3.4.40 \(\int \frac {1}{(a-i b \text {ArcSin}(1+i d x^2))^{5/2}} \, dx\) [340]

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

[Out]

-1/3*x*FresnelS((a-I*b*arcsin(1+I*d*x^2))^(1/2)/(-I*b)^(1/2)/Pi^(1/2))*(cosh(1/2*a/b)+I*sinh(1/2*a/b))*Pi^(1/2
)/b^2/(cos(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))/(-I*b)^(1/2)-1/3*x*FresnelC((a-I*b*arcsin(1+I*d*
x^2))^(1/2)/(-I*b)^(1/2)/Pi^(1/2))*(I*cosh(1/2*a/b)+sinh(1/2*a/b))*(-I*b)^(1/2)*Pi^(1/2)/b^3/(cos(1/2*arcsin(1
+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))-1/3*(-2*I*d*x^2+d^2*x^4)^(1/2)/b/d/x/(a-I*b*arcsin(1+I*d*x^2))^(3/2)-1/
3*x/b^2/(a-I*b*arcsin(1+I*d*x^2))^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 326, normalized size of antiderivative = 1.00, 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 = {4912, 4903} \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-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 {a-i b \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi } \sqrt {-i b}}\right )}{3 b^3 \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 {\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 \text {ArcSin}\left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )}{3 \sqrt {-i b} b^2 \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}{3 b^2 \sqrt {a-i b \text {ArcSin}\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 \text {ArcSin}\left (1+i d x^2\right )\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4903

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[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] - 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]

Rule 4912

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]

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 {-i 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 )}{3 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*}

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Mathematica [A]
time = 0.56, size = 308, normalized size = 0.94 \begin {gather*} -\frac {\frac {b \sqrt {d x^2 \left (-2 i+d x^2\right )}}{d x \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{3/2}}+\frac {x}{\sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}}+\frac {\sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {a-i b \text {ArcSin}\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} \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 {\sqrt {\pi } x S\left (\frac {\sqrt {a-i b \text {ArcSin}\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} \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 )}}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((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*FresnelC[Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]/(Sqrt[(-I)*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] - I*Si
nh[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*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]*(C
os[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])))/b^2

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Maple [F]
time = 0.90, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsinh \left (d \,x^{2}-i\right )\right )^{\frac {5}{2}}}\, dx\]

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 = 0.00, 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))^(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 = 0.00, 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))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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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))**(5/2),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))^(5/2),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 = -0.00 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2-\mathrm {i}\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(d*x^2 - 1i))^(5/2),x)

[Out]

int(1/(a + b*asinh(d*x^2 - 1i))^(5/2), x)

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