\(\int \frac {1}{(a+i b \arcsin (1-i d x^2))^{5/2}} \, dx\) [333]

Optimal result

Integrand size = 22, antiderivative size = 326 \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=-\frac {\sqrt {2 i d x^2+d^2 x^4}}{3 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^{3/2}}-\frac {x}{3 b^2 \sqrt {a+i b \arcsin \left (1-i d x^2\right )}}-\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a+i b \arcsin \left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{3 \sqrt {i b} b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a+i b \arcsin \left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{3 \sqrt {i b} b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )} \]

[Out]

-1/3*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
)/b^2/(cos(1/2*arcsin(-1+I*d*x^2))+sin(1/2*arcsin(-1+I*d*x^2)))/(I*b)^(1/2)-1/3*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)/b^2/(cos(1/2*arcsin(-1+I*d*x^2))+
sin(1/2*arcsin(-1+I*d*x^2)))/(I*b)^(1/2)-1/3*(2*I*d*x^2+d^2*x^4)^(1/2)/b/d/x/(a-I*b*arcsin(-1+I*d*x^2))^(3/2)-
1/3*x/b^2/(a-I*b*arcsin(-1+I*d*x^2))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4912, 4903} \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=-\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+i b \arcsin \left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right )}{3 \sqrt {i b} b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {\sqrt {\pi } x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+i b \arcsin \left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right )}{3 \sqrt {i b} b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {x}{3 b^2 \sqrt {a+i b \arcsin \left (1-i d x^2\right )}}-\frac {\sqrt {d^2 x^4+2 i d x^2}}{3 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^{3/2}} \]

[In]

Int[(a + I*b*ArcSin[1 - I*d*x^2])^(-5/2),x]

[Out]

-1/3*Sqrt[(2*I)*d*x^2 + d^2*x^4]/(b*d*x*(a + I*b*ArcSin[1 - I*d*x^2])^(3/2)) - x/(3*b^2*Sqrt[a + I*b*ArcSin[1
- I*d*x^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)]))/(3*Sqrt[I*b]*b^2*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) - (Sqrt[Pi]*x*Fre
snelC[Sqrt[a + I*b*ArcSin[1 - I*d*x^2]]/(Sqrt[I*b]*Sqrt[Pi])]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(3*Sqrt[I*b]*
b^2*(Cos[ArcSin[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]

Rule 4912

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*((a + b*ArcSin[c + d*x^2])^(n + 2)/(
4*b^2*(n + 1)*(n + 2))), x] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcSin[c + d*x^2])^(n + 2), x], x]
+ Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]*((a + b*ArcSin[c + d*x^2])^(n + 1)/(2*b*d*(n + 1)*x)), x]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {2 i d x^2+d^2 x^4}}{3 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^{3/2}}-\frac {x}{3 b^2 \sqrt {a+i b \arcsin \left (1-i d x^2\right )}}+\frac {\int \frac {1}{\sqrt {a+i b \arcsin \left (1-i d x^2\right )}} \, dx}{3 b^2} \\ & = -\frac {\sqrt {2 i d x^2+d^2 x^4}}{3 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^{3/2}}-\frac {x}{3 b^2 \sqrt {a+i b \arcsin \left (1-i d x^2\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 )}{3 \sqrt {i b} b^2 \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 {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 )}{3 \sqrt {i b} b^2 \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 = 0.59 (sec) , antiderivative size = 308, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=-\frac {\frac {b \sqrt {d x^2 \left (2 i+d x^2\right )}}{d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^{3/2}}+\frac {x}{\sqrt {a+i b \arcsin \left (1-i d x^2\right )}}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a+i b \arcsin \left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {i b} \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a+i b \arcsin \left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {i b} \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}}{3 b^2} \]

[In]

Integrate[(a + I*b*ArcSin[1 - I*d*x^2])^(-5/2),x]

[Out]

-1/3*((b*Sqrt[d*x^2*(2*I + d*x^2)])/(d*x*(a + I*b*ArcSin[1 - I*d*x^2])^(3/2)) + x/Sqrt[a + I*b*ArcSin[1 - I*d*
x^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[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) + (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])))/b^2

Maple [F]

\[\int \frac {1}{{\left (a +b \,\operatorname {arcsinh}\left (d \,x^{2}+i\right )\right )}^{\frac {5}{2}}}d x\]

[In]

int(1/(a+b*arcsinh(I+d*x^2))^(5/2),x)

[Out]

int(1/(a+b*arcsinh(I+d*x^2))^(5/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arcsinh(I+d*x^2))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*asinh(I+d*x**2))**(5/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x^{2} + i\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arcsinh(I+d*x^2))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(d*x^2 + I) + a)^(-5/2), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arcsinh(I+d*x^2))^(5/2),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. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^{5/2}} \,d x \]

[In]

int(1/(a + b*asinh(d*x^2 + 1i))^(5/2),x)

[Out]

int(1/(a + b*asinh(d*x^2 + 1i))^(5/2), x)