3.4.41 \(\int \frac {1}{(a-i b \text {ArcSin}(1+i d x^2))^{7/2}} \, dx\) [341]

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

[Out]

-1/15*x/b^2/(a-I*b*arcsin(1+I*d*x^2))^(3/2)-1/15*(I/b)^(3/2)*x*FresnelC((I/b)^(1/2)*(a-I*b*arcsin(1+I*d*x^2))^
(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)))+1/15*x*FresnelS((I/b)^(1/2)*(a-I*b*arcsin(1+I*d*x^2))^(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)))-1/5*(-2*I*d*x^2+d^2*x^4)^(1/2)
/b/d/x/(a-I*b*arcsin(1+I*d*x^2))^(5/2)-1/15*(-2*I*d*x^2+d^2*x^4)^(1/2)/b^3/d/x/(a-I*b*arcsin(1+I*d*x^2))^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 389, 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, 4906} \begin {gather*} -\frac {\sqrt {d^2 x^4-2 i d x^2}}{15 b^3 d x \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\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 \text {ArcSin}\left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{15 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 } \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 \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right )}{15 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}{15 b^2 \left (a-i b \text {ArcSin}\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 \text {ArcSin}\left (1+i d x^2\right )\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*Sqrt[(-2*I)*d*x^2 + d^2*x^4]/(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)*
Sqrt[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 4906

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]

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 )^{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*}

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Mathematica [A]
time = 0.65, size = 370, normalized size = 0.95 \begin {gather*} \frac {\frac {-\frac {3 b \sqrt {d x^2 \left (-2 i+d x^2\right )}}{d}-x^2 \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )+\frac {\sqrt {d x^2 \left (-2 i+d x^2\right )} \left (i a+b \text {ArcSin}\left (1+i d x^2\right )\right )^2}{b d}}{x \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{5/2}}+\frac {\sqrt {\frac {i}{b}} \sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\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 )}{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 {\frac {i}{b}} \sqrt {\pi } x S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\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 )}{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 )}}{15 b^2} \end {gather*}

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 = 0.92, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsinh \left (d \,x^{2}-i\right )\right )^{\frac {7}{2}}}\, dx\]

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 = 0.00, 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))^(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 = 0.00, 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))^(7/2),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))**(7/2),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))^(7/2),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 )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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