\(\int x^2 (d+e x^2) (a+b \arccos (c x)) \, dx\) [17]

Optimal result

Integrand size = 19, antiderivative size = 120 \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=-\frac {b \left (5 c^2 d+3 e\right ) \sqrt {1-c^2 x^2}}{15 c^5}+\frac {b \left (5 c^2 d+6 e\right ) \left (1-c^2 x^2\right )^{3/2}}{45 c^5}-\frac {b e \left (1-c^2 x^2\right )^{5/2}}{25 c^5}+\frac {1}{3} d x^3 (a+b \arccos (c x))+\frac {1}{5} e x^5 (a+b \arccos (c x)) \]

[Out]

1/45*b*(5*c^2*d+6*e)*(-c^2*x^2+1)^(3/2)/c^5-1/25*b*e*(-c^2*x^2+1)^(5/2)/c^5+1/3*d*x^3*(a+b*arccos(c*x))+1/5*e*
x^5*(a+b*arccos(c*x))-1/15*b*(5*c^2*d+3*e)*(-c^2*x^2+1)^(1/2)/c^5

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 4816, 12, 457, 78} \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {1}{3} d x^3 (a+b \arccos (c x))+\frac {1}{5} e x^5 (a+b \arccos (c x))+\frac {b \left (1-c^2 x^2\right )^{3/2} \left (5 c^2 d+6 e\right )}{45 c^5}-\frac {b \sqrt {1-c^2 x^2} \left (5 c^2 d+3 e\right )}{15 c^5}-\frac {b e \left (1-c^2 x^2\right )^{5/2}}{25 c^5} \]

[In]

Int[x^2*(d + e*x^2)*(a + b*ArcCos[c*x]),x]

[Out]

-1/15*(b*(5*c^2*d + 3*e)*Sqrt[1 - c^2*x^2])/c^5 + (b*(5*c^2*d + 6*e)*(1 - c^2*x^2)^(3/2))/(45*c^5) - (b*e*(1 -
c^2*x^2)^(5/2))/(25*c^5) + (d*x^3*(a + b*ArcCos[c*x]))/3 + (e*x^5*(a + b*ArcCos[c*x]))/5

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4816

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} d x^3 (a+b \arccos (c x))+\frac {1}{5} e x^5 (a+b \arccos (c x))+(b c) \int \frac {x^3 \left (5 d+3 e x^2\right )}{15 \sqrt {1-c^2 x^2}} \, dx \\ & = \frac {1}{3} d x^3 (a+b \arccos (c x))+\frac {1}{5} e x^5 (a+b \arccos (c x))+\frac {1}{15} (b c) \int \frac {x^3 \left (5 d+3 e x^2\right )}{\sqrt {1-c^2 x^2}} \, dx \\ & = \frac {1}{3} d x^3 (a+b \arccos (c x))+\frac {1}{5} e x^5 (a+b \arccos (c x))+\frac {1}{30} (b c) \text {Subst}\left (\int \frac {x (5 d+3 e x)}{\sqrt {1-c^2 x}} \, dx,x,x^2\right ) \\ & = \frac {1}{3} d x^3 (a+b \arccos (c x))+\frac {1}{5} e x^5 (a+b \arccos (c x))+\frac {1}{30} (b c) \text {Subst}\left (\int \left (\frac {5 c^2 d+3 e}{c^4 \sqrt {1-c^2 x}}+\frac {\left (-5 c^2 d-6 e\right ) \sqrt {1-c^2 x}}{c^4}+\frac {3 e \left (1-c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b \left (5 c^2 d+3 e\right ) \sqrt {1-c^2 x^2}}{15 c^5}+\frac {b \left (5 c^2 d+6 e\right ) \left (1-c^2 x^2\right )^{3/2}}{45 c^5}-\frac {b e \left (1-c^2 x^2\right )^{5/2}}{25 c^5}+\frac {1}{3} d x^3 (a+b \arccos (c x))+\frac {1}{5} e x^5 (a+b \arccos (c x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.04 \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {1}{3} a d x^3+\frac {1}{5} a e x^5+b d \left (-\frac {2}{9 c^3}-\frac {x^2}{9 c}\right ) \sqrt {1-c^2 x^2}+b e \sqrt {1-c^2 x^2} \left (-\frac {8}{75 c^5}-\frac {4 x^2}{75 c^3}-\frac {x^4}{25 c}\right )+\frac {1}{3} b d x^3 \arccos (c x)+\frac {1}{5} b e x^5 \arccos (c x) \]

[In]

Integrate[x^2*(d + e*x^2)*(a + b*ArcCos[c*x]),x]

[Out]

(a*d*x^3)/3 + (a*e*x^5)/5 + b*d*(-2/(9*c^3) - x^2/(9*c))*Sqrt[1 - c^2*x^2] + b*e*Sqrt[1 - c^2*x^2]*(-8/(75*c^5
) - (4*x^2)/(75*c^3) - x^4/(25*c)) + (b*d*x^3*ArcCos[c*x])/3 + (b*e*x^5*ArcCos[c*x])/5

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.28

method	result	size
parts	\(a \left (\frac {1}{5} e \,x^{5}+\frac {1}{3} d \,x^{3}\right )+\frac {b \left (\frac {c^{3} \arccos \left (c x \right ) x^{5} e}{5}+\frac {\arccos \left (c x \right ) c^{3} x^{3} d}{3}+\frac {3 e \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )+5 d \,c^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{15 c^{2}}\right )}{c^{3}}\)	\(154\)
derivativedivides	\(\frac {\frac {a \left (\frac {1}{3} d \,c^{5} x^{3}+\frac {1}{5} e \,c^{5} x^{5}\right )}{c^{2}}+\frac {b \left (\frac {\arccos \left (c x \right ) d \,c^{5} x^{3}}{3}+\frac {\arccos \left (c x \right ) e \,c^{5} x^{5}}{5}+\frac {e \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )}{5}+\frac {d \,c^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c^{3}}\)	\(161\)
default	\(\frac {\frac {a \left (\frac {1}{3} d \,c^{5} x^{3}+\frac {1}{5} e \,c^{5} x^{5}\right )}{c^{2}}+\frac {b \left (\frac {\arccos \left (c x \right ) d \,c^{5} x^{3}}{3}+\frac {\arccos \left (c x \right ) e \,c^{5} x^{5}}{5}+\frac {e \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )}{5}+\frac {d \,c^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c^{3}}\)	\(161\)

[In]

int(x^2*(e*x^2+d)*(a+b*arccos(c*x)),x,method=_RETURNVERBOSE)

[Out]

a*(1/5*e*x^5+1/3*d*x^3)+b/c^3*(1/5*c^3*arccos(c*x)*x^5*e+1/3*arccos(c*x)*c^3*x^3*d+1/15/c^2*(3*e*(-1/5*c^4*x^4
*(-c^2*x^2+1)^(1/2)-4/15*c^2*x^2*(-c^2*x^2+1)^(1/2)-8/15*(-c^2*x^2+1)^(1/2))+5*d*c^2*(-1/3*c^2*x^2*(-c^2*x^2+1
)^(1/2)-2/3*(-c^2*x^2+1)^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.90 \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {45 \, a c^{5} e x^{5} + 75 \, a c^{5} d x^{3} + 15 \, {\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3}\right )} \arccos \left (c x\right ) - {\left (9 \, b c^{4} e x^{4} + 50 \, b c^{2} d + {\left (25 \, b c^{4} d + 12 \, b c^{2} e\right )} x^{2} + 24 \, b e\right )} \sqrt {-c^{2} x^{2} + 1}}{225 \, c^{5}} \]

[In]

integrate(x^2*(e*x^2+d)*(a+b*arccos(c*x)),x, algorithm="fricas")

[Out]

1/225*(45*a*c^5*e*x^5 + 75*a*c^5*d*x^3 + 15*(3*b*c^5*e*x^5 + 5*b*c^5*d*x^3)*arccos(c*x) - (9*b*c^4*e*x^4 + 50*
b*c^2*d + (25*b*c^4*d + 12*b*c^2*e)*x^2 + 24*b*e)*sqrt(-c^2*x^2 + 1))/c^5

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.48 \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{5}}{5} + \frac {b d x^{3} \operatorname {acos}{\left (c x \right )}}{3} + \frac {b e x^{5} \operatorname {acos}{\left (c x \right )}}{5} - \frac {b d x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} - \frac {b e x^{4} \sqrt {- c^{2} x^{2} + 1}}{25 c} - \frac {2 b d \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} - \frac {4 b e x^{2} \sqrt {- c^{2} x^{2} + 1}}{75 c^{3}} - \frac {8 b e \sqrt {- c^{2} x^{2} + 1}}{75 c^{5}} & \text {for}\: c \neq 0 \\\left (a + \frac {\pi b}{2}\right ) \left (\frac {d x^{3}}{3} + \frac {e x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(e*x**2+d)*(a+b*acos(c*x)),x)

[Out]

Piecewise((a*d*x**3/3 + a*e*x**5/5 + b*d*x**3*acos(c*x)/3 + b*e*x**5*acos(c*x)/5 - b*d*x**2*sqrt(-c**2*x**2 +
1)/(9*c) - b*e*x**4*sqrt(-c**2*x**2 + 1)/(25*c) - 2*b*d*sqrt(-c**2*x**2 + 1)/(9*c**3) - 4*b*e*x**2*sqrt(-c**2*
x**2 + 1)/(75*c**3) - 8*b*e*sqrt(-c**2*x**2 + 1)/(75*c**5), Ne(c, 0)), ((a + pi*b/2)*(d*x**3/3 + e*x**5/5), Tr
ue))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.20 \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {1}{5} \, a e x^{5} + \frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \arccos \left (c x\right ) - c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d + \frac {1}{75} \, {\left (15 \, x^{5} \arccos \left (c x\right ) - {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b e \]

[In]

integrate(x^2*(e*x^2+d)*(a+b*arccos(c*x)),x, algorithm="maxima")

[Out]

1/5*a*e*x^5 + 1/3*a*d*x^3 + 1/9*(3*x^3*arccos(c*x) - c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4)
)*b*d + 1/75*(15*x^5*arccos(c*x) - (3*sqrt(-c^2*x^2 + 1)*x^4/c^2 + 4*sqrt(-c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(-c^2*
x^2 + 1)/c^6)*c)*b*e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.18 \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {1}{5} \, b e x^{5} \arccos \left (c x\right ) + \frac {1}{5} \, a e x^{5} + \frac {1}{3} \, b d x^{3} \arccos \left (c x\right ) - \frac {\sqrt {-c^{2} x^{2} + 1} b e x^{4}}{25 \, c} + \frac {1}{3} \, a d x^{3} - \frac {\sqrt {-c^{2} x^{2} + 1} b d x^{2}}{9 \, c} - \frac {4 \, \sqrt {-c^{2} x^{2} + 1} b e x^{2}}{75 \, c^{3}} - \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b d}{9 \, c^{3}} - \frac {8 \, \sqrt {-c^{2} x^{2} + 1} b e}{75 \, c^{5}} \]

[In]

integrate(x^2*(e*x^2+d)*(a+b*arccos(c*x)),x, algorithm="giac")

[Out]

1/5*b*e*x^5*arccos(c*x) + 1/5*a*e*x^5 + 1/3*b*d*x^3*arccos(c*x) - 1/25*sqrt(-c^2*x^2 + 1)*b*e*x^4/c + 1/3*a*d*
x^3 - 1/9*sqrt(-c^2*x^2 + 1)*b*d*x^2/c - 4/75*sqrt(-c^2*x^2 + 1)*b*e*x^2/c^3 - 2/9*sqrt(-c^2*x^2 + 1)*b*d/c^3
- 8/75*sqrt(-c^2*x^2 + 1)*b*e/c^5

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\int x^2\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,\left (e\,x^2+d\right ) \,d x \]

[In]

int(x^2*(a + b*acos(c*x))*(d + e*x^2),x)

[Out]

int(x^2*(a + b*acos(c*x))*(d + e*x^2), x)