3.4.29 \(\int (a+i b \text {ArcSin}(1-i d x^2))^{3/2} \, dx\) [329]

Optimal. Leaf size=312 \[ -\frac {3 b \sqrt {2 i d x^2+d^2 x^4} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{d x}+x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{3/2}+\frac {3 \sqrt {i b} b \sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )}-\frac {3 b^2 \sqrt {\pi } x S\left (\frac {\sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {i b} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )} \]

[Out]

x*(a-I*b*arcsin(-1+I*d*x^2))^(3/2)-3*b^2*x*FresnelS((a-I*b*arcsin(-1+I*d*x^2))^(1/2)/(I*b)^(1/2)/Pi^(1/2))*(co
sh(1/2*a/b)-I*sinh(1/2*a/b))*Pi^(1/2)/(cos(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))/(I*b)^(1/2)+3*
b*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)/(cos(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))-3*b*(2*I*d*x^2+d^2*x^4)^(1/2)*(a-I*b*arcs
in(-1+I*d*x^2))^(1/2)/d/x

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Rubi [A]
time = 0.08, antiderivative size = 312, 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 = {4898, 4903} \begin {gather*} -\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 \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\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 {3 b \sqrt {d^2 x^4+2 i d x^2} \sqrt {a+i b \text {ArcSin}\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 \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi } \sqrt {i b}}\right )}{\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 )}+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])^(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*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]) - (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[Arc
Sin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))

Rule 4898

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 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]

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 \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 )}-\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 )}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 258, normalized size = 0.83 \begin {gather*} -\frac {3 b \sqrt {d x^2 \left (2 i+d x^2\right )} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{d x}+x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{3/2}+\frac {3 b^2 \sqrt {\pi } x \left (-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 )-\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 )\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 )} \end {gather*}

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*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)])) - 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]))

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Maple [F]
time = 0.92, size = 0, normalized size = 0.00 \[\int \left (a +b \arcsinh \left (d \,x^{2}+i\right )\right )^{\frac {3}{2}}\, dx\]

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 = 0.00, 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((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 = 0.00, 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((a+b*arcsinh(I+d*x^2))^(3/2),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((a+b*asinh(I+d*x**2))**(3/2),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((a+b*arcsinh(I+d*x^2))^(3/2),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 {\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(d*x^2 + 1i))^(3/2),x)

[Out]

int((a + b*asinh(d*x^2 + 1i))^(3/2), x)

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