\(\int (a-i b \arcsin (1+i d x^2))^{5/2} \, dx\) [270]

Optimal result

Integrand size = 22, antiderivative size = 348 \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2} \, dx=15 b^2 x \sqrt {a-i b \arcsin \left (1+i d x^2\right )}-\frac {5 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2}+\frac {15 b^2 \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )}-\frac {15 b^2 \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \] Output:

15*b^2*x*(a-I*b*arcsin(1+I*d*x^2))^(1/2)-5*b*(-2*I*d*x^2+d^2*x^4)^(1/2)*(a 
-I*b*arcsin(1+I*d*x^2))^(3/2)/d/x+x*(a-I*b*arcsin(1+I*d*x^2))^(5/2)+15*b^2 
*Pi^(1/2)*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))/(I/b)^(1/2)/(cos(1/2*arcsin(1+I*d*x^2))-s 
in(1/2*arcsin(1+I*d*x^2)))-15*b^2*Pi^(1/2)*x*FresnelC((I/b)^(1/2)*(a-I*b*a 
rcsin(1+I*d*x^2))^(1/2)/Pi^(1/2))*(cosh(1/2*a/b)+I*sinh(1/2*a/b))/(I/b)^(1 
/2)/(cos(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.97 \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2} \, dx=-\frac {5 b \sqrt {d x^2 \left (-2 i+d x^2\right )} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2}+\frac {15 b^2 x \left (\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )+\sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )-\sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \arcsin \left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+i d x^2\right )\right )\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 343, 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 = {5313, 5310}

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.

\(\displaystyle \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^{5/2} \, dx\)

\(\Big \downarrow \) 5313

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

\(\Big \downarrow \) 5310

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

Input:

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

Output:

(-5*b*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*(a - I*b*ArcSin[1 + I*d*x^2])^(3/2))/(d 
*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^(5/2) + 15*b^2*(x*Sqrt[a - I*b*ArcSi 
n[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[Ar 
cSin[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])))
 

Defintions of rubi rules used

rule 5310

Int[Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[x*Sq 
rt[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*A 
rcSin[c + d*x^2]]]/(Sqrt[c/b]*(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 5313

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a - I*b*asin(I*d*x**2 + 1))**(5/2), x)
 

Maxima [F]

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

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

Output:

integrate((-I*b*arcsin(I*d*x^2 + 1) + a)^(5/2), x)
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt( - asin(d*i*x**2 + 1)*b*i + a),x)*a**2 - 2*int(sqrt( - asin(d*i*x 
**2 + 1)*b*i + a)*asin(d*i*x**2 + 1),x)*a*b*i - int(sqrt( - asin(d*i*x**2 
+ 1)*b*i + a)*asin(d*i*x**2 + 1)**2,x)*b**2