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

Optimal. Leaf size=161 \[ \frac{1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b d^2 \left (1-c^2 x^2\right )^{7/2}}{49 c^3}+\frac{b d^2 \left (1-c^2 x^2\right )^{5/2}}{175 c^3}+\frac{4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{315 c^3}+\frac{8 b d^2 \sqrt{1-c^2 x^2}}{105 c^3} \]

[Out]

(8*b*d^2*Sqrt[1 - c^2*x^2])/(105*c^3) + (4*b*d^2*(1 - c^2*x^2)^(3/2))/(315*c^3) + (b*d^2*(1 - c^2*x^2)^(5/2))/
(175*c^3) - (b*d^2*(1 - c^2*x^2)^(7/2))/(49*c^3) + (d^2*x^3*(a + b*ArcSin[c*x]))/3 - (2*c^2*d^2*x^5*(a + b*Arc
Sin[c*x]))/5 + (c^4*d^2*x^7*(a + b*ArcSin[c*x]))/7

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Rubi [A]  time = 0.169966, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {270, 4687, 12, 1251, 771} \[ \frac{1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b d^2 \left (1-c^2 x^2\right )^{7/2}}{49 c^3}+\frac{b d^2 \left (1-c^2 x^2\right )^{5/2}}{175 c^3}+\frac{4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{315 c^3}+\frac{8 b d^2 \sqrt{1-c^2 x^2}}{105 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^2*Sqrt[1 - c^2*x^2])/(105*c^3) + (4*b*d^2*(1 - c^2*x^2)^(3/2))/(315*c^3) + (b*d^2*(1 - c^2*x^2)^(5/2))/
(175*c^3) - (b*d^2*(1 - c^2*x^2)^(7/2))/(49*c^3) + (d^2*x^3*(a + b*ArcSin[c*x]))/3 - (2*c^2*d^2*x^5*(a + b*Arc
Sin[c*x]))/5 + (c^4*d^2*x^7*(a + b*ArcSin[c*x]))/7

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int x^2 \left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{d^2 x^3 \left (35-42 c^2 x^2+15 c^4 x^4\right )}{105 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{105} \left (b c d^2\right ) \int \frac{x^3 \left (35-42 c^2 x^2+15 c^4 x^4\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{210} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{x \left (35-42 c^2 x+15 c^4 x^2\right )}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{210} \left (b c d^2\right ) \operatorname{Subst}\left (\int \left (\frac{8}{c^2 \sqrt{1-c^2 x}}+\frac{4 \sqrt{1-c^2 x}}{c^2}+\frac{3 \left (1-c^2 x\right )^{3/2}}{c^2}-\frac{15 \left (1-c^2 x\right )^{5/2}}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac{8 b d^2 \sqrt{1-c^2 x^2}}{105 c^3}+\frac{4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{315 c^3}+\frac{b d^2 \left (1-c^2 x^2\right )^{5/2}}{175 c^3}-\frac{b d^2 \left (1-c^2 x^2\right )^{7/2}}{49 c^3}+\frac{1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0922021, size = 111, normalized size = 0.69 \[ \frac{d^2 \left (105 a c^3 x^3 \left (15 c^4 x^4-42 c^2 x^2+35\right )+b \sqrt{1-c^2 x^2} \left (225 c^6 x^6-612 c^4 x^4+409 c^2 x^2+818\right )+105 b c^3 x^3 \left (15 c^4 x^4-42 c^2 x^2+35\right ) \sin ^{-1}(c x)\right )}{11025 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(105*a*c^3*x^3*(35 - 42*c^2*x^2 + 15*c^4*x^4) + b*Sqrt[1 - c^2*x^2]*(818 + 409*c^2*x^2 - 612*c^4*x^4 + 22
5*c^6*x^6) + 105*b*c^3*x^3*(35 - 42*c^2*x^2 + 15*c^4*x^4)*ArcSin[c*x]))/(11025*c^3)

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Maple [A]  time = 0.006, size = 152, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{3}} \left ({d}^{2}a \left ({\frac{{c}^{7}{x}^{7}}{7}}-{\frac{2\,{c}^{5}{x}^{5}}{5}}+{\frac{{c}^{3}{x}^{3}}{3}} \right ) +{d}^{2}b \left ({\frac{\arcsin \left ( cx \right ){c}^{7}{x}^{7}}{7}}-{\frac{2\,\arcsin \left ( cx \right ){c}^{5}{x}^{5}}{5}}+{\frac{{c}^{3}{x}^{3}\arcsin \left ( cx \right ) }{3}}+{\frac{{c}^{6}{x}^{6}}{49}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{68\,{c}^{4}{x}^{4}}{1225}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{409\,{c}^{2}{x}^{2}}{11025}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{818}{11025}\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)^2*(a+b*arcsin(c*x)),x)

[Out]

1/c^3*(d^2*a*(1/7*c^7*x^7-2/5*c^5*x^5+1/3*c^3*x^3)+d^2*b*(1/7*arcsin(c*x)*c^7*x^7-2/5*arcsin(c*x)*c^5*x^5+1/3*
c^3*x^3*arcsin(c*x)+1/49*c^6*x^6*(-c^2*x^2+1)^(1/2)-68/1225*c^4*x^4*(-c^2*x^2+1)^(1/2)+409/11025*c^2*x^2*(-c^2
*x^2+1)^(1/2)+818/11025*(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.58702, size = 360, normalized size = 2.24 \begin{align*} \frac{1}{7} \, a c^{4} d^{2} x^{7} - \frac{2}{5} \, a c^{2} d^{2} 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^{4} d^{2} - \frac{2}{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^{2} + \frac{1}{3} \, a d^{2} 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^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^4*d^2*x^7 - 2/5*a*c^2*d^2*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^4*d^2 - 2/75*(15*x^5*arc
sin(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
^2 + 1/3*a*d^2*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^2

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Fricas [A]  time = 2.55373, size = 329, normalized size = 2.04 \begin{align*} \frac{1575 \, a c^{7} d^{2} x^{7} - 4410 \, a c^{5} d^{2} x^{5} + 3675 \, a c^{3} d^{2} x^{3} + 105 \,{\left (15 \, b c^{7} d^{2} x^{7} - 42 \, b c^{5} d^{2} x^{5} + 35 \, b c^{3} d^{2} x^{3}\right )} \arcsin \left (c x\right ) +{\left (225 \, b c^{6} d^{2} x^{6} - 612 \, b c^{4} d^{2} x^{4} + 409 \, b c^{2} d^{2} x^{2} + 818 \, b d^{2}\right )} \sqrt{-c^{2} x^{2} + 1}}{11025 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11025*(1575*a*c^7*d^2*x^7 - 4410*a*c^5*d^2*x^5 + 3675*a*c^3*d^2*x^3 + 105*(15*b*c^7*d^2*x^7 - 42*b*c^5*d^2*x
^5 + 35*b*c^3*d^2*x^3)*arcsin(c*x) + (225*b*c^6*d^2*x^6 - 612*b*c^4*d^2*x^4 + 409*b*c^2*d^2*x^2 + 818*b*d^2)*s
qrt(-c^2*x^2 + 1))/c^3

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Sympy [A]  time = 14.9122, size = 202, normalized size = 1.25 \begin{align*} \begin{cases} \frac{a c^{4} d^{2} x^{7}}{7} - \frac{2 a c^{2} d^{2} x^{5}}{5} + \frac{a d^{2} x^{3}}{3} + \frac{b c^{4} d^{2} x^{7} \operatorname{asin}{\left (c x \right )}}{7} + \frac{b c^{3} d^{2} x^{6} \sqrt{- c^{2} x^{2} + 1}}{49} - \frac{2 b c^{2} d^{2} x^{5} \operatorname{asin}{\left (c x \right )}}{5} - \frac{68 b c d^{2} x^{4} \sqrt{- c^{2} x^{2} + 1}}{1225} + \frac{b d^{2} x^{3} \operatorname{asin}{\left (c x \right )}}{3} + \frac{409 b d^{2} x^{2} \sqrt{- c^{2} x^{2} + 1}}{11025 c} + \frac{818 b d^{2} \sqrt{- c^{2} x^{2} + 1}}{11025 c^{3}} & \text{for}\: c \neq 0 \\\frac{a d^{2} 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)**2*(a+b*asin(c*x)),x)

[Out]

Piecewise((a*c**4*d**2*x**7/7 - 2*a*c**2*d**2*x**5/5 + a*d**2*x**3/3 + b*c**4*d**2*x**7*asin(c*x)/7 + b*c**3*d
**2*x**6*sqrt(-c**2*x**2 + 1)/49 - 2*b*c**2*d**2*x**5*asin(c*x)/5 - 68*b*c*d**2*x**4*sqrt(-c**2*x**2 + 1)/1225
+ b*d**2*x**3*asin(c*x)/3 + 409*b*d**2*x**2*sqrt(-c**2*x**2 + 1)/(11025*c) + 818*b*d**2*sqrt(-c**2*x**2 + 1)/
(11025*c**3), Ne(c, 0)), (a*d**2*x**3/3, True))

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Giac [A]  time = 1.30501, size = 306, normalized size = 1.9 \begin{align*} \frac{1}{7} \, a c^{4} d^{2} x^{7} - \frac{2}{5} \, a c^{2} d^{2} x^{5} + \frac{1}{3} \, a d^{2} x^{3} + \frac{{\left (c^{2} x^{2} - 1\right )}^{3} b d^{2} x \arcsin \left (c x\right )}{7 \, c^{2}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{2} b d^{2} x \arcsin \left (c x\right )}{35 \, c^{2}} - \frac{4 \,{\left (c^{2} x^{2} - 1\right )} b d^{2} x \arcsin \left (c x\right )}{105 \, c^{2}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{3} \sqrt{-c^{2} x^{2} + 1} b d^{2}}{49 \, c^{3}} + \frac{8 \, b d^{2} x \arcsin \left (c x\right )}{105 \, c^{2}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt{-c^{2} x^{2} + 1} b d^{2}}{175 \, c^{3}} + \frac{4 \,{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b d^{2}}{315 \, c^{3}} + \frac{8 \, \sqrt{-c^{2} x^{2} + 1} b d^{2}}{105 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^4*d^2*x^7 - 2/5*a*c^2*d^2*x^5 + 1/3*a*d^2*x^3 + 1/7*(c^2*x^2 - 1)^3*b*d^2*x*arcsin(c*x)/c^2 + 1/35*(c^
2*x^2 - 1)^2*b*d^2*x*arcsin(c*x)/c^2 - 4/105*(c^2*x^2 - 1)*b*d^2*x*arcsin(c*x)/c^2 + 1/49*(c^2*x^2 - 1)^3*sqrt
(-c^2*x^2 + 1)*b*d^2/c^3 + 8/105*b*d^2*x*arcsin(c*x)/c^2 + 1/175*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d^2/c^3
+ 4/315*(-c^2*x^2 + 1)^(3/2)*b*d^2/c^3 + 8/105*sqrt(-c^2*x^2 + 1)*b*d^2/c^3