\(\int \frac {1}{(a+b \arcsin (1+d x^2))^{5/2}} \, dx\) [422]

Optimal result

Integrand size = 16, antiderivative size = 261 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{3 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a+b \arcsin \left (1+d x^2\right )}}+\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]

[Out]

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))*(
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*(-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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 261, 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 = {4912, 4903} \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arcsin \left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}+\frac {x}{3 b^2 \sqrt {a+b \arcsin \left (d x^2+1\right )}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^{3/2}} \]

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(-5/2),x]

[Out]

-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)]))/(3*b
^(5/2)*(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)]))/(3*b^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 +
d*x^2]/2]))

Rule 4903

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]

Rule 4912

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] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcSin[c + d*x^2])^(n + 2), x], 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]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps \begin{align*} \text {integral}& = -\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 {\int \frac {1}{\sqrt {a+b \arcsin \left (1+d x^2\right )}} \, dx}{3 b^2} \\ & = -\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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\frac {x \left (\frac {b \left (2+d x^2\right )}{\sqrt {-d x^2 \left (2+d x^2\right )} \left (a+b \arcsin \left (1+d x^2\right )\right )^{3/2}}+\frac {1}{\sqrt {a+b \arcsin \left (1+d x^2\right )}}+\frac {\sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {\sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arcsin \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}\right )}{3 b^2} \]

[In]

Integrate[(a + b*ArcSin[1 + d*x^2])^(-5/2),x]

[Out]

(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*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]*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)

Maple [F]

\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )}^{\frac {5}{2}}}d x\]

[In]

int(1/(a+b*arcsin(d*x^2+1))^(5/2),x)

[Out]

int(1/(a+b*arcsin(d*x^2+1))^(5/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

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

[In]

integrate(1/(a+b*asin(d*x**2+1))**(5/2),x)

[Out]

Integral((a + b*asin(d*x**2 + 1))**(-5/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: sign: argument cannot be imaginary; found sqrt((-_SAGE_VAR_d*_SAGE
_VAR_x^2)-2)

Giac [F]

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

[In]

integrate(1/(a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 + 1) + a)^(-5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^{5/2}} \,d x \]

[In]

int(1/(a + b*asin(d*x^2 + 1))^(5/2),x)

[Out]

int(1/(a + b*asin(d*x^2 + 1))^(5/2), x)