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

Optimal. Leaf size=231 \[ -\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 231, 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 = {4819} \[ -\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt {\pi } \sqrt {-i b}}\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[Pi]*x*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]))) - (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 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 \frac {1}{\sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}} \, dx &=-\frac {\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 (\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 )}-\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 )}{\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 = 0.00, size = 180, normalized size = 0.78 \[ \frac {\sqrt {\pi } x \left (\left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \left (-C\left (\frac {\sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )\right )-\left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi]*x*(-(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)])) - 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[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(1/(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(1/(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]=[92,-31]schur row 1 2.27097e-10Francis algorithm not precise enough for[1.0,0.0,-46
900090472,-5.52870773363e+15,-1.83301540649e+20]Bad conditionned root j= 1 value -88413.1703987 ratio 0.928933
994168 mindist 1.4753899225Warning, 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, nee
d to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done ass
uming [d,t_nostep]=[69,-52]Bad conditionned root j= 3 value -186564.652407 ratio 2.53779195402 mindist 12.5891
719132schur row 1 1.93713e-09Francis algorithm not precise enough for[1.0,0.0,-137282971022,-2.76877787308e+16
,-1.57055117809e+21]Bad conditionned root j= 1 value -151275.671615 ratio 2.85445476241 mindist 13.6978000362W
arning, 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: 1.31sym2poly/r2sym(con
st gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a +b \arcsinh \left (d \,x^{2}-i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 \[ \int \frac {1}{\sqrt {b \operatorname {arsinh}\left (d x^{2} - i\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*arcsinh(d*x^2 - I) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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(1/(a+b*asinh(-I+d*x**2))**(1/2),x)

[Out]

Exception raised: TypeError

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