3.333 \(\int \frac{1}{(a+i b \sin ^{-1}(1-i d x^2))^{5/2}} \, dx\)

Optimal. Leaf size=326 \[ -\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi } \sqrt{i b}}\right )}{3 \sqrt{i b} b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right )}{3 \sqrt{i b} b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x}{3 b^2 \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}-\frac{\sqrt{d^2 x^4+2 i d x^2}}{3 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}} \]

[Out]

-Sqrt[(2*I)*d*x^2 + d^2*x^4]/(3*b*d*x*(a + I*b*ArcSin[1 - I*d*x^2])^(3/2)) - x/(3*b^2*Sqrt[a + I*b*ArcSin[1 -
I*d*x^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*S
inh[a/(2*b)]))/(3*Sqrt[I*b]*b^2*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) - (Sqrt[Pi]*x*Fresn
elC[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(3*Sqrt[I*b]*b^
2*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))

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Rubi [A]  time = 0.0722763, antiderivative size = 326, 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 = {4828, 4819} \[ -\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi } \sqrt{i b}}\right )}{3 \sqrt{i b} b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right )}{3 \sqrt{i b} b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x}{3 b^2 \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}-\frac{\sqrt{d^2 x^4+2 i d x^2}}{3 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*b*ArcSin[1 - I*d*x^2])^(-5/2),x]

[Out]

-Sqrt[(2*I)*d*x^2 + d^2*x^4]/(3*b*d*x*(a + I*b*ArcSin[1 - I*d*x^2])^(3/2)) - x/(3*b^2*Sqrt[a + I*b*ArcSin[1 -
I*d*x^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*S
inh[a/(2*b)]))/(3*Sqrt[I*b]*b^2*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) - (Sqrt[Pi]*x*Fresn
elC[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(3*Sqrt[I*b]*b^
2*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))

Rule 4828

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

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 \frac{1}{\left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{5/2}} \, dx &=-\frac{\sqrt{2 i d x^2+d^2 x^4}}{3 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}-\frac{x}{3 b^2 \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}+\frac{\int \frac{1}{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}} \, dx}{3 b^2}\\ &=-\frac{\sqrt{2 i d x^2+d^2 x^4}}{3 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}-\frac{x}{3 b^2 \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\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 )}{3 \sqrt{i b} b^2 \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 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 )}{3 \sqrt{i b} b^2 \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 = 0.77374, size = 308, normalized size = 0.94 \[ -\frac{\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi } \sqrt{i b}}\right )}{\sqrt{i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+\frac{\sqrt{\pi } x \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{i b} \sqrt{\pi }}\right )}{\sqrt{i b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+\frac{b \sqrt{d x^2 \left (d x^2+2 i\right )}}{d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}+\frac{x}{\sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*b*ArcSin[1 - I*d*x^2])^(-5/2),x]

[Out]

-((b*Sqrt[d*x^2*(2*I + d*x^2)])/(d*x*(a + I*b*ArcSin[1 - I*d*x^2])^(3/2)) + x/Sqrt[a + I*b*ArcSin[1 - I*d*x^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])) + (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])))/(3*b^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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsinh(I+d*x^2))^(5/2),x)

[Out]

int(1/(a+b*arcsinh(I+d*x^2))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsinh}\left (d x^{2} + i\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(I+d*x^2))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(d*x^2 + I) + a)^(-5/2), 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(1/(a+b*arcsinh(I+d*x^2))^(5/2),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(1/(a+b*asinh(I+d*x**2))**(5/2),x)

[Out]

Exception raised: TypeError

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(I+d*x^2))^(5/2),x, algorithm="giac")

[Out]

Timed out