3.336 \(\int (a-i b \sin ^{-1}(1+i d x^2))^{3/2} \, dx\)

Optimal. Leaf size=310 \[ -\frac{3 \sqrt{\pi } b^2 x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{-i b} \sqrt{\pi }}\right )}{\sqrt{-i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac{3 b \sqrt{d^2 x^4-2 i d x^2} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{d x}-\frac{3 \sqrt{\pi } \sqrt{-i b} b x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi } \sqrt{-i b}}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2} \]

[Out]

(-3*b*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/(d*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^
(3/2) - (3*b^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[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])) - (3*Sqrt[(-I)*b]*
b*Sqrt[Pi]*x*FresnelC[Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]/(Sqrt[(-I)*b]*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])

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Rubi [A]  time = 0.0962973, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4814, 4819} \[ -\frac{3 \sqrt{\pi } b^2 x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{-i b} \sqrt{\pi }}\right )}{\sqrt{-i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac{3 b \sqrt{d^2 x^4-2 i d x^2} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{d x}-\frac{3 \sqrt{\pi } \sqrt{-i b} b x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi } \sqrt{-i b}}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/(d*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^
(3/2) - (3*b^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[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])) - (3*Sqrt[(-I)*b]*
b*Sqrt[Pi]*x*FresnelC[Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]/(Sqrt[(-I)*b]*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 4814

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcSin[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcSin[c + d*x^2])^(n - 2), x], x] + Simp[(2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*(
a + b*ArcSin[c + d*x^2])^(n - 1))/(d*x), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rule 4819

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[a + b*ArcSin[c + d*x^2]])/(Sqrt[b*c]*Sqrt[Pi])])/(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 \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2} \, dx &=-\frac{3 b \sqrt{-2 i d x^2+d^2 x^4} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}+\left (3 b^2\right ) \int \frac{1}{\sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}} \, dx\\ &=-\frac{3 b \sqrt{-2 i d x^2+d^2 x^4} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}-\frac{3 b^2 \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 )}-\frac{3 \sqrt{-i b} b \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 (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*}

Mathematica [A]  time = 0.252598, size = 255, normalized size = 0.82 \[ -\frac{3 \sqrt{\pi } (-i b)^{3/2} x \left (\left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) \left (-\text{FresnelC}\left (\frac{\sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi } \sqrt{-i b}}\right )\right )-\left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{-i b} \sqrt{\pi }}\right )\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}-\frac{3 b \sqrt{d x^2 \left (d x^2-2 i\right )} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*Sqrt[d*x^2*(-2*I + d*x^2)]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/(d*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^(3
/2) - (3*((-I)*b)^(3/2)*Sqrt[Pi]*x*(-(FresnelC[Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]/(Sqrt[(-I)*b]*Sqrt[Pi])]*(Cos
h[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)])))/(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( -i+d{x}^{2} \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x^{2} - i\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(-I+d*x^2))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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((a+b*asinh(-I+d*x**2))**(3/2),x)

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x^{2} - i\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(-I+d*x^2))^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x^2 - I) + a)^(3/2), x)