3.90 \(\int \sin ^{-1}(a x)^{5/2} \, dx\)

Optimal. Leaf size=88 \[ \frac{5 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^{3/2}}{2 a}+\frac{15 \sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\sin ^{-1}(a x)}\right )}{4 a}+x \sin ^{-1}(a x)^{5/2}-\frac{15}{4} x \sqrt{\sin ^{-1}(a x)} \]

[Out]

(-15*x*Sqrt[ArcSin[a*x]])/4 + (5*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^(3/2))/(2*a) + x*ArcSin[a*x]^(5/2) + (15*Sqrt[P
i/2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcSin[a*x]]])/(4*a)

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Rubi [A]  time = 0.164965, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4619, 4677, 4723, 3305, 3351} \[ \frac{5 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^{3/2}}{2 a}+\frac{15 \sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\sin ^{-1}(a x)}\right )}{4 a}+x \sin ^{-1}(a x)^{5/2}-\frac{15}{4} x \sqrt{\sin ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^(5/2),x]

[Out]

(-15*x*Sqrt[ArcSin[a*x]])/4 + (5*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^(3/2))/(2*a) + x*ArcSin[a*x]^(5/2) + (15*Sqrt[P
i/2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcSin[a*x]]])/(4*a)

Rule 4619

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSin[c*x])^n, x] - Dist[b*c*n, Int[
(x*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \sin ^{-1}(a x)^{5/2} \, dx &=x \sin ^{-1}(a x)^{5/2}-\frac{1}{2} (5 a) \int \frac{x \sin ^{-1}(a x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{5 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^{3/2}}{2 a}+x \sin ^{-1}(a x)^{5/2}-\frac{15}{4} \int \sqrt{\sin ^{-1}(a x)} \, dx\\ &=-\frac{15}{4} x \sqrt{\sin ^{-1}(a x)}+\frac{5 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^{3/2}}{2 a}+x \sin ^{-1}(a x)^{5/2}+\frac{1}{8} (15 a) \int \frac{x}{\sqrt{1-a^2 x^2} \sqrt{\sin ^{-1}(a x)}} \, dx\\ &=-\frac{15}{4} x \sqrt{\sin ^{-1}(a x)}+\frac{5 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^{3/2}}{2 a}+x \sin ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{x}} \, dx,x,\sin ^{-1}(a x)\right )}{8 a}\\ &=-\frac{15}{4} x \sqrt{\sin ^{-1}(a x)}+\frac{5 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^{3/2}}{2 a}+x \sin ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt{\sin ^{-1}(a x)}\right )}{4 a}\\ &=-\frac{15}{4} x \sqrt{\sin ^{-1}(a x)}+\frac{5 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^{3/2}}{2 a}+x \sin ^{-1}(a x)^{5/2}+\frac{15 \sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\sin ^{-1}(a x)}\right )}{4 a}\\ \end{align*}

Mathematica [C]  time = 0.0354744, size = 68, normalized size = 0.77 \[ \frac{-\sqrt{-i \sin ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},-i \sin ^{-1}(a x)\right )-\sqrt{i \sin ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},i \sin ^{-1}(a x)\right )}{2 a \sqrt{\sin ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a*x]^(5/2),x]

[Out]

(-(Sqrt[(-I)*ArcSin[a*x]]*Gamma[7/2, (-I)*ArcSin[a*x]]) - Sqrt[I*ArcSin[a*x]]*Gamma[7/2, I*ArcSin[a*x]])/(2*a*
Sqrt[ArcSin[a*x]])

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Maple [A]  time = 0.037, size = 88, normalized size = 1. \begin{align*}{\frac{\sqrt{2}}{8\,a\sqrt{\pi }} \left ( 4\, \left ( \arcsin \left ( ax \right ) \right ) ^{5/2}\sqrt{2}\sqrt{\pi }xa+10\, \left ( \arcsin \left ( ax \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }\sqrt{-{a}^{2}{x}^{2}+1}-15\,\sqrt{2}\sqrt{\arcsin \left ( ax \right ) }\sqrt{\pi }xa+15\,\pi \,{\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{\arcsin \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)^(5/2),x)

[Out]

1/8/a*2^(1/2)/Pi^(1/2)*(4*arcsin(a*x)^(5/2)*2^(1/2)*Pi^(1/2)*x*a+10*arcsin(a*x)^(3/2)*2^(1/2)*Pi^(1/2)*(-a^2*x
^2+1)^(1/2)-15*2^(1/2)*arcsin(a*x)^(1/2)*Pi^(1/2)*x*a+15*Pi*FresnelS(2^(1/2)/Pi^(1/2)*arcsin(a*x)^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)**(5/2),x)

[Out]

Timed out

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Giac [C]  time = 1.51359, size = 209, normalized size = 2.38 \begin{align*} -\frac{i \, \arcsin \left (a x\right )^{\frac{5}{2}} e^{\left (i \, \arcsin \left (a x\right )\right )}}{2 \, a} + \frac{i \, \arcsin \left (a x\right )^{\frac{5}{2}} e^{\left (-i \, \arcsin \left (a x\right )\right )}}{2 \, a} + \frac{5 \, \arcsin \left (a x\right )^{\frac{3}{2}} e^{\left (i \, \arcsin \left (a x\right )\right )}}{4 \, a} + \frac{5 \, \arcsin \left (a x\right )^{\frac{3}{2}} e^{\left (-i \, \arcsin \left (a x\right )\right )}}{4 \, a} + \frac{\left (15 i - 15\right ) \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\arcsin \left (a x\right )}\right )}{32 \, a} - \frac{\left (15 i + 15\right ) \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\arcsin \left (a x\right )}\right )}{32 \, a} + \frac{15 i \, \sqrt{\arcsin \left (a x\right )} e^{\left (i \, \arcsin \left (a x\right )\right )}}{8 \, a} - \frac{15 i \, \sqrt{\arcsin \left (a x\right )} e^{\left (-i \, \arcsin \left (a x\right )\right )}}{8 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*arcsin(a*x)^(5/2)*e^(I*arcsin(a*x))/a + 1/2*I*arcsin(a*x)^(5/2)*e^(-I*arcsin(a*x))/a + 5/4*arcsin(a*x)^
(3/2)*e^(I*arcsin(a*x))/a + 5/4*arcsin(a*x)^(3/2)*e^(-I*arcsin(a*x))/a + (15/32*I - 15/32)*sqrt(2)*sqrt(pi)*er
f((1/2*I - 1/2)*sqrt(2)*sqrt(arcsin(a*x)))/a - (15/32*I + 15/32)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*s
qrt(arcsin(a*x)))/a + 15/8*I*sqrt(arcsin(a*x))*e^(I*arcsin(a*x))/a - 15/8*I*sqrt(arcsin(a*x))*e^(-I*arcsin(a*x
))/a