3.1 \(\int x^4 (d-c^2 d x^2) (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=128 \[ -\frac{1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{b d \left (1-c^2 x^2\right )^{7/2}}{49 c^5}-\frac{8 b d \left (1-c^2 x^2\right )^{5/2}}{175 c^5}+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{105 c^5}+\frac{2 b d \sqrt{1-c^2 x^2}}{35 c^5} \]

[Out]

(2*b*d*Sqrt[1 - c^2*x^2])/(35*c^5) + (b*d*(1 - c^2*x^2)^(3/2))/(105*c^5) - (8*b*d*(1 - c^2*x^2)^(5/2))/(175*c^
5) + (b*d*(1 - c^2*x^2)^(7/2))/(49*c^5) + (d*x^5*(a + b*ArcSin[c*x]))/5 - (c^2*d*x^7*(a + b*ArcSin[c*x]))/7

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Rubi [A]  time = 0.120213, antiderivative size = 128, 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} \[ -\frac{1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{b d \left (1-c^2 x^2\right )^{7/2}}{49 c^5}-\frac{8 b d \left (1-c^2 x^2\right )^{5/2}}{175 c^5}+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{105 c^5}+\frac{2 b d \sqrt{1-c^2 x^2}}{35 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(d - c^2*d*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(2*b*d*Sqrt[1 - c^2*x^2])/(35*c^5) + (b*d*(1 - c^2*x^2)^(3/2))/(105*c^5) - (8*b*d*(1 - c^2*x^2)^(5/2))/(175*c^
5) + (b*d*(1 - c^2*x^2)^(7/2))/(49*c^5) + (d*x^5*(a + b*ArcSin[c*x]))/5 - (c^2*d*x^7*(a + b*ArcSin[c*x]))/7

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^4 \left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{d x^5 \left (7-5 c^2 x^2\right )}{35 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{35} (b c d) \int \frac{x^5 \left (7-5 c^2 x^2\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{70} (b c d) \operatorname{Subst}\left (\int \frac{x^2 \left (7-5 c^2 x\right )}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{70} (b c d) \operatorname{Subst}\left (\int \left (\frac{2}{c^4 \sqrt{1-c^2 x}}+\frac{\sqrt{1-c^2 x}}{c^4}-\frac{8 \left (1-c^2 x\right )^{3/2}}{c^4}+\frac{5 \left (1-c^2 x\right )^{5/2}}{c^4}\right ) \, dx,x,x^2\right )\\ &=\frac{2 b d \sqrt{1-c^2 x^2}}{35 c^5}+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{105 c^5}-\frac{8 b d \left (1-c^2 x^2\right )^{5/2}}{175 c^5}+\frac{b d \left (1-c^2 x^2\right )^{7/2}}{49 c^5}+\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.135931, size = 87, normalized size = 0.68 \[ \frac{d \left (-105 a x^5 \left (5 c^2 x^2-7\right )+\frac{b \sqrt{1-c^2 x^2} \left (-75 c^6 x^6+57 c^4 x^4+76 c^2 x^2+152\right )}{c^5}-105 b x^5 \left (5 c^2 x^2-7\right ) \sin ^{-1}(c x)\right )}{3675} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(d - c^2*d*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(d*(-105*a*x^5*(-7 + 5*c^2*x^2) + (b*Sqrt[1 - c^2*x^2]*(152 + 76*c^2*x^2 + 57*c^4*x^4 - 75*c^6*x^6))/c^5 - 105
*b*x^5*(-7 + 5*c^2*x^2)*ArcSin[c*x]))/3675

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Maple [A]  time = 0.008, size = 130, normalized size = 1. \begin{align*}{\frac{1}{{c}^{5}} \left ( -da \left ({\frac{{c}^{7}{x}^{7}}{7}}-{\frac{{c}^{5}{x}^{5}}{5}} \right ) -db \left ({\frac{\arcsin \left ( cx \right ){c}^{7}{x}^{7}}{7}}-{\frac{\arcsin \left ( cx \right ){c}^{5}{x}^{5}}{5}}+{\frac{{c}^{6}{x}^{6}}{49}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{19\,{c}^{4}{x}^{4}}{1225}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{76\,{c}^{2}{x}^{2}}{3675}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{152}{3675}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(-c^2*d*x^2+d)*(a+b*arcsin(c*x)),x)

[Out]

1/c^5*(-d*a*(1/7*c^7*x^7-1/5*c^5*x^5)-d*b*(1/7*arcsin(c*x)*c^7*x^7-1/5*arcsin(c*x)*c^5*x^5+1/49*c^6*x^6*(-c^2*
x^2+1)^(1/2)-19/1225*c^4*x^4*(-c^2*x^2+1)^(1/2)-76/3675*c^2*x^2*(-c^2*x^2+1)^(1/2)-152/3675*(-c^2*x^2+1)^(1/2)
))

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Maxima [A]  time = 1.62084, size = 255, normalized size = 1.99 \begin{align*} -\frac{1}{7} \, a c^{2} d x^{7} + \frac{1}{5} \, a d x^{5} - \frac{1}{245} \,{\left (35 \, x^{7} \arcsin \left (c x\right ) +{\left (\frac{5 \, \sqrt{-c^{2} x^{2} + 1} x^{6}}{c^{2}} + \frac{6 \, \sqrt{-c^{2} x^{2} + 1} x^{4}}{c^{4}} + \frac{8 \, \sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{6}} + \frac{16 \, \sqrt{-c^{2} x^{2} + 1}}{c^{8}}\right )} c\right )} b c^{2} d + \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 d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-c^2*d*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

-1/7*a*c^2*d*x^7 + 1/5*a*d*x^5 - 1/245*(35*x^7*arcsin(c*x) + (5*sqrt(-c^2*x^2 + 1)*x^6/c^2 + 6*sqrt(-c^2*x^2 +
1)*x^4/c^4 + 8*sqrt(-c^2*x^2 + 1)*x^2/c^6 + 16*sqrt(-c^2*x^2 + 1)/c^8)*c)*b*c^2*d + 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*d

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Fricas [A]  time = 2.13976, size = 244, normalized size = 1.91 \begin{align*} -\frac{525 \, a c^{7} d x^{7} - 735 \, a c^{5} d x^{5} + 105 \,{\left (5 \, b c^{7} d x^{7} - 7 \, b c^{5} d x^{5}\right )} \arcsin \left (c x\right ) +{\left (75 \, b c^{6} d x^{6} - 57 \, b c^{4} d x^{4} - 76 \, b c^{2} d x^{2} - 152 \, b d\right )} \sqrt{-c^{2} x^{2} + 1}}{3675 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-c^2*d*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

-1/3675*(525*a*c^7*d*x^7 - 735*a*c^5*d*x^5 + 105*(5*b*c^7*d*x^7 - 7*b*c^5*d*x^5)*arcsin(c*x) + (75*b*c^6*d*x^6
- 57*b*c^4*d*x^4 - 76*b*c^2*d*x^2 - 152*b*d)*sqrt(-c^2*x^2 + 1))/c^5

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Sympy [A]  time = 12.6751, size = 151, normalized size = 1.18 \begin{align*} \begin{cases} - \frac{a c^{2} d x^{7}}{7} + \frac{a d x^{5}}{5} - \frac{b c^{2} d x^{7} \operatorname{asin}{\left (c x \right )}}{7} - \frac{b c d x^{6} \sqrt{- c^{2} x^{2} + 1}}{49} + \frac{b d x^{5} \operatorname{asin}{\left (c x \right )}}{5} + \frac{19 b d x^{4} \sqrt{- c^{2} x^{2} + 1}}{1225 c} + \frac{76 b d x^{2} \sqrt{- c^{2} x^{2} + 1}}{3675 c^{3}} + \frac{152 b d \sqrt{- c^{2} x^{2} + 1}}{3675 c^{5}} & \text{for}\: c \neq 0 \\\frac{a d x^{5}}{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(-c**2*d*x**2+d)*(a+b*asin(c*x)),x)

[Out]

Piecewise((-a*c**2*d*x**7/7 + a*d*x**5/5 - b*c**2*d*x**7*asin(c*x)/7 - b*c*d*x**6*sqrt(-c**2*x**2 + 1)/49 + b*
d*x**5*asin(c*x)/5 + 19*b*d*x**4*sqrt(-c**2*x**2 + 1)/(1225*c) + 76*b*d*x**2*sqrt(-c**2*x**2 + 1)/(3675*c**3)
+ 152*b*d*sqrt(-c**2*x**2 + 1)/(3675*c**5), Ne(c, 0)), (a*d*x**5/5, True))

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Giac [A]  time = 1.42701, size = 263, normalized size = 2.05 \begin{align*} -\frac{1}{7} \, a c^{2} d x^{7} + \frac{1}{5} \, a d x^{5} - \frac{{\left (c^{2} x^{2} - 1\right )}^{3} b d x \arcsin \left (c x\right )}{7 \, c^{4}} - \frac{8 \,{\left (c^{2} x^{2} - 1\right )}^{2} b d x \arcsin \left (c x\right )}{35 \, c^{4}} - \frac{{\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right )}{35 \, c^{4}} - \frac{{\left (c^{2} x^{2} - 1\right )}^{3} \sqrt{-c^{2} x^{2} + 1} b d}{49 \, c^{5}} + \frac{2 \, b d x \arcsin \left (c x\right )}{35 \, c^{4}} - \frac{8 \,{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt{-c^{2} x^{2} + 1} b d}{175 \, c^{5}} + \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b d}{105 \, c^{5}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1} b d}{35 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-c^2*d*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

-1/7*a*c^2*d*x^7 + 1/5*a*d*x^5 - 1/7*(c^2*x^2 - 1)^3*b*d*x*arcsin(c*x)/c^4 - 8/35*(c^2*x^2 - 1)^2*b*d*x*arcsin
(c*x)/c^4 - 1/35*(c^2*x^2 - 1)*b*d*x*arcsin(c*x)/c^4 - 1/49*(c^2*x^2 - 1)^3*sqrt(-c^2*x^2 + 1)*b*d/c^5 + 2/35*
b*d*x*arcsin(c*x)/c^4 - 8/175*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d/c^5 + 1/105*(-c^2*x^2 + 1)^(3/2)*b*d/c^5
+ 2/35*sqrt(-c^2*x^2 + 1)*b*d/c^5