Integrand size = 16, antiderivative size = 317 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{5 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}+\frac {\sqrt {-2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \arcsin \left (1+d x^2\right )}}-\frac {\left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {\left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{15 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]
1/15*x/b^2/(a+b*arcsin(d*x^2+1))^(3/2)-1/15*(1/b)^(7/2)*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))*Pi^ (1/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))+1/15*(1/b)^(7/2) *x*FresnelC((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))*Pi^(1/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1) ))-1/5*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2+1))^(5/2)+1/15*(-d ^2*x^4-2*d*x^2)^(1/2)/b^3/d/x/(a+b*arcsin(d*x^2+1))^(1/2)
Time = 0.57 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\frac {\frac {-\frac {3 b \sqrt {-d x^2 \left (2+d x^2\right )}}{d}+x^2 \left (a+b \arcsin \left (1+d x^2\right )\right )+\frac {\sqrt {-d x^2 \left (2+d x^2\right )} \left (a+b \arcsin \left (1+d x^2\right )\right )^2}{b d}}{x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}}-\frac {\left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )}+\frac {\left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )}}{15 b^2} \]
(((-3*b*Sqrt[-(d*x^2*(2 + d*x^2))])/d + x^2*(a + b*ArcSin[1 + d*x^2]) + (S qrt[-(d*x^2*(2 + d*x^2))]*(a + b*ArcSin[1 + d*x^2])^2)/(b*d))/(x*(a + b*Ar cSin[1 + d*x^2])^(5/2)) - ((b^(-1))^(3/2)*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)]))/ (Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]) + ((b^(-1))^(3/2)*Sq rt[Pi]*x*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*( Cos[a/(2*b)] + Sin[a/(2*b)]))/(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d *x^2]/2]))/(15*b^2)
Time = 0.38 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.
\(\displaystyle \int \frac {1}{\left (a+b \arcsin \left (d x^2+1\right )\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 5327 |
\(\displaystyle -\frac {\int \frac {1}{\left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}dx}{15 b^2}+\frac {x}{15 b^2 \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{5 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^{5/2}}\) |
\(\Big \downarrow \) 5321 |
\(\displaystyle -\frac {-\frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \arcsin \left (d x^2+1\right )}}-\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )}+\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )}}{15 b^2}+\frac {x}{15 b^2 \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{5 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^{5/2}}\) |
-1/5*Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*(a + b*ArcSin[1 + d*x^2])^(5/2)) + x/ (15*b^2*(a + b*ArcSin[1 + d*x^2])^(3/2)) - (-(Sqrt[-2*d*x^2 - d^2*x^4]/(b* d*x*Sqrt[a + b*ArcSin[1 + d*x^2]])) + ((b^(-1))^(3/2)*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)]))/(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]) - ((b^(- 1))^(3/2)*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)]))/(Cos[ArcSin[1 + d*x^2]/2] - Sin[ ArcSin[1 + d*x^2]/2]))/(15*b^2)
3.5.23.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )}^{\frac {7}{2}}}d x\]
Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{\frac {7}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi nary; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)
\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^{7/2}} \,d x \]