\(\int \frac {1}{(a-i b \arcsin (1+i d x^2))^{7/2}} \, dx\) [276]

Optimal result

Integrand size = 22, antiderivative size = 389 \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=-\frac {\sqrt {-2 i d x^2+d^2 x^4}}{5 b d x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2}}-\frac {x}{15 b^2 \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{3/2}}-\frac {\sqrt {-2 i d x^2+d^2 x^4}}{15 b^3 d x \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}-\frac {\left (\frac {i}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{15 b^2 \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}+\frac {\sqrt {\frac {i}{b}} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right )}{15 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \] Output:

-1/5*(-2*I*d*x^2+d^2*x^4)^(1/2)/b/d/x/(a-I*b*arcsin(1+I*d*x^2))^(5/2)-1/15 
*x/b^2/(a-I*b*arcsin(1+I*d*x^2))^(3/2)-1/15*(-2*I*d*x^2+d^2*x^4)^(1/2)/b^3 
/d/x/(a-I*b*arcsin(1+I*d*x^2))^(1/2)-1/15*(I/b)^(3/2)*Pi^(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*sin 
h(1/2*a/b))/b^2/(cos(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))+1/ 
15*(I/b)^(1/2)*Pi^(1/2)*x*FresnelS((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))/b^3/(cos(1/2*arcsin(1+I*d*x 
^2))-sin(1/2*arcsin(1+I*d*x^2)))
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 370, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=\frac {\frac {-\frac {3 b \sqrt {d x^2 \left (-2 i+d x^2\right )}}{d}-x^2 \left (a-i b \arcsin \left (1+i d x^2\right )\right )+\frac {\sqrt {d x^2 \left (-2 i+d x^2\right )} \left (i a+b \arcsin \left (1+i d x^2\right )\right )^2}{b d}}{x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2}}+\frac {\sqrt {\frac {i}{b}} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (-i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right )}{b \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}+\frac {\sqrt {\frac {i}{b}} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right )}{b \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}}{15 b^2} \] Input:

Integrate[(a - I*b*ArcSin[1 + I*d*x^2])^(-7/2),x]
 

Output:

(((-3*b*Sqrt[d*x^2*(-2*I + d*x^2)])/d - x^2*(a - I*b*ArcSin[1 + I*d*x^2]) 
+ (Sqrt[d*x^2*(-2*I + d*x^2)]*(I*a + b*ArcSin[1 + I*d*x^2])^2)/(b*d))/(x*( 
a - I*b*ArcSin[1 + I*d*x^2])^(5/2)) + (Sqrt[I/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)] + S 
inh[a/(2*b)]))/(b*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2] 
)) + (Sqrt[I/b]*Sqrt[Pi]*x*FresnelS[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d 
*x^2]])/Sqrt[Pi]]*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)]))/(b*(Cos[ArcSin[1 + I* 
d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])))/(15*b^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 384, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5327, 5321}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 5327

\(\displaystyle \frac {\int \frac {1}{\left (a-i b \arcsin \left (i d x^2+1\right )\right )^{3/2}}dx}{15 b^2}-\frac {x}{15 b^2 \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{3/2}}-\frac {\sqrt {d^2 x^4-2 i d x^2}}{5 b d x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2}}\)

\(\Big \downarrow \) 5321

\(\displaystyle \frac {-\frac {\sqrt {d^2 x^4-2 i d x^2}}{b d x \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}-\frac {\sqrt {\pi } \left (\frac {i}{b}\right )^{3/2} x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )}+\frac {\sqrt {\pi } \left (\frac {i}{b}\right )^{3/2} x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )}}{15 b^2}-\frac {x}{15 b^2 \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{3/2}}-\frac {\sqrt {d^2 x^4-2 i d x^2}}{5 b d x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2}}\)

Input:

Int[(a - I*b*ArcSin[1 + I*d*x^2])^(-7/2),x]
 

Output:

-1/5*Sqrt[(-2*I)*d*x^2 + d^2*x^4]/(b*d*x*(a - I*b*ArcSin[1 + I*d*x^2])^(5/ 
2)) - x/(15*b^2*(a - I*b*ArcSin[1 + I*d*x^2])^(3/2)) + (-(Sqrt[(-2*I)*d*x^ 
2 + d^2*x^4]/(b*d*x*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])) + ((I/b)^(3/2)*Sqr 
t[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)]))/(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin 
[1 + I*d*x^2]/2]) - ((I/b)^(3/2)*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)]))/(Cos 
[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2]))/(15*b^2)
 

Defintions of rubi rules used

rule 5321

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[- 
Sqrt[-2*c*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcSin[c + d*x^2]]), x] + (-Si 
mp[(c/b)^(3/2)*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]]]/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Si 
n[ArcSin[c + d*x^2]/2])), x] + Simp[(c/b)^(3/2)*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]]]/(Co 
s[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2])), x]) /; FreeQ[{a, 
 b, c, d}, x] && EqQ[c^2, 1]
 

rule 5327

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] + (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] - Simp[1/(4*b^2*(n + 1)*(n + 2))   Int[(a + b*ArcSin[c + d*x^2])^(n + 
 2), x], x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[ 
n, -2]
 

Maple [F]

\[\int \frac {1}{{\left (a -i b \arcsin \left (i d \,x^{2}+1\right )\right )}^{\frac {7}{2}}}d x\]

Input:

int(1/(a-I*b*arcsin(1+I*d*x^2))^(7/2),x)
 

Output:

int(1/(a-I*b*arcsin(1+I*d*x^2))^(7/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a-I*b*arcsin(1+I*d*x^2))^(7/2),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{\left (a - i b \operatorname {asin}{\left (i d x^{2} + 1 \right )}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(1/(a-I*b*asin(1+I*d*x**2))**(7/2),x)
 

Output:

Integral((a - I*b*asin(I*d*x**2 + 1))**(-7/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=\int { \frac {1}{{\left (-i \, b \arcsin \left (i \, d x^{2} + 1\right ) + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/(a-I*b*arcsin(1+I*d*x^2))^(7/2),x, algorithm="maxima")
 

Output:

integrate((-I*b*arcsin(I*d*x^2 + 1) + a)^(-7/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a-I*b*arcsin(1+I*d*x^2))^(7/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{{\left (a-b\,\mathrm {asin}\left (1{}\mathrm {i}\,d\,x^2+1\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \] Input:

int(1/(a - b*asin(d*x^2*1i + 1)*1i)^(7/2),x)
 

Output:

int(1/(a - b*asin(d*x^2*1i + 1)*1i)^(7/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a-i b \arcsin \left (1+i d x^2\right )\right )^{7/2}} \, dx=\left (\int \frac {\sqrt {-\mathit {asin} \left (d i \,x^{2}+1\right ) b i +a}}{\mathit {asin} \left (d i \,x^{2}+1\right )^{5} b^{5} i -3 \mathit {asin} \left (d i \,x^{2}+1\right )^{4} a \,b^{4}-2 \mathit {asin} \left (d i \,x^{2}+1\right )^{3} a^{2} b^{3} i -2 \mathit {asin} \left (d i \,x^{2}+1\right )^{2} a^{3} b^{2}-3 \mathit {asin} \left (d i \,x^{2}+1\right ) a^{4} b i +a^{5}}d x \right ) a +\left (\int \frac {\sqrt {-\mathit {asin} \left (d i \,x^{2}+1\right ) b i +a}\, \mathit {asin} \left (d i \,x^{2}+1\right )}{\mathit {asin} \left (d i \,x^{2}+1\right )^{5} b^{5} i -3 \mathit {asin} \left (d i \,x^{2}+1\right )^{4} a \,b^{4}-2 \mathit {asin} \left (d i \,x^{2}+1\right )^{3} a^{2} b^{3} i -2 \mathit {asin} \left (d i \,x^{2}+1\right )^{2} a^{3} b^{2}-3 \mathit {asin} \left (d i \,x^{2}+1\right ) a^{4} b i +a^{5}}d x \right ) b i \] Input:

int(1/(a-I*b*asin(1+I*d*x^2))^(7/2),x)
 

Output:

int(sqrt( - asin(d*i*x**2 + 1)*b*i + a)/(asin(d*i*x**2 + 1)**5*b**5*i - 3* 
asin(d*i*x**2 + 1)**4*a*b**4 - 2*asin(d*i*x**2 + 1)**3*a**2*b**3*i - 2*asi 
n(d*i*x**2 + 1)**2*a**3*b**2 - 3*asin(d*i*x**2 + 1)*a**4*b*i + a**5),x)*a 
+ int((sqrt( - asin(d*i*x**2 + 1)*b*i + a)*asin(d*i*x**2 + 1))/(asin(d*i*x 
**2 + 1)**5*b**5*i - 3*asin(d*i*x**2 + 1)**4*a*b**4 - 2*asin(d*i*x**2 + 1) 
**3*a**2*b**3*i - 2*asin(d*i*x**2 + 1)**2*a**3*b**2 - 3*asin(d*i*x**2 + 1) 
*a**4*b*i + a**5),x)*b*i