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

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

[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

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Rubi [A]  time = 0.0750953, antiderivative size = 61, normalized size of antiderivative = 1., 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 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 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 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]

Rule 12

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

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]

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*}

Mathematica [A]  time = 0.057314, size = 47, normalized size = 0.77 \[ -\frac{x \left (75 x^6-168 x^4+35 x^2+210\right )}{3675}-\frac{1}{35} \left (5 x^2+2\right ) \left (1-x^2\right )^{5/2} \cos ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.28, size = 430, normalized size = 7.1 \begin{align*}{\frac{i+7\,\arccos \left ( x \right ) }{6272} \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\,ix+\sqrt{-{x}^{2}+1} \right ) }-{\frac{i+5\,\arccos \left ( x \right ) }{3200} \left ( 16\,i{x}^{5}-16\,\sqrt{-{x}^{2}+1}{x}^{4}-20\,i{x}^{3}+12\,\sqrt{-{x}^{2}+1}{x}^{2}+5\,ix-\sqrt{-{x}^{2}+1} \right ) }-{\frac{i+3\,\arccos \left ( x \right ) }{384} \left ( 4\,i{x}^{3}-4\,\sqrt{-{x}^{2}+1}{x}^{2}-3\,ix+\sqrt{-{x}^{2}+1} \right ) }+{\frac{3\,\arccos \left ( x \right ) +3\,i}{128} \left ( ix-\sqrt{-{x}^{2}+1} \right ) }-{\frac{3\,\arccos \left ( x \right ) -3\,i}{128} \left ( ix+\sqrt{-{x}^{2}+1} \right ) }+{\frac{-i+3\,\arccos \left ( x \right ) }{384} \left ( 4\,i{x}^{3}+4\,\sqrt{-{x}^{2}+1}{x}^{2}-3\,ix-\sqrt{-{x}^{2}+1} \right ) }+{\frac{-i+5\,\arccos \left ( x \right ) }{3200} \left ( 16\,i{x}^{5}+16\,\sqrt{-{x}^{2}+1}{x}^{4}-20\,i{x}^{3}-12\,\sqrt{-{x}^{2}+1}{x}^{2}+5\,ix+\sqrt{-{x}^{2}+1} \right ) }-{\frac{-i+7\,\arccos \left ( x \right ) }{6272} \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\,ix-\sqrt{-{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))-1/3200*(I+5*arccos(x))*(16*I*x^5-16*(-x^2+1)^(1/2)*x^4-20*I*x^3+12*(-x^2+1)^(1/2
)*x^2+5*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))+3/128*(a
rccos(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))+1/3200*(-I+5*arccos(x))*(16*I*x^5+16*(-x^2+1)^(1/2)*x^4-20*I*x^3-12*(-x
^2+1)^(1/2)*x^2+5*I*x+(-x^2+1)^(1/2))-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))

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Maxima [A]  time = 1.40885, size = 66, normalized size = 1.08 \begin{align*} -\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 \left (x\right ) - \frac{2}{35} \, x \end{align*}

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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Fricas [A]  time = 2.45917, size = 138, normalized size = 2.26 \begin{align*} -\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 \left (x\right ) - \frac{2}{35} \, x \end{align*}

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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Sympy [A]  time = 153.986, size = 88, normalized size = 1.44 \begin{align*} - \frac{x^{7}}{49} - \frac{x^{6} \sqrt{1 - x^{2}} \operatorname{acos}{\left (x \right )}}{7} + \frac{8 x^{5}}{175} + \frac{8 x^{4} \sqrt{1 - x^{2}} \operatorname{acos}{\left (x \right )}}{35} - \frac{x^{3}}{105} - \frac{x^{2} \sqrt{1 - x^{2}} \operatorname{acos}{\left (x \right )}}{35} - \frac{2 x}{35} - \frac{2 \sqrt{1 - x^{2}} \operatorname{acos}{\left (x \right )}}{35} \end{align*}

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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Giac [A]  time = 1.07375, size = 81, normalized size = 1.33 \begin{align*} -\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 \left (x\right ) - \frac{2}{35} \, x \end{align*}

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