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

Optimal. Leaf size=102 \[ \frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac{4 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac{4 b^2 x}{9 c^2}-\frac{2}{27} b^2 x^3 \]

[Out]

(-4*b^2*x)/(9*c^2) - (2*b^2*x^3)/27 + (4*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c^3) + (2*b*x^2*Sqrt[1 -
c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c) + (x^3*(a + b*ArcSin[c*x])^2)/3

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Rubi [A]  time = 0.153869, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4627, 4707, 4677, 8, 30} \[ \frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac{4 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac{4 b^2 x}{9 c^2}-\frac{2}{27} b^2 x^3 \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSin[c*x])^2,x]

[Out]

(-4*b^2*x)/(9*c^2) - (2*b^2*x^3)/27 + (4*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c^3) + (2*b*x^2*Sqrt[1 -
c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c) + (x^3*(a + b*ArcSin[c*x])^2)/3

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4707

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcSin[c*x])^n)/Sqrt[d + e*x^2], x], x] + Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^2 \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac{1}{3} (2 b c) \int \frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac{1}{9} \left (2 b^2\right ) \int x^2 \, dx-\frac{(4 b) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{9 c}\\ &=-\frac{2}{27} b^2 x^3+\frac{4 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac{\left (4 b^2\right ) \int 1 \, dx}{9 c^2}\\ &=-\frac{4 b^2 x}{9 c^2}-\frac{2 b^2 x^3}{27}+\frac{4 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}(c x)\right )^2\\ \end{align*}

Mathematica [A]  time = 0.188787, size = 95, normalized size = 0.93 \[ \frac{1}{3} \left (x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac{2 b \left (-3 c^2 x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-6 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+b c^3 x^3+6 b c x\right )}{9 c^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(a + b*ArcSin[c*x])^2 - (2*b*(6*b*c*x + b*c^3*x^3 - 6*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) - 3*c^2*x^2*S
qrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])))/(9*c^3))/3

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Maple [A]  time = 0.025, size = 126, normalized size = 1.2 \begin{align*}{\frac{1}{{c}^{3}} \left ({\frac{{a}^{2}{c}^{3}{x}^{3}}{3}}+{b}^{2} \left ({\frac{{c}^{3}{x}^{3} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{3}}+{\frac{2\,\arcsin \left ( cx \right ) \left ({c}^{2}{x}^{2}+2 \right ) }{9}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{2\,{c}^{3}{x}^{3}}{27}}-{\frac{4\,cx}{9}} \right ) +2\,ab \left ( 1/3\,{c}^{3}{x}^{3}\arcsin \left ( cx \right ) +1/9\,{c}^{2}{x}^{2}\sqrt{-{c}^{2}{x}^{2}+1}+2/9\,\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(1/3*a^2*c^3*x^3+b^2*(1/3*c^3*x^3*arcsin(c*x)^2+2/9*arcsin(c*x)*(c^2*x^2+2)*(-c^2*x^2+1)^(1/2)-2/27*c^3*
x^3-4/9*c*x)+2*a*b*(1/3*c^3*x^3*arcsin(c*x)+1/9*c^2*x^2*(-c^2*x^2+1)^(1/2)+2/9*(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.54948, size = 192, normalized size = 1.88 \begin{align*} \frac{1}{3} \, b^{2} x^{3} \arcsin \left (c x\right )^{2} + \frac{1}{3} \, a^{2} x^{3} + \frac{2}{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 )} a b + \frac{2}{27} \,{\left (3 \, c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{c^{4}}\right )} \arcsin \left (c x\right ) - \frac{c^{2} x^{3} + 6 \, x}{c^{2}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3*arcsin(c*x)^2 + 1/3*a^2*x^3 + 2/9*(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))*a*b + 2/27*(3*c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4)*arcsin(c*x) - (c^2*x^3
+ 6*x)/c^2)*b^2

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Fricas [A]  time = 1.43564, size = 255, normalized size = 2.5 \begin{align*} \frac{9 \, b^{2} c^{3} x^{3} \arcsin \left (c x\right )^{2} + 18 \, a b c^{3} x^{3} \arcsin \left (c x\right ) +{\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{3} x^{3} - 12 \, b^{2} c x + 6 \,{\left (a b c^{2} x^{2} + 2 \, a b +{\left (b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} x^{2} + 1}}{27 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(9*b^2*c^3*x^3*arcsin(c*x)^2 + 18*a*b*c^3*x^3*arcsin(c*x) + (9*a^2 - 2*b^2)*c^3*x^3 - 12*b^2*c*x + 6*(a*b
*c^2*x^2 + 2*a*b + (b^2*c^2*x^2 + 2*b^2)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))/c^3

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Sympy [A]  time = 2.05873, size = 170, normalized size = 1.67 \begin{align*} \begin{cases} \frac{a^{2} x^{3}}{3} + \frac{2 a b x^{3} \operatorname{asin}{\left (c x \right )}}{3} + \frac{2 a b x^{2} \sqrt{- c^{2} x^{2} + 1}}{9 c} + \frac{4 a b \sqrt{- c^{2} x^{2} + 1}}{9 c^{3}} + \frac{b^{2} x^{3} \operatorname{asin}^{2}{\left (c x \right )}}{3} - \frac{2 b^{2} x^{3}}{27} + \frac{2 b^{2} x^{2} \sqrt{- c^{2} x^{2} + 1} \operatorname{asin}{\left (c x \right )}}{9 c} - \frac{4 b^{2} x}{9 c^{2}} + \frac{4 b^{2} \sqrt{- c^{2} x^{2} + 1} \operatorname{asin}{\left (c x \right )}}{9 c^{3}} & \text{for}\: c \neq 0 \\\frac{a^{2} x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x**3/3 + 2*a*b*x**3*asin(c*x)/3 + 2*a*b*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + 4*a*b*sqrt(-c**2*x**
2 + 1)/(9*c**3) + b**2*x**3*asin(c*x)**2/3 - 2*b**2*x**3/27 + 2*b**2*x**2*sqrt(-c**2*x**2 + 1)*asin(c*x)/(9*c)
- 4*b**2*x/(9*c**2) + 4*b**2*sqrt(-c**2*x**2 + 1)*asin(c*x)/(9*c**3), Ne(c, 0)), (a**2*x**3/3, True))

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Giac [B]  time = 1.34679, size = 262, normalized size = 2.57 \begin{align*} \frac{1}{3} \, a^{2} x^{3} + \frac{{\left (c^{2} x^{2} - 1\right )} b^{2} x \arcsin \left (c x\right )^{2}}{3 \, c^{2}} + \frac{2 \,{\left (c^{2} x^{2} - 1\right )} a b x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac{b^{2} x \arcsin \left (c x\right )^{2}}{3 \, c^{2}} - \frac{2 \,{\left (c^{2} x^{2} - 1\right )} b^{2} x}{27 \, c^{2}} + \frac{2 \, a b x \arcsin \left (c x\right )}{3 \, c^{2}} - \frac{2 \,{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b^{2} \arcsin \left (c x\right )}{9 \, c^{3}} - \frac{14 \, b^{2} x}{27 \, c^{2}} - \frac{2 \,{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} a b}{9 \, c^{3}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1} b^{2} \arcsin \left (c x\right )}{3 \, c^{3}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1} a b}{3 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*x^3 + 1/3*(c^2*x^2 - 1)*b^2*x*arcsin(c*x)^2/c^2 + 2/3*(c^2*x^2 - 1)*a*b*x*arcsin(c*x)/c^2 + 1/3*b^2*x*
arcsin(c*x)^2/c^2 - 2/27*(c^2*x^2 - 1)*b^2*x/c^2 + 2/3*a*b*x*arcsin(c*x)/c^2 - 2/9*(-c^2*x^2 + 1)^(3/2)*b^2*ar
csin(c*x)/c^3 - 14/27*b^2*x/c^2 - 2/9*(-c^2*x^2 + 1)^(3/2)*a*b/c^3 + 2/3*sqrt(-c^2*x^2 + 1)*b^2*arcsin(c*x)/c^
3 + 2/3*sqrt(-c^2*x^2 + 1)*a*b/c^3