Integrand size = 17, antiderivative size = 61 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {2 x}{35}-\frac {x^3}{105}+\frac {8 x^5}{175}-\frac {x^7}{49}-\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x) \] Output:
-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)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {x \left (210+35 x^2-168 x^4+75 x^6\right )}{3675}-\frac {1}{35} \left (1-x^2\right )^{5/2} \left (2+5 x^2\right ) \arccos (x) \] Input:
Integrate[x^3*(1 - x^2)^(3/2)*ArcCos[x],x]
Output:
-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
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {5195, 27, 290, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx\) |
\(\Big \downarrow \) 5195 |
\(\displaystyle \int -\frac {1}{35} \left (1-x^2\right )^2 \left (5 x^2+2\right )dx+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x)-\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{35} \int \left (1-x^2\right )^2 \left (5 x^2+2\right )dx+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x)-\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)\) |
\(\Big \downarrow \) 290 |
\(\displaystyle -\frac {1}{35} \int \left (5 x^6-8 x^4+x^2+2\right )dx+\frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x)-\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{7} \left (1-x^2\right )^{7/2} \arccos (x)-\frac {1}{5} \left (1-x^2\right )^{5/2} \arccos (x)+\frac {1}{35} \left (-\frac {5 x^7}{7}+\frac {8 x^5}{5}-\frac {x^3}{3}-2 x\right )\) |
Input:
Int[x^3*(1 - x^2)^(3/2)*ArcCos[x],x]
Output:
(-2*x - x^3/3 + (8*x^5)/5 - (5*x^7)/7)/35 - ((1 - x^2)^(5/2)*ArcCos[x])/5 + ((1 - x^2)^(7/2)*ArcCos[x])/7
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
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]}, Simp[(a + b*ArcCos [c*x]) u, x] + Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int[Sim plifyIntegrand[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])
Time = 0.44 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26
method | result | size |
default | \(-\frac {\left (x^{2}-1\right )^{2} \sqrt {-x^{2}+1}\, \arccos \left (x \right )}{5}-\frac {\left (3 x^{4}-10 x^{2}+15\right ) x}{75}-\frac {\left (x^{2}-1\right )^{3} \sqrt {-x^{2}+1}\, \arccos \left (x \right )}{7}-\frac {\left (5 x^{6}-21 x^{4}+35 x^{2}-35\right ) x}{245}\) | \(77\) |
orering | \(\frac {\left (325 x^{8}-866 x^{6}+553 x^{4}+420 x^{2}-280\right ) \left (-x^{2}+1\right )^{\frac {3}{2}} \arccos \left (x \right )}{1225 \left (-1+x \right ) \left (1+x \right ) \left (x^{2}-1\right )}-\frac {\left (75 x^{6}-168 x^{4}+35 x^{2}+210\right ) \left (3 x^{2} \left (-x^{2}+1\right )^{\frac {3}{2}} \arccos \left (x \right )-3 x^{4} \sqrt {-x^{2}+1}\, \arccos \left (x \right )-x^{3} \left (-x^{2}+1\right )\right )}{3675 x^{2} \left (-1+x \right ) \left (1+x \right )}\) | \(131\) |
Input:
int(x^3*(-x^2+1)^(3/2)*arccos(x),x,method=_RETURNVERBOSE)
Output:
-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
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, x^{6} - 8 \, x^{4} + x^{2} + 2\right )} \sqrt {-x^{2} + 1} \arccos \left (x\right ) - \frac {2}{35} \, x \] Input:
integrate(x^3*(-x^2+1)^(3/2)*arccos(x),x, algorithm="fricas")
Output:
-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
Time = 3.98 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=- \frac {x^{7}}{49} - \frac {x^{6} \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{7} + \frac {8 x^{5}}{175} + \frac {8 x^{4} \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{35} - \frac {x^{3}}{105} - \frac {x^{2} \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{35} - \frac {2 x}{35} - \frac {2 \sqrt {1 - x^{2}} \operatorname {acos}{\left (x \right )}}{35} \] Input:
integrate(x**3*(-x**2+1)**(3/2)*acos(x),x)
Output:
-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*sqrt(1 - x**2)*acos(x)/35 - 2*x/35 - 2*sq rt(1 - x**2)*acos(x)/35
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}} x^{2} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}}\right )} \arccos \left (x\right ) - \frac {2}{35} \, x \] Input:
integrate(x^3*(-x^2+1)^(3/2)*arccos(x),x, algorithm="maxima")
Output:
-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
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\frac {1}{49} \, x^{7} + \frac {8}{175} \, x^{5} - \frac {1}{105} \, x^{3} - \frac {1}{35} \, {\left (5 \, {\left (x^{2} - 1\right )}^{3} \sqrt {-x^{2} + 1} + 7 \, {\left (x^{2} - 1\right )}^{2} \sqrt {-x^{2} + 1}\right )} \arccos \left (x\right ) - \frac {2}{35} \, x \] Input:
integrate(x^3*(-x^2+1)^(3/2)*arccos(x),x, algorithm="giac")
Output:
-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
Timed out. \[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=\int x^3\,\mathrm {acos}\left (x\right )\,{\left (1-x^2\right )}^{3/2} \,d x \] Input:
int(x^3*acos(x)*(1 - x^2)^(3/2),x)
Output:
int(x^3*acos(x)*(1 - x^2)^(3/2), x)
\[ \int x^3 \left (1-x^2\right )^{3/2} \arccos (x) \, dx=-\left (\int \sqrt {-x^{2}+1}\, \mathit {acos} \left (x \right ) x^{5}d x \right )+\int \sqrt {-x^{2}+1}\, \mathit {acos} \left (x \right ) x^{3}d x \] Input:
int(x^3*(-x^2+1)^(3/2)*acos(x),x)
Output:
- int(sqrt( - x**2 + 1)*acos(x)*x**5,x) + int(sqrt( - x**2 + 1)*acos(x)*x **3,x)