3.329 \(\int (a+i b \sin ^{-1}(1-i d x^2))^{3/2} \, dx\)

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

[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 = 0.11, 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 = {4814, 4819} \[ -\frac {3 \sqrt {\pi } b^2 x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right )}{\sqrt {i b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac {3 b \sqrt {d^2 x^4+2 i d x^2} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{d x}+\frac {3 \sqrt {\pi } \sqrt {i b} b x \left (-\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {\pi } \sqrt {i b}}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*b*ArcSin[1 - I*d*x^2])^(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 4814

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 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 \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 = 0.23, size = 258, normalized size = 0.83 \[ \frac {3 \sqrt {\pi } b^2 x \left (\left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \left (-S\left (\frac {\sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right )\right )-\left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right )\right )}{\sqrt {i b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac {3 b \sqrt {d x^2 \left (d x^2+2 i\right )} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2} \]

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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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [d,t_nostep]=[33,44]Bad conditionned root j= 2 value -63897.972732 ratio 2.32090080903 mindist 1
0.1217065137Warning, choosing root of [1,0,%%%{-6,[2,4]%%%}+%%%{-8,[0,0]%%%},%%%{-8,[3,6]%%%}+%%%{-32,[1,2]%%%
},%%%{-3,[4,8]%%%}+%%%{-24,[2,4]%%%}+%%%{16,[0,0]%%%}] at parameters values [7,-27]Warning, need to choose a b
ranch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming [d,t_nost
ep]=[50,45]Francis algorithm failure for[1.0,0.0,-61509375008,-8.30376562824e+15,-3.1528360132e+20]schur row 1
1.29492e-09Francis algorithm not precise enough for[1.0,0.0,-61509375008,-8.30376562824e+15,-3.1528360132e+20
]Bad conditionned root j= 2 value -101243.096423 ratio 1.67716116885 mindist 7.19967768138Francis algorithm fa
ilure for[1.0,0.0,-137282971022,-2.76877787308e+16,-1.57055117809e+21]Warning, choosing root of [1,0,%%%{-6,[2
,4]%%%}+%%%{-8,[0,0]%%%},%%%{-8,[3,6]%%%}+%%%{-32,[1,2]%%%},%%%{-3,[4,8]%%%}+%%%{-24,[2,4]%%%}+%%%{16,[0,0]%%%
}] at parameters values [63,-49]Evaluation time: 1.36sym2poly/r2sym(const gen & e,const index_m & i,const vect
eur & l) Error: Bad Argument Value

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maple [F]  time = 0.10, 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 \[ \int {\left (b \operatorname {arsinh}\left (d x^{2} + i\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^{3/2} \,d x \]

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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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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

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