Integrand size = 16, antiderivative size = 277 \[ \int \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2} \, dx=-15 b^2 x \sqrt {a+b \arcsin \left (1+d x^2\right )}+\frac {5 b \sqrt {-2 d x^2-d^2 x^4} \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}-\frac {15 \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {15 \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]
x*(a+b*arcsin(d*x^2+1))^(5/2)-15*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)/(1/b)^(5/2)/(cos( 1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))+15*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)/ (1/b)^(5/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))+5*b*(a+b*a rcsin(d*x^2+1))^(3/2)*(-d^2*x^4-2*d*x^2)^(1/2)/d/x-15*b^2*x*(a+b*arcsin(d* x^2+1))^(1/2)
Time = 0.19 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.97 \[ \int \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2} \, dx=\frac {5 b \sqrt {-d x^2 \left (2+d x^2\right )} \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}-\frac {15 x \left (\sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )-\sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )+\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (1+d x^2\right )} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]
(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*x*(Sqrt[Pi]*FresnelS[(Sqrt[b^(-1)]* Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]) - S qrt[Pi]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(C os[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))^(5/2)*(Co s[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))
Time = 0.38 (sec) , antiderivative size = 278, 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 = {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.
\(\displaystyle \int \left (a+b \arcsin \left (d x^2+1\right )\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 5313 |
\(\displaystyle -15 b^2 \int \sqrt {a+b \arcsin \left (d x^2+1\right )}dx+\frac {5 b \sqrt {-d^2 x^4-2 d x^2} \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}{d x}+x \left (a+b \arcsin \left (d x^2+1\right )\right )^{5/2}\) |
\(\Big \downarrow \) 5310 |
\(\displaystyle -15 b^2 \left (-\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {1}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {1}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}+x \sqrt {a+b \arcsin \left (d x^2+1\right )}\right )+\frac {5 b \sqrt {-d^2 x^4-2 d x^2} \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}}{d x}+x \left (a+b \arcsin \left (d x^2+1\right )\right )^{5/2}\) |
(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*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])) - (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])))
3.5.17.3.1 Defintions of rubi rules used
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]
\[\int {\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )}^{\frac {5}{2}}d x\]
Exception generated. \[ \int \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2} \, dx=\int \left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{\frac {5}{2}}\, dx \]
Exception generated. \[ \int \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi nary; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)
\[ \int \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2} \, dx=\int { {\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2} \, dx=\int {\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^{5/2} \,d x \]