\(\int \sqrt {a-i b \arcsin (1+i d x^2)} \, dx\) [337]

Optimal result

Integrand size = 22, antiderivative size = 262 \[ \int \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \, dx=x \sqrt {a-i b \arcsin \left (1+i d x^2\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \]

[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)/(cos
(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))/(I/b)^(1/2)-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)/(cos(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x
^2)))/(I/b)^(1/2)+x*(a-I*b*arcsin(1+I*d*x^2))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {4895} \[ \int \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \, dx=-\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}+x \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \]

[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]])/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*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[I/b]*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2]))

Rule 4895

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]]]/(Sqrt
[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[A
rcSin[c + d*x^2]/2]))), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]

Rubi steps \begin{align*} \text {integral}& = x \sqrt {a-i b \arcsin \left (1+i d x^2\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \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 )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \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 )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.99 \[ \int \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \, dx=\frac {x \left (\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )+\sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )-\sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \]

[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]) + Sq
rt[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)]) - Sq
rt[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)])))/(S
qrt[I/b]*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2]))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \, dx=\text {Exception raised: TypeError} \]

[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)

Sympy [F(-2)]

Exception generated. \[ \int \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real -I

Maxima [F]

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

[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)

Giac [F(-2)]

Exception generated. \[ \int \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \, dx=\text {Exception raised: TypeError} \]

[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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \, dx=\int \sqrt {a+b\,\mathrm {asinh}\left (d\,x^2-\mathrm {i}\right )} \,d x \]

[In]

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

[Out]

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