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

Optimal. Leaf size=191 \[ \frac{x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{Shi}\left (\frac{a-i b \sin ^{-1}\left (i d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac{x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{i \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0228964, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {4816} \[ \frac{x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{Shi}\left (\frac{a-i b \sin ^{-1}\left (i d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac{x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{i \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4816

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> -Simp[(x*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*Co
sIntegral[(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[ArcS
in[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 (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{Shi}\left (\frac{a-i b \sin ^{-1}\left (1+i d x^2\right )}{2 b}\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*}

Mathematica [A]  time = 0.639787, size = 146, normalized size = 0.76 \[ \frac{x \left (\left (-\sinh \left (\frac{a}{2 b}\right )-i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\frac{i a}{b}+\sin ^{-1}\left (1+i d x^2\right )\right )\right )+\left (\sinh \left (\frac{a}{2 b}\right )-i \cosh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{1}{2} \left (\frac{i a}{b}+\sin ^{-1}\left (i d x^2+1\right )\right )\right )\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[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*a)/b + ArcSin[1 + I*d*x^2])/2]))/(2*b*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSi
n[1 + I*d*x^2]/2]))

________________________________________________________________________________________

Maple [F]  time = 0.058, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( -i+d{x}^{2} \right ) \right ) ^{-1}\, dx \end{align*}

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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{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(1/(a+b*arcsinh(-I+d*x^2)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b \log \left (d x^{2} + \sqrt{d^{2} x^{4} - 2 i \, d x^{2}} - i\right ) + a}, x\right ) \end{align*}

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^4 - 2*I*d*x^2) - I) + a), x)

________________________________________________________________________________________

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(1/(a+b*asinh(-I+d*x**2)),x)

[Out]

Exception raised: TypeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{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(1/(a+b*arcsinh(-I+d*x^2)),x, algorithm="giac")

[Out]

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