3.656 \(\int x^3 (1-x^2)^{3/2} \cos ^{-1}(x) \, dx\)

Optimal. Leaf size=61 \[ -\frac {x^7}{49}+\frac {8 x^5}{175}-\frac {x^3}{105}+\frac {1}{7} \left (1-x^2\right )^{7/2} \cos ^{-1}(x)-\frac {1}{5} \left (1-x^2\right )^{5/2} \cos ^{-1}(x)-\frac {2 x}{35} \]

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Rubi [A]  time = 0.08, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {266, 43, 4690, 12, 373} \[ -\frac {x^7}{49}+\frac {8 x^5}{175}-\frac {x^3}{105}+\frac {1}{7} \left (1-x^2\right )^{7/2} \cos ^{-1}(x)-\frac {1}{5} \left (1-x^2\right )^{5/2} \cos ^{-1}(x)-\frac {2 x}{35} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(1 - x^2)^(3/2)*ArcCos[x],x]

[Out]

(-2*x)/35 - x^3/105 + (8*x^5)/175 - x^7/49 - ((1 - x^2)^(5/2)*ArcCos[x])/5 + ((1 - x^2)^(7/2)*ArcCos[x])/7

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 4690

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcCos[c*x]), u, x] + Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 - c
^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2
, 0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rubi steps

\begin {align*} \int x^3 \left (1-x^2\right )^{3/2} \cos ^{-1}(x) \, dx &=-\frac {1}{5} \left (1-x^2\right )^{5/2} \cos ^{-1}(x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \cos ^{-1}(x)+\int \frac {1}{35} \left (-2-5 x^2\right ) \left (1-x^2\right )^2 \, dx\\ &=-\frac {1}{5} \left (1-x^2\right )^{5/2} \cos ^{-1}(x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \cos ^{-1}(x)+\frac {1}{35} \int \left (-2-5 x^2\right ) \left (1-x^2\right )^2 \, dx\\ &=-\frac {1}{5} \left (1-x^2\right )^{5/2} \cos ^{-1}(x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \cos ^{-1}(x)+\frac {1}{35} \int \left (-2-x^2+8 x^4-5 x^6\right ) \, dx\\ &=-\frac {2 x}{35}-\frac {x^3}{105}+\frac {8 x^5}{175}-\frac {x^7}{49}-\frac {1}{5} \left (1-x^2\right )^{5/2} \cos ^{-1}(x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \cos ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 47, normalized size = 0.77 \[ -\frac {1}{35} \left (5 x^2+2\right ) \left (1-x^2\right )^{5/2} \cos ^{-1}(x)-\frac {x \left (75 x^6-168 x^4+35 x^2+210\right )}{3675} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3675*(x*(210 + 35*x^2 - 168*x^4 + 75*x^6)) - ((1 - x^2)^(5/2)*(2 + 5*x^2)*ArcCos[x])/35

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

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 0.97, size = 47, normalized size = 0.77 \[ -\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, x^{6} - 8 \, x^{4} + x^{2} + 2\right )} \sqrt {-x^{2} + 1} \arccos \relax (x) - \frac {2}{35} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/49*x^7 + 8/175*x^5 - 1/105*x^3 - 1/35*(5*x^6 - 8*x^4 + x^2 + 2)*sqrt(-x^2 + 1)*arccos(x) - 2/35*x

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giac [A]  time = 1.18, size = 60, normalized size = 0.98 \[ -\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, {\left (x^{2} - 1\right )}^{3} \sqrt {-x^{2} + 1} + 7 \, {\left (x^{2} - 1\right )}^{2} \sqrt {-x^{2} + 1}\right )} \arccos \relax (x) - \frac {2}{35} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/49*x^7 + 8/175*x^5 - 1/105*x^3 - 1/35*(5*(x^2 - 1)^3*sqrt(-x^2 + 1) + 7*(x^2 - 1)^2*sqrt(-x^2 + 1))*arccos(
x) - 2/35*x

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maple [C]  time = 0.49, size = 286, normalized size = 4.69

method	result	size
default	\(\frac {\left (i+7 \arccos \relax (x )\right ) \left (64 i x^{7}-64 \sqrt {-x^{2}+1}\, x^{6}-112 i x^{5}+80 \sqrt {-x^{2}+1}\, x^{4}+56 i x^{3}-24 \sqrt {-x^{2}+1}\, x^{2}-7 i x +\sqrt {-x^{2}+1}\right )}{6272}+\frac {3 \left (\arccos \relax (x )+i\right ) \left (i x -\sqrt {-x^{2}+1}\right )}{128}-\frac {3 \left (\arccos \relax (x )-i\right ) \left (i x +\sqrt {-x^{2}+1}\right )}{128}+\frac {\left (-i+3 \arccos \relax (x )\right ) \left (4 i x^{3}+4 \sqrt {-x^{2}+1}\, x^{2}-3 i x -\sqrt {-x^{2}+1}\right )}{384}-\frac {3 \cos \left (6 \arccos \relax (x )\right ) \left (2 i+35 \arccos \relax (x )\right ) \left (i x +\sqrt {-x^{2}+1}\right )}{39200}+\frac {\sin \left (6 \arccos \relax (x )\right ) \left (37 i+35 \arccos \relax (x )\right ) \left (-i \sqrt {-x^{2}+1}+x \right )}{78400}-\frac {\cos \left (4 \arccos \relax (x )\right ) \left (7 i+15 \arccos \relax (x )\right ) \left (i x +\sqrt {-x^{2}+1}\right )}{2400}+\frac {\sin \left (4 \arccos \relax (x )\right ) \left (11 i+45 \arccos \relax (x )\right ) \left (-i \sqrt {-x^{2}+1}+x \right )}{4800}\)	\(286\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6272*(I+7*arccos(x))*(64*I*x^7-64*(-x^2+1)^(1/2)*x^6-112*I*x^5+80*(-x^2+1)^(1/2)*x^4+56*I*x^3-24*(-x^2+1)^(1
/2)*x^2-7*I*x+(-x^2+1)^(1/2))+3/128*(arccos(x)+I)*(I*x-(-x^2+1)^(1/2))-3/128*(arccos(x)-I)*(I*x+(-x^2+1)^(1/2)
)+1/384*(-I+3*arccos(x))*(4*I*x^3+4*(-x^2+1)^(1/2)*x^2-3*I*x-(-x^2+1)^(1/2))-3/39200*cos(6*arccos(x))*(2*I+35*
arccos(x))*(I*x+(-x^2+1)^(1/2))+1/78400*sin(6*arccos(x))*(37*I+35*arccos(x))*(-I*(-x^2+1)^(1/2)+x)-1/2400*cos(
4*arccos(x))*(7*I+15*arccos(x))*(I*x+(-x^2+1)^(1/2))+1/4800*sin(4*arccos(x))*(11*I+45*arccos(x))*(-I*(-x^2+1)^
(1/2)+x)

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maxima [A]  time = 0.96, size = 49, normalized size = 0.80 \[ -\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}} x^{2} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}}\right )} \arccos \relax (x) - \frac {2}{35} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/49*x^7 + 8/175*x^5 - 1/105*x^3 - 1/35*(5*(-x^2 + 1)^(5/2)*x^2 + 2*(-x^2 + 1)^(5/2))*arccos(x) - 2/35*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(x)*(1 - x^2)^(3/2),x)

[Out]

int(x^3*acos(x)*(1 - x^2)^(3/2), x)

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sympy [A]  time = 162.01, size = 88, normalized size = 1.44 \[ - \frac {x^{7}}{49} - \frac {x^{6} \sqrt {1 - x^{2}} \operatorname {acos}{\relax (x )}}{7} + \frac {8 x^{5}}{175} + \frac {8 x^{4} \sqrt {1 - x^{2}} \operatorname {acos}{\relax (x )}}{35} - \frac {x^{3}}{105} - \frac {x^{2} \sqrt {1 - x^{2}} \operatorname {acos}{\relax (x )}}{35} - \frac {2 x}{35} - \frac {2 \sqrt {1 - x^{2}} \operatorname {acos}{\relax (x )}}{35} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**7/49 - x**6*sqrt(1 - x**2)*acos(x)/7 + 8*x**5/175 + 8*x**4*sqrt(1 - x**2)*acos(x)/35 - x**3/105 - x**2*sqr
t(1 - x**2)*acos(x)/35 - 2*x/35 - 2*sqrt(1 - x**2)*acos(x)/35

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