∫ sin−1 (x)_ (1−x2)5∕2 dx

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

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Rubi [A] time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, 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} \[ -\frac {1}{6 \left (1-x^2\right )}+\frac {1}{3} \log \left (1-x^2\right )+\frac {2 x \sin ^{-1}(x)}{3 \sqrt {1-x^2}}+\frac {x \sin ^{-1}(x)}{3 \left (1-x^2\right )^{3/2}} \]

-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

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

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]

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]

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]

\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*}

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Mathematica [A] time = 0.10, size = 45, normalized size = 0.73 \[ \frac {1}{6} \left (\frac {1}{x^2-1}+2 \log \left (1-x^2\right )-\frac {2 x \left (2 x^2-3\right ) \sin ^{-1}(x)}{\left (1-x^2\right )^{3/2}}\right ) \]

((-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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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{-1}(x)}{\left (1-x^2\right )^{5/2}} \, dx \]

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fricas [A] time = 1.16, size = 61, normalized size = 0.98 \[ -\frac {2 \, {\left (2 \, x^{3} - 3 \, x\right )} \sqrt {-x^{2} + 1} \arcsin \relax (x) - x^{2} - 2 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x^{2} - 1\right ) + 1}{6 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \]

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

-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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giac [A] time = 1.38, size = 54, normalized size = 0.87 \[ -\frac {{\left (2 \, x^{2} - 3\right )} \sqrt {-x^{2} + 1} x \arcsin \relax (x)}{3 \, {\left (x^{2} - 1\right )}^{2}} - \frac {2 \, x^{2} - 3}{6 \, {\left (x^{2} - 1\right )}} + \frac {1}{3} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

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

-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))

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method	result	size

default	\(\frac {1}{6 x^{2}-6}+\frac {x \arcsin \relax (x ) \sqrt {-x^{2}+1}}{3 \left (x^{2}-1\right )^{2}}+\frac {\ln \left (-x^{2}+1\right )}{3}-\frac {2 \sqrt {-x^{2}+1}\, \arcsin \relax (x ) x}{3 \left (x^{2}-1\right )}\)	\(63\)

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

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

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maxima [A] time = 0.96, size = 48, normalized size = 0.77 \[ \frac {1}{3} \, {\left (\frac {2 \, x}{\sqrt {-x^{2} + 1}} + \frac {x}{{\left (-x^{2} + 1\right )}^{\frac {3}{2}}}\right )} \arcsin \relax (x) + \frac {1}{6 \, {\left (x^{2} - 1\right )}} + \frac {1}{3} \, \log \left (-3 \, x^{2} + 3\right ) \]

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

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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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asin}\relax (x)}{{\left (1-x^2\right )}^{5/2}} \,d x \]

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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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