\(\int \frac {1}{\sqrt {a-i b \arcsin (1+i d x^2)}} \, dx\) [338]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F(-2)]
   Sympy [F(-2)]
   Maxima [F]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 231 \[ \int \frac {1}{\sqrt {a-i b \arcsin \left (1+i d x^2\right )}} \, dx=-\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {-i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {-i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \]

[Out]

-x*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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 231, 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 = {4903} \[ \int \frac {1}{\sqrt {a-i b \arcsin \left (1+i d x^2\right )}} \, dx=-\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a-i b \arcsin \left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a-i b \arcsin \left (i d x^2+1\right )}}{\sqrt {-i b} \sqrt {\pi }}\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \]

[In]

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 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*} \text {integral}& = -\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a-i b \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} \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 {FresnelS}\left (\frac {\sqrt {a-i b \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} \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 = 0.00 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\sqrt {a-i b \arcsin \left (1+i d x^2\right )}} \, dx=\frac {\sqrt {\pi } x \left (-\operatorname {FresnelC}\left (\frac {\sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {-i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )-\operatorname {FresnelS}\left (\frac {\sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {-i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )\right )}{\sqrt {-i b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \]

[In]

Integrate[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]))

Maple [F]

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

[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)

Fricas [F(-2)]

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

[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)

Sympy [F(-2)]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F(-2)]

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

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

Mupad [F(-1)]

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

[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)