Integrand size = 18, antiderivative size = 299 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=-15 b^2 x \sqrt {a-b \arcsin \left (1-d x^2\right )}+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}+\frac {15 \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {15 \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \] Output:
-15*b^2*x*(a+b*arcsin(d*x^2-1))^(1/2)+5*b*(-d^2*x^4+2*d*x^2)^(1/2)*(a+b*ar csin(d*x^2-1))^(3/2)/d/x+x*(a+b*arcsin(d*x^2-1))^(5/2)+15*Pi^(1/2)*x*Fresn elC((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)-sin(1 /2*a/b))/(-1/b)^(5/2)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))- 15*Pi^(1/2)*x*FresnelS((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/Pi^(1/2))* (cos(1/2*a/b)+sin(1/2*a/b))/(-1/b)^(5/2)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2 *arcsin(d*x^2-1)))
Time = 0.23 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.98 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\frac {5 b \sqrt {-d x^2 \left (-2+d x^2\right )} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}+\frac {15 b x \left (-\sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )+\sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )+\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )\right )}{\left (-\frac {1}{b}\right )^{3/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \] Input:
Integrate[(a - b*ArcSin[1 - d*x^2])^(5/2),x]
Output:
(5*b*Sqrt[-(d*x^2*(-2 + d*x^2))]*(a - b*ArcSin[1 - d*x^2])^(3/2))/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^(5/2) + (15*b*x*(-(Sqrt[Pi]*FresnelC[(Sqrt[-b^ (-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b) ])) + Sqrt[Pi]*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt [Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]) + Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]]*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])))/((-b^(-1)) ^(3/2)*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]))
Time = 0.50 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 5313 |
\(\displaystyle -15 b^2 \int \sqrt {a-b \arcsin \left (1-d x^2\right )}dx+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}\) |
\(\Big \downarrow \) 5310 |
\(\displaystyle -15 b^2 \left (-\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+x \sqrt {a-b \arcsin \left (1-d x^2\right )}\right )+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}\) |
Input:
Int[(a - b*ArcSin[1 - d*x^2])^(5/2),x]
Output:
(5*b*Sqrt[2*d*x^2 - d^2*x^4]*(a - b*ArcSin[1 - d*x^2])^(3/2))/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^(5/2) - 15*b^2*(x*Sqrt[a - b*ArcSin[1 - d*x^2]] - (Sqrt[Pi]*x*FresnelC[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi ]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Sqrt[-b^(-1)]*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])) + (Sqrt[Pi]*x*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Sqrt[-b ^(-1)]*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - 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]
\[\int {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{\frac {5}{2}}d x\]
Input:
int((a+b*arcsin(d*x^2-1))^(5/2),x)
Output:
int((a+b*arcsin(d*x^2-1))^(5/2),x)
Exception generated. \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*arcsin(d*x^2-1))^(5/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\int \left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((a+b*asin(d*x**2-1))**(5/2),x)
Output:
Integral((a + b*asin(d*x**2 - 1))**(5/2), x)
\[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\int { {\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+b*arcsin(d*x^2-1))^(5/2),x, algorithm="maxima")
Output:
integrate((b*arcsin(d*x^2 - 1) + a)^(5/2), x)
\[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\int { {\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+b*arcsin(d*x^2-1))^(5/2),x, algorithm="giac")
Output:
integrate((b*arcsin(d*x^2 - 1) + a)^(5/2), x)
Timed out. \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\int {\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^{5/2} \,d x \] Input:
int((a + b*asin(d*x^2 - 1))^(5/2),x)
Output:
int((a + b*asin(d*x^2 - 1))^(5/2), x)
\[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\left (\int \sqrt {\mathit {asin} \left (d \,x^{2}-1\right ) b +a}d x \right ) a^{2}+2 \left (\int \sqrt {\mathit {asin} \left (d \,x^{2}-1\right ) b +a}\, \mathit {asin} \left (d \,x^{2}-1\right )d x \right ) a b +\left (\int \sqrt {\mathit {asin} \left (d \,x^{2}-1\right ) b +a}\, \mathit {asin} \left (d \,x^{2}-1\right )^{2}d x \right ) b^{2} \] Input:
int((a+b*asin(d*x^2-1))^(5/2),x)
Output:
int(sqrt(asin(d*x**2 - 1)*b + a),x)*a**2 + 2*int(sqrt(asin(d*x**2 - 1)*b + a)*asin(d*x**2 - 1),x)*a*b + int(sqrt(asin(d*x**2 - 1)*b + a)*asin(d*x**2 - 1)**2,x)*b**2