3.3 \(\int x^2 (d-c^2 d x^2) (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=105 \[ -\frac{1}{5} c^2 d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b d \left (1-c^2 x^2\right )^{5/2}}{25 c^3}+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{45 c^3}+\frac{2 b d \sqrt{1-c^2 x^2}}{15 c^3} \]

[Out]

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

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Rubi [A]  time = 0.103338, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {14, 4687, 12, 446, 77} \[ -\frac{1}{5} c^2 d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b d \left (1-c^2 x^2\right )^{5/2}}{25 c^3}+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{45 c^3}+\frac{2 b d \sqrt{1-c^2 x^2}}{15 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d - c^2*d*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

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

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 4687

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = I
ntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[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] && EqQ[c^2*d + e, 0] && IGtQ[p, 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 446

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 77

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]))))

Rubi steps

\begin{align*} \int x^2 \left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{d x^3 \left (5-3 c^2 x^2\right )}{15 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{3} d x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{15} (b c d) \int \frac{x^3 \left (5-3 c^2 x^2\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{3} d x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{30} (b c d) \operatorname{Subst}\left (\int \frac{x \left (5-3 c^2 x\right )}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{3} d x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{30} (b c d) \operatorname{Subst}\left (\int \left (\frac{2}{c^2 \sqrt{1-c^2 x}}+\frac{\sqrt{1-c^2 x}}{c^2}-\frac{3 \left (1-c^2 x\right )^{3/2}}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac{2 b d \sqrt{1-c^2 x^2}}{15 c^3}+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{45 c^3}-\frac{b d \left (1-c^2 x^2\right )^{5/2}}{25 c^3}+\frac{1}{3} d x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0962014, size = 85, normalized size = 0.81 \[ \frac{d \left (a \left (75 c^3 x^3-45 c^5 x^5\right )+b \sqrt{1-c^2 x^2} \left (-9 c^4 x^4+13 c^2 x^2+26\right )+15 b c^3 x^3 \left (5-3 c^2 x^2\right ) \sin ^{-1}(c x)\right )}{225 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(d - c^2*d*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(d*(b*Sqrt[1 - c^2*x^2]*(26 + 13*c^2*x^2 - 9*c^4*x^4) + a*(75*c^3*x^3 - 45*c^5*x^5) + 15*b*c^3*x^3*(5 - 3*c^2*
x^2)*ArcSin[c*x]))/(225*c^3)

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Maple [A]  time = 0.006, size = 110, normalized size = 1.1 \begin{align*}{\frac{1}{{c}^{3}} \left ( -da \left ({\frac{{c}^{5}{x}^{5}}{5}}-{\frac{{c}^{3}{x}^{3}}{3}} \right ) -db \left ({\frac{\arcsin \left ( cx \right ){c}^{5}{x}^{5}}{5}}-{\frac{{c}^{3}{x}^{3}\arcsin \left ( cx \right ) }{3}}+{\frac{{c}^{4}{x}^{4}}{25}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{13\,{c}^{2}{x}^{2}}{225}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{26}{225}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-c^2*d*x^2+d)*(a+b*arcsin(c*x)),x)

[Out]

1/c^3*(-d*a*(1/5*c^5*x^5-1/3*c^3*x^3)-d*b*(1/5*arcsin(c*x)*c^5*x^5-1/3*c^3*x^3*arcsin(c*x)+1/25*c^4*x^4*(-c^2*
x^2+1)^(1/2)-13/225*c^2*x^2*(-c^2*x^2+1)^(1/2)-26/225*(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.52165, size = 200, normalized size = 1.9 \begin{align*} -\frac{1}{5} \, a c^{2} d x^{5} - \frac{1}{75} \,{\left (15 \, x^{5} \arcsin \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 c^{2} d + \frac{1}{3} \, a d x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \arcsin \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.14777, size = 213, normalized size = 2.03 \begin{align*} -\frac{45 \, a c^{5} d x^{5} - 75 \, a c^{3} d x^{3} + 15 \,{\left (3 \, b c^{5} d x^{5} - 5 \, b c^{3} d x^{3}\right )} \arcsin \left (c x\right ) +{\left (9 \, b c^{4} d x^{4} - 13 \, b c^{2} d x^{2} - 26 \, b d\right )} \sqrt{-c^{2} x^{2} + 1}}{225 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/225*(45*a*c^5*d*x^5 - 75*a*c^3*d*x^3 + 15*(3*b*c^5*d*x^5 - 5*b*c^3*d*x^3)*arcsin(c*x) + (9*b*c^4*d*x^4 - 13
*b*c^2*d*x^2 - 26*b*d)*sqrt(-c^2*x^2 + 1))/c^3

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Sympy [A]  time = 3.55446, size = 126, normalized size = 1.2 \begin{align*} \begin{cases} - \frac{a c^{2} d x^{5}}{5} + \frac{a d x^{3}}{3} - \frac{b c^{2} d x^{5} \operatorname{asin}{\left (c x \right )}}{5} - \frac{b c d x^{4} \sqrt{- c^{2} x^{2} + 1}}{25} + \frac{b d x^{3} \operatorname{asin}{\left (c x \right )}}{3} + \frac{13 b d x^{2} \sqrt{- c^{2} x^{2} + 1}}{225 c} + \frac{26 b d \sqrt{- c^{2} x^{2} + 1}}{225 c^{3}} & \text{for}\: c \neq 0 \\\frac{a d x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-c**2*d*x**2+d)*(a+b*asin(c*x)),x)

[Out]

Piecewise((-a*c**2*d*x**5/5 + a*d*x**3/3 - b*c**2*d*x**5*asin(c*x)/5 - b*c*d*x**4*sqrt(-c**2*x**2 + 1)/25 + b*
d*x**3*asin(c*x)/3 + 13*b*d*x**2*sqrt(-c**2*x**2 + 1)/(225*c) + 26*b*d*sqrt(-c**2*x**2 + 1)/(225*c**3), Ne(c,
0)), (a*d*x**3/3, True))

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Giac [A]  time = 1.35016, size = 192, normalized size = 1.83 \begin{align*} -\frac{1}{5} \, a c^{2} d x^{5} + \frac{1}{3} \, a d x^{3} - \frac{{\left (c^{2} x^{2} - 1\right )}^{2} b d x \arcsin \left (c x\right )}{5 \, c^{2}} - \frac{{\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right )}{15 \, c^{2}} + \frac{2 \, b d x \arcsin \left (c x\right )}{15 \, c^{2}} - \frac{{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt{-c^{2} x^{2} + 1} b d}{25 \, c^{3}} + \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b d}{45 \, c^{3}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1} b d}{15 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a*c^2*d*x^5 + 1/3*a*d*x^3 - 1/5*(c^2*x^2 - 1)^2*b*d*x*arcsin(c*x)/c^2 - 1/15*(c^2*x^2 - 1)*b*d*x*arcsin(c
*x)/c^2 + 2/15*b*d*x*arcsin(c*x)/c^2 - 1/25*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d/c^3 + 1/45*(-c^2*x^2 + 1)^(
3/2)*b*d/c^3 + 2/15*sqrt(-c^2*x^2 + 1)*b*d/c^3