosh[a/(2*b)] - Sinh[a/(2*b)]))/(2*b*(Cos[ArcSin
[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) - (x*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b
- ArcSin[1 - I*d*x^2]/2])/(2*b*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))

Rule 4900

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(-x)*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*(
CosIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))
), x] - Simp[x*(c*Cos[a/(2*b)] + Sin[a/(2*b)])*(SinIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[Arc
Sin[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}{a+i b \sin ^{-1}\left (1-i d x^2\right )} \, dx &=\frac {x \text {Ci}\left (-\frac {i \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )}{2 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 {x \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}{2 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*}

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Mathematica [A]
time = 0.52, size = 150, normalized size = 0.77 \begin {gather*} \frac {x \left (\text {CosIntegral}\left (\frac {1}{2} \left (-\frac {i a}{b}+\text {ArcSin}\left (1-i d x^2\right )\right )\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )+\left (-i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )}{2 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[(a + I*b*ArcSin[1 - I*d*x^2])^(-1),x]

[Out]

(x*(CosIntegral[(((-I)*a)/b + ArcSin[1 - I*d*x^2])/2]*(I*Cosh[a/(2*b)] - Sinh[a/(2*b)]) + ((-I)*Cosh[a/(2*b)]
- Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b - ArcSin[1 - I*d*x^2]/2]))/(2*b*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[Arc
Sin[1 - I*d*x^2]/2]))

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Maple [F]
time = 0.92, size = 0, normalized size = 0.00 \[\int \frac {1}{a +b \arcsinh \left (d \,x^{2}+i\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(b*log(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I) + a), x)

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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)),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)),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 = -0.01 \begin {gather*} \int \frac {1}{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(1/(a + b*asinh(d*x^2 + 1i)),x)

[Out]

int(1/(a + b*asinh(d*x^2 + 1i)), x)

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