Integrand size = 16, antiderivative size = 261 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{3 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a+b \arcsin \left (1+d x^2\right )}}+\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]
1/3*x*FresnelC((a+b*arcsin(d*x^2+1))^(1/2)/b^(1/2)/Pi^(1/2))*(cos(1/2*a/b) -sin(1/2*a/b))*Pi^(1/2)/b^(5/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d *x^2+1)))+1/3*x*FresnelS((a+b*arcsin(d*x^2+1))^(1/2)/b^(1/2)/Pi^(1/2))*(co s(1/2*a/b)+sin(1/2*a/b))*Pi^(1/2)/b^(5/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/ 2*arcsin(d*x^2+1)))-1/3*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2+1 ))^(3/2)+1/3*x/b^2/(a+b*arcsin(d*x^2+1))^(1/2)
Time = 0.36 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\frac {x \left (\frac {b \left (2+d x^2\right )}{\sqrt {-d x^2 \left (2+d x^2\right )} \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}+\frac {1}{\sqrt {a+b \arcsin \left (1+d x^2\right )}}+\frac {\sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {\sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}\right )}{3 b^2} \]
(x*((b*(2 + d*x^2))/(Sqrt[-(d*x^2*(2 + d*x^2))]*(a + b*ArcSin[1 + d*x^2])^ (3/2)) + 1/Sqrt[a + b*ArcSin[1 + d*x^2]] + (Sqrt[Pi]*FresnelC[Sqrt[a + b*A rcSin[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Sqrt [b]*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])) + (Sqrt[Pi]*Fre snelS[Sqrt[a + b*ArcSin[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] + Si n[a/(2*b)]))/(Sqrt[b]*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2] ))))/(3*b^2)
Time = 0.34 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5327, 5318}
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.
\(\displaystyle \int \frac {1}{\left (a+b \arcsin \left (d x^2+1\right )\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5327 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {a+b \arcsin \left (d x^2+1\right )}}dx}{3 b^2}+\frac {x}{3 b^2 \sqrt {a+b \arcsin \left (d x^2+1\right )}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}\) |
\(\Big \downarrow \) 5318 |
\(\displaystyle -\frac {-\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}-\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}}{3 b^2}+\frac {x}{3 b^2 \sqrt {a+b \arcsin \left (d x^2+1\right )}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}\) |
-1/3*Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*(a + b*ArcSin[1 + d*x^2])^(3/2)) + x/ (3*b^2*Sqrt[a + b*ArcSin[1 + d*x^2]]) - (-((Sqrt[Pi]*x*FresnelC[Sqrt[a + b *ArcSin[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Sq rt[b]*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))) - (Sqrt[Pi]* x*FresnelS[Sqrt[a + b*ArcSin[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Sqrt[b]*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^ 2]/2])))/(3*b^2)
3.5.22.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )}^{\frac {5}{2}}}d x\]
Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi nary; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)
\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^{5/2}} \,d x \]