\(\int x^3 (1-x^2)^{3/2} \arccos (x) \, dx\) [656]

Optimal result

Integrand size = 17, antiderivative size = 61 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {2 x}{35}-\frac {x^3}{105}+\frac {8 x^5}{175}-\frac {x^7}{49}-\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {272, 45, 4780, 12, 380} \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x)-\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)-\frac {x^7}{49}+\frac {8 x^5}{175}-\frac {x^3}{105}-\frac {2 x}{35} \]

[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 45

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 272

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 380

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 4780

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

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (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} \arccos (x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (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} \arccos (x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (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} \arccos (x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {x \left (210+35 x^2-168 x^4+75 x^6\right )}{3675}-\frac {1}{35} \left (1-x^2\right )^{5/2} \left (2+5 x^2\right ) \arccos (x) \]

[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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\left (x^{2}-1\right )^{2} \sqrt {-x^{2}+1}\, \arccos \left (x \right )}{5}-\frac {\left (3 x^{4}-10 x^{2}+15\right ) x}{75}-\frac {\left (x^{2}-1\right )^{3} \sqrt {-x^{2}+1}\, \arccos \left (x \right )}{7}-\frac {\left (5 x^{6}-21 x^{4}+35 x^{2}-35\right ) x}{245}\)	\(77\)

[In]

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

[Out]

-1/5*(x^2-1)^2*(-x^2+1)^(1/2)*arccos(x)-1/75*(3*x^4-10*x^2+15)*x-1/7*(x^2-1)^3*(-x^2+1)^(1/2)*arccos(x)-1/245*
(5*x^6-21*x^4+35*x^2-35)*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, x^{6} - 8 \, x^{4} + x^{2} + 2\right )} \sqrt {-x^{2} + 1} \arccos \left (x\right ) - \frac {2}{35} \, x \]

[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

Sympy [A] (verification not implemented)

Time = 6.92 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=- \frac {x^{7}}{49} - \frac {x^{6} \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{7} + \frac {8 x^{5}}{175} + \frac {8 x^{4} \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{35} - \frac {x^{3}}{105} - \frac {x^{2} \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{35} - \frac {2 x}{35} - \frac {2 \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{35} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}} x^{2} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}}\right )} \arccos \left (x\right ) - \frac {2}{35} \, x \]

[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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, {\left (x^{2} - 1\right )}^{3} \sqrt {-x^{2} + 1} + 7 \, {\left (x^{2} - 1\right )}^{2} \sqrt {-x^{2} + 1}\right )} \arccos \left (x\right ) - \frac {2}{35} \, x \]

[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

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=\int x^3\,\mathrm {acos}\left (x\right )\,{\left (1-x^2\right )}^{3/2} \,d x \]

[In]

int(x^3*acos(x)*(1 - x^2)^(3/2),x)

[Out]

int(x^3*acos(x)*(1 - x^2)^(3/2), x)