3.330 \(\int \sqrt {a+i b \sin ^{-1}(1-i d x^2)} \, dx\)

Optimal. Leaf size=263 \[ -\frac {\sqrt {\pi } \sqrt {-\frac {i}{b}} b x \left (\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}+\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+x \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )} \]

[Out]

x*FresnelS((-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)/(c
os(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))/(-I/b)^(1/2)-b*x*FresnelC((-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)/(cos(1/2*arcsin(-1+I*d*x^2))
+sin(1/2*arcsin(-1+I*d*x^2)))+x*(a-I*b*arcsin(-1+I*d*x^2))^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {4811} \[ -\frac {\sqrt {\pi } \sqrt {-\frac {i}{b}} b x \left (\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}+\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+x \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]],x]

[Out]

x*Sqrt[a + I*b*ArcSin[1 - I*d*x^2]] + (Sqrt[Pi]*x*FresnelS[(Sqrt[(-I)/b]*Sqrt[a + I*b*ArcSin[1 - I*d*x^2]])/Sq
rt[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])) - (Sqrt[(-I)/b]*b*Sqrt[Pi]*x*FresnelC[(Sqrt[(-I)/b]*Sqrt[a + I*b*ArcSin[1 - I*d*x^2]])/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])

Rule 4811

Int[Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[x*Sqrt[a + b*ArcSin[c + d*x^2]], x] + (
-Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqr
t[c/b]*(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[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[c/b]*(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 \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )} \, dx &=x \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}+\frac {\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 (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {-\frac {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 {\sqrt {-\frac {i}{b}} b \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 (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 )}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 259, normalized size = 0.98 \[ \frac {x \left (-\sqrt {\pi } \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right )+\sqrt {\pi } \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right )+\sqrt {-\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right ) \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}\right )}{\sqrt {-\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]],x]

[Out]

(x*(Sqrt[(-I)/b]*Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]) -
Sqrt[Pi]*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)]
) + Sqrt[Pi]*FresnelS[(Sqrt[(-I)/b]*Sqrt[a + I*b*ArcSin[1 - I*d*x^2]])/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))^(1/2),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))^(1/2),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]=[84,56]Bad conditionned root j= 0 value -263411.197936 ratio 0.852649578726 mindist
22.2638977566Bad conditionned root j= 1 value -263430.401038-11.2659672545*i ratio 0.734861252591 mindist 22.
2638977566Bad conditionned root j= 2 value -263430.401038+11.2659672545*i ratio 0.73486125259 mindist 22.26389
77566schur row 1 7.1988e-07Francis algorithm not precise enough for[1.0,0.0,-156243662.0,-1.06308198511e+12,-2
.03434057625e+15]Bad conditionned root j= 0 value -5110.30395302 ratio 1.76012200073 mindist 7.30402974054Warn
ing, 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 branch for the ro
ot of a polynomial with parameters. This might be wrong.The choice was done assuming [d,t_nostep]=[2,-51]schur
row 1 4.28283e-07Francis algorithm not precise enough for[1.0,0.0,-162364832.0,-1.12616258573e+12,-2.19686198
884e+15]Bad conditionned root j= 0 value -5207.61031281 ratio 1.42050071801 mindist 5.61479175428Francis algor
ithm failure 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: 1.27sym2poly/r2sym(const gen & e,const index_m & i,con
st vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \sqrt {a +b \arcsinh \left (d \,x^{2}+i\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {arsinh}\left (d x^{2} + i\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(I+d*x^2))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*arcsinh(d*x^2 + I) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError

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