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