3.341 \(\int \frac{1}{(a-i b \sin ^{-1}(1+i d x^2))^{7/2}} \, dx\)

Optimal. Leaf size=389 \[ -\frac{\sqrt{d^2 x^4-2 i d x^2}}{15 b^3 d x \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}-\frac{\sqrt{\pi } \left (\frac{i}{b}\right )^{3/2} x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right )}{15 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac{\sqrt{\pi } \sqrt{\frac{i}{b}} x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{\pi }}\right )}{15 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac{x}{15 b^2 \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}-\frac{\sqrt{d^2 x^4-2 i d x^2}}{5 b d x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}} \]

[Out]

-Sqrt[(-2*I)*d*x^2 + d^2*x^4]/(5*b*d*x*(a - I*b*ArcSin[1 + I*d*x^2])^(5/2)) - x/(15*b^2*(a - I*b*ArcSin[1 + I*
d*x^2])^(3/2)) - Sqrt[(-2*I)*d*x^2 + d^2*x^4]/(15*b^3*d*x*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]) - ((I/b)^(3/2)*Sq
rt[Pi]*x*FresnelC[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(
15*b^2*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])) + (Sqrt[I/b]*Sqrt[Pi]*x*FresnelS[(Sqrt[I/b]*
Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)]))/(15*b^3*(Cos[ArcSin[1 + I*d*x^
2]/2] - Sin[ArcSin[1 + I*d*x^2]/2]))

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Rubi [A]  time = 0.0813284, antiderivative size = 389, 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, 4822} \[ -\frac{\sqrt{d^2 x^4-2 i d x^2}}{15 b^3 d x \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}-\frac{\sqrt{\pi } \left (\frac{i}{b}\right )^{3/2} x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right )}{15 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac{\sqrt{\pi } \sqrt{\frac{i}{b}} x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{\pi }}\right )}{15 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac{x}{15 b^2 \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}-\frac{\sqrt{d^2 x^4-2 i d x^2}}{5 b d x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a - I*b*ArcSin[1 + I*d*x^2])^(-7/2),x]

[Out]

-Sqrt[(-2*I)*d*x^2 + d^2*x^4]/(5*b*d*x*(a - I*b*ArcSin[1 + I*d*x^2])^(5/2)) - x/(15*b^2*(a - I*b*ArcSin[1 + I*
d*x^2])^(3/2)) - Sqrt[(-2*I)*d*x^2 + d^2*x^4]/(15*b^3*d*x*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]) - ((I/b)^(3/2)*Sq
rt[Pi]*x*FresnelC[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(
15*b^2*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])) + (Sqrt[I/b]*Sqrt[Pi]*x*FresnelS[(Sqrt[I/b]*
Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)]))/(15*b^3*(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 4822

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> -Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]/(b*d*x*S
qrt[a + b*ArcSin[c + d*x^2]]), x] + (-Simp[((c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sq
rt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2]), x] +
Simp[((c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a/(2*b)])*FresnelS[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*
x^2]]])/(Cos[(1/2)*ArcSin[c + d*x^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 )^{7/2}} \, dx &=-\frac{\sqrt{-2 i d x^2+d^2 x^4}}{5 b d x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}}-\frac{x}{15 b^2 \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}+\frac{\int \frac{1}{\left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}} \, dx}{15 b^2}\\ &=-\frac{\sqrt{-2 i d x^2+d^2 x^4}}{5 b d x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}}-\frac{x}{15 b^2 \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}-\frac{\sqrt{-2 i d x^2+d^2 x^4}}{15 b^3 d x \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}-\frac{\left (\frac{i}{b}\right )^{3/2} \sqrt{\pi } x C\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right ) \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right )}{15 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{\frac{i}{b}} \sqrt{\pi } x S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right ) \left (i \cosh \left (\frac{a}{2 b}\right )+\sinh \left (\frac{a}{2 b}\right )\right )}{15 b^3 \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 = 0.914323, size = 370, normalized size = 0.95 \[ \frac{\frac{\sqrt{\pi } \sqrt{\frac{i}{b}} x \left (\sinh \left (\frac{a}{2 b}\right )-i \cosh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt{\pi }}\right )}{b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac{\sqrt{\pi } \sqrt{\frac{i}{b}} x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{i}{b}} \sqrt{a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt{\pi }}\right )}{b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac{x^2 \left (-\left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )\right )+\frac{\sqrt{d x^2 \left (d x^2-2 i\right )} \left (i a+b \sin ^{-1}\left (1+i d x^2\right )\right )^2}{b d}-\frac{3 b \sqrt{d x^2 \left (d x^2-2 i\right )}}{d}}{x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}}}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - I*b*ArcSin[1 + I*d*x^2])^(-7/2),x]

[Out]

(((-3*b*Sqrt[d*x^2*(-2*I + d*x^2)])/d - x^2*(a - I*b*ArcSin[1 + I*d*x^2]) + (Sqrt[d*x^2*(-2*I + d*x^2)]*(I*a +
b*ArcSin[1 + I*d*x^2])^2)/(b*d))/(x*(a - I*b*ArcSin[1 + I*d*x^2])^(5/2)) + (Sqrt[I/b]*Sqrt[Pi]*x*FresnelC[(Sq
rt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*((-I)*Cosh[a/(2*b)] + Sinh[a/(2*b)]))/(b*(Cos[ArcSin[1 +
I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])) + (Sqrt[I/b]*Sqrt[Pi]*x*FresnelS[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 +
I*d*x^2]])/Sqrt[Pi]]*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)]))/(b*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x
^2]/2])))/(15*b^2)

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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( -i+d{x}^{2} \right ) \right ) ^{-{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsinh(-I+d*x^2))^(7/2),x)

[Out]

int(1/(a+b*arcsinh(-I+d*x^2))^(7/2),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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(d*x^2 - I) + a)^(-7/2), 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))^(7/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [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*asinh(-I+d*x**2))**(7/2),x)

[Out]

Timed out

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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))^(7/2),x, algorithm="giac")

[Out]

Timed out