3.658 \(\int \frac {(1-x^2)^{3/2} \sin ^{-1}(x)}{x^6} \, dx\)

Optimal. Leaf size=41 \[ -\frac {1}{20 x^4}+\frac {1}{5 x^2}-\frac {\left (1-x^2\right )^{5/2} \sin ^{-1}(x)}{5 x^5}+\frac {\log (x)}{5} \]

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Rubi [A]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4681, 266, 43} \[ \frac {1}{5 x^2}-\frac {1}{20 x^4}-\frac {\left (1-x^2\right )^{5/2} \sin ^{-1}(x)}{5 x^5}+\frac {\log (x)}{5} \]

Antiderivative was successfully verified.

[In]

Int[((1 - x^2)^(3/2)*ArcSin[x])/x^6,x]

[Out]

-1/(20*x^4) + 1/(5*x^2) - ((1 - x^2)^(5/2)*ArcSin[x])/(5*x^5) + Log[x]/5

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rule 4681

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

Rubi steps

\begin {align*} \int \frac {\left (1-x^2\right )^{3/2} \sin ^{-1}(x)}{x^6} \, dx &=-\frac {\left (1-x^2\right )^{5/2} \sin ^{-1}(x)}{5 x^5}+\frac {1}{5} \int \frac {\left (1-x^2\right )^2}{x^5} \, dx\\ &=-\frac {\left (1-x^2\right )^{5/2} \sin ^{-1}(x)}{5 x^5}+\frac {1}{10} \operatorname {Subst}\left (\int \frac {(1-x)^2}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (1-x^2\right )^{5/2} \sin ^{-1}(x)}{5 x^5}+\frac {1}{10} \operatorname {Subst}\left (\int \left (\frac {1}{x^3}-\frac {2}{x^2}+\frac {1}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{20 x^4}+\frac {1}{5 x^2}-\frac {\left (1-x^2\right )^{5/2} \sin ^{-1}(x)}{5 x^5}+\frac {\log (x)}{5}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 36, normalized size = 0.88 \[ -\frac {-4 x^5 \log (x)-4 x^3+4 \left (1-x^2\right )^{5/2} \sin ^{-1}(x)+x}{20 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - x^2)^(3/2)*ArcSin[x])/x^6,x]

[Out]

-1/20*(x - 4*x^3 + 4*(1 - x^2)^(5/2)*ArcSin[x] - 4*x^5*Log[x])/x^5

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (1-x^2\right )^{3/2} \sin ^{-1}(x)}{x^6} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[((1 - x^2)^(3/2)*ArcSin[x])/x^6,x]

[Out]

Could not integrate

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fricas [A]  time = 1.09, size = 44, normalized size = 1.07 \[ \frac {4 \, x^{5} \log \relax (x) + 4 \, x^{3} - 4 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \sqrt {-x^{2} + 1} \arcsin \relax (x) - x}{20 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+1)^(3/2)*arcsin(x)/x^6,x, algorithm="fricas")

[Out]

1/20*(4*x^5*log(x) + 4*x^3 - 4*(x^4 - 2*x^2 + 1)*sqrt(-x^2 + 1)*arcsin(x) - x)/x^5

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giac [B]  time = 0.99, size = 135, normalized size = 3.29 \[ -\frac {1}{160} \, {\left (\frac {x^{5} {\left (\frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - \frac {10 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{4}}{x^{4}} - 1\right )}}{{\left (\sqrt {-x^{2} + 1} - 1\right )}^{5}} + \frac {10 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{x^{3}} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{5}}{x^{5}}\right )} \arcsin \relax (x) - \frac {3 \, x^{4} - 4 \, x^{2} + 1}{20 \, x^{4}} + \frac {1}{10} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+1)^(3/2)*arcsin(x)/x^6,x, algorithm="giac")

[Out]

-1/160*(x^5*(5*(sqrt(-x^2 + 1) - 1)^2/x^2 - 10*(sqrt(-x^2 + 1) - 1)^4/x^4 - 1)/(sqrt(-x^2 + 1) - 1)^5 + 10*(sq
rt(-x^2 + 1) - 1)/x - 5*(sqrt(-x^2 + 1) - 1)^3/x^3 + (sqrt(-x^2 + 1) - 1)^5/x^5)*arcsin(x) - 1/20*(3*x^4 - 4*x
^2 + 1)/x^4 + 1/10*log(x^2)

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maple [C]  time = 1.00, size = 201, normalized size = 4.90

method	result	size
default	\(-\frac {2 i \arcsin \relax (x )}{5}+\frac {\left (-\sqrt {-x^{2}+1}\, x^{4}+i x^{5}+2 \sqrt {-x^{2}+1}\, x^{2}-\sqrt {-x^{2}+1}\right ) \left (20 \arcsin \relax (x ) x^{8}-4 i x^{8}-4 \sqrt {-x^{2}+1}\, x^{7}-40 \arcsin \relax (x ) x^{6}+i x^{6}+9 \sqrt {-x^{2}+1}\, x^{5}+40 \arcsin \relax (x ) x^{4}-6 \sqrt {-x^{2}+1}\, x^{3}-20 \arcsin \relax (x ) x^{2}+\sqrt {-x^{2}+1}\, x +4 \arcsin \relax (x )\right )}{20 \left (5 x^{8}-10 x^{6}+10 x^{4}-5 x^{2}+1\right ) x^{5}}+\frac {\ln \left (\left (i x +\sqrt {-x^{2}+1}\right )^{2}-1\right )}{5}\)	\(201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+1)^(3/2)*arcsin(x)/x^6,x,method=_RETURNVERBOSE)

[Out]

-2/5*I*arcsin(x)+1/20*(-(-x^2+1)^(1/2)*x^4+I*x^5+2*(-x^2+1)^(1/2)*x^2-(-x^2+1)^(1/2))*(20*arcsin(x)*x^8-4*I*x^
8-4*(-x^2+1)^(1/2)*x^7-40*arcsin(x)*x^6+I*x^6+9*(-x^2+1)^(1/2)*x^5+40*arcsin(x)*x^4-6*(-x^2+1)^(1/2)*x^3-20*ar
csin(x)*x^2+(-x^2+1)^(1/2)*x+4*arcsin(x))/(5*x^8-10*x^6+10*x^4-5*x^2+1)/x^5+1/5*ln((I*x+(-x^2+1)^(1/2))^2-1)

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maxima [A]  time = 0.97, size = 35, normalized size = 0.85 \[ -\frac {{\left (-x^{2} + 1\right )}^{\frac {5}{2}} \arcsin \relax (x)}{5 \, x^{5}} + \frac {4 \, x^{2} - 1}{20 \, x^{4}} + \frac {1}{10} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+1)^(3/2)*arcsin(x)/x^6,x, algorithm="maxima")

[Out]

-1/5*(-x^2 + 1)^(5/2)*arcsin(x)/x^5 + 1/20*(4*x^2 - 1)/x^4 + 1/10*log(x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asin}\relax (x)\,{\left (1-x^2\right )}^{3/2}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((asin(x)*(1 - x^2)^(3/2))/x^6,x)

[Out]

int((asin(x)*(1 - x^2)^(3/2))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (x - 1\right ) \left (x + 1\right )\right )^{\frac {3}{2}} \operatorname {asin}{\relax (x )}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+1)**(3/2)*asin(x)/x**6,x)

[Out]

Integral((-(x - 1)*(x + 1))**(3/2)*asin(x)/x**6, x)

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