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

Optimal. Leaf size=62 \[ -\frac{1}{6 \left (1-x^2\right )}+\frac{1}{3} \log \left (1-x^2\right )+\frac{2 x \sin ^{-1}(x)}{3 \sqrt{1-x^2}}+\frac{x \sin ^{-1}(x)}{3 \left (1-x^2\right )^{3/2}} \]

[Out]

-1/(6*(1 - x^2)) + (x*ArcSin[x])/(3*(1 - x^2)^(3/2)) + (2*x*ArcSin[x])/(3*Sqrt[1 - x^2]) + Log[1 - x^2]/3

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Rubi [A]  time = 0.0391695, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4655, 4651, 260, 261} \[ -\frac{1}{6 \left (1-x^2\right )}+\frac{1}{3} \log \left (1-x^2\right )+\frac{2 x \sin ^{-1}(x)}{3 \sqrt{1-x^2}}+\frac{x \sin ^{-1}(x)}{3 \left (1-x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[x]/(1 - x^2)^(5/2),x]

[Out]

-1/(6*(1 - x^2)) + (x*ArcSin[x])/(3*(1 - x^2)^(3/2)) + (2*x*ArcSin[x])/(3*Sqrt[1 - x^2]) + Log[1 - x^2]/3

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a + b*
ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\sin ^{-1}(x)}{\left (1-x^2\right )^{5/2}} \, dx &=\frac{x \sin ^{-1}(x)}{3 \left (1-x^2\right )^{3/2}}-\frac{1}{3} \int \frac{x}{\left (1-x^2\right )^2} \, dx+\frac{2}{3} \int \frac{\sin ^{-1}(x)}{\left (1-x^2\right )^{3/2}} \, dx\\ &=-\frac{1}{6 \left (1-x^2\right )}+\frac{x \sin ^{-1}(x)}{3 \left (1-x^2\right )^{3/2}}+\frac{2 x \sin ^{-1}(x)}{3 \sqrt{1-x^2}}-\frac{2}{3} \int \frac{x}{1-x^2} \, dx\\ &=-\frac{1}{6 \left (1-x^2\right )}+\frac{x \sin ^{-1}(x)}{3 \left (1-x^2\right )^{3/2}}+\frac{2 x \sin ^{-1}(x)}{3 \sqrt{1-x^2}}+\frac{1}{3} \log \left (1-x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0879138, size = 45, normalized size = 0.73 \[ \frac{1}{6} \left (\frac{1}{x^2-1}+2 \log \left (1-x^2\right )-\frac{2 x \left (2 x^2-3\right ) \sin ^{-1}(x)}{\left (1-x^2\right )^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[x]/(1 - x^2)^(5/2),x]

[Out]

((-1 + x^2)^(-1) - (2*x*(-3 + 2*x^2)*ArcSin[x])/(1 - x^2)^(3/2) + 2*Log[1 - x^2])/6

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Maple [A]  time = 0.037, size = 63, normalized size = 1. \begin{align*}{\frac{\arcsin \left ( x \right ) x}{3\, \left ({x}^{2}-1 \right ) ^{2}}\sqrt{-{x}^{2}+1}}+{\frac{1}{6\,{x}^{2}-6}}-{\frac{2\,\arcsin \left ( x \right ) x}{3\,{x}^{2}-3}\sqrt{-{x}^{2}+1}}+{\frac{\ln \left ( -{x}^{2}+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(x)/(-x^2+1)^(5/2),x)

[Out]

1/3*x*arcsin(x)*(-x^2+1)^(1/2)/(x^2-1)^2+1/6/(x^2-1)-2/3*(-x^2+1)^(1/2)/(x^2-1)*arcsin(x)*x+1/3*ln(-x^2+1)

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Maxima [A]  time = 1.41139, size = 65, normalized size = 1.05 \begin{align*} \frac{1}{3} \,{\left (\frac{2 \, x}{\sqrt{-x^{2} + 1}} + \frac{x}{{\left (-x^{2} + 1\right )}^{\frac{3}{2}}}\right )} \arcsin \left (x\right ) + \frac{1}{6 \,{\left (x^{2} - 1\right )}} + \frac{1}{3} \, \log \left (-3 \, x^{2} + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(-x^2+1)^(5/2),x, algorithm="maxima")

[Out]

1/3*(2*x/sqrt(-x^2 + 1) + x/(-x^2 + 1)^(3/2))*arcsin(x) + 1/6/(x^2 - 1) + 1/3*log(-3*x^2 + 3)

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Fricas [A]  time = 2.53875, size = 151, normalized size = 2.44 \begin{align*} -\frac{2 \,{\left (2 \, x^{3} - 3 \, x\right )} \sqrt{-x^{2} + 1} \arcsin \left (x\right ) - x^{2} - 2 \,{\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x^{2} - 1\right ) + 1}{6 \,{\left (x^{4} - 2 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(-x^2+1)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(2*(2*x^3 - 3*x)*sqrt(-x^2 + 1)*arcsin(x) - x^2 - 2*(x^4 - 2*x^2 + 1)*log(x^2 - 1) + 1)/(x^4 - 2*x^2 + 1)

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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(x)/(-x**2+1)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.08977, size = 73, normalized size = 1.18 \begin{align*} -\frac{{\left (2 \, x^{2} - 3\right )} \sqrt{-x^{2} + 1} x \arcsin \left (x\right )}{3 \,{\left (x^{2} - 1\right )}^{2}} - \frac{2 \, x^{2} - 3}{6 \,{\left (x^{2} - 1\right )}} + \frac{1}{3} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(-x^2+1)^(5/2),x, algorithm="giac")

[Out]

-1/3*(2*x^2 - 3)*sqrt(-x^2 + 1)*x*arcsin(x)/(x^2 - 1)^2 - 1/6*(2*x^2 - 3)/(x^2 - 1) + 1/3*log(abs(x^2 - 1))