3.236 \(\int \frac{x^3 (a+b \sin ^{-1}(c x))^2}{\sqrt{d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=277 \[ \frac{4 a b x \sqrt{1-c^2 x^2}}{3 c^3 \sqrt{d-c^2 d x^2}}+\frac{2 b x^3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c \sqrt{d-c^2 d x^2}}-\frac{x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d}-\frac{2 b^2 \left (1-c^2 x^2\right )^2}{27 c^4 \sqrt{d-c^2 d x^2}}+\frac{14 b^2 \left (1-c^2 x^2\right )}{9 c^4 \sqrt{d-c^2 d x^2}}+\frac{4 b^2 x \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{3 c^3 \sqrt{d-c^2 d x^2}} \]

[Out]

(4*a*b*x*Sqrt[1 - c^2*x^2])/(3*c^3*Sqrt[d - c^2*d*x^2]) + (14*b^2*(1 - c^2*x^2))/(9*c^4*Sqrt[d - c^2*d*x^2]) -
(2*b^2*(1 - c^2*x^2)^2)/(27*c^4*Sqrt[d - c^2*d*x^2]) + (4*b^2*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(3*c^3*Sqrt[d
- c^2*d*x^2]) + (2*b*x^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c*Sqrt[d - c^2*d*x^2]) - (2*Sqrt[d - c^2*d*
x^2]*(a + b*ArcSin[c*x])^2)/(3*c^4*d) - (x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(3*c^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.329423, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {4707, 4677, 4619, 261, 4627, 266, 43} \[ \frac{4 a b x \sqrt{1-c^2 x^2}}{3 c^3 \sqrt{d-c^2 d x^2}}+\frac{2 b x^3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c \sqrt{d-c^2 d x^2}}-\frac{x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d}-\frac{2 b^2 \left (1-c^2 x^2\right )^2}{27 c^4 \sqrt{d-c^2 d x^2}}+\frac{14 b^2 \left (1-c^2 x^2\right )}{9 c^4 \sqrt{d-c^2 d x^2}}+\frac{4 b^2 x \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{3 c^3 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSin[c*x])^2)/Sqrt[d - c^2*d*x^2],x]

[Out]

(4*a*b*x*Sqrt[1 - c^2*x^2])/(3*c^3*Sqrt[d - c^2*d*x^2]) + (14*b^2*(1 - c^2*x^2))/(9*c^4*Sqrt[d - c^2*d*x^2]) -
(2*b^2*(1 - c^2*x^2)^2)/(27*c^4*Sqrt[d - c^2*d*x^2]) + (4*b^2*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(3*c^3*Sqrt[d
- c^2*d*x^2]) + (2*b*x^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c*Sqrt[d - c^2*d*x^2]) - (2*Sqrt[d - c^2*d*
x^2]*(a + b*ArcSin[c*x])^2)/(3*c^4*d) - (x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(3*c^2*d)

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 4619

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{d-c^2 d x^2}} \, dx &=-\frac{x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}+\frac{2 \int \frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{d-c^2 d x^2}} \, dx}{3 c^2}+\frac{\left (2 b \sqrt{1-c^2 x^2}\right ) \int x^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx}{3 c \sqrt{d-c^2 d x^2}}\\ &=\frac{2 b x^3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c \sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d}-\frac{x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{\left (2 b^2 \sqrt{1-c^2 x^2}\right ) \int \frac{x^3}{\sqrt{1-c^2 x^2}} \, dx}{9 \sqrt{d-c^2 d x^2}}+\frac{\left (4 b \sqrt{1-c^2 x^2}\right ) \int \left (a+b \sin ^{-1}(c x)\right ) \, dx}{3 c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{4 a b x \sqrt{1-c^2 x^2}}{3 c^3 \sqrt{d-c^2 d x^2}}+\frac{2 b x^3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c \sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d}-\frac{x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{\left (b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )}{9 \sqrt{d-c^2 d x^2}}+\frac{\left (4 b^2 \sqrt{1-c^2 x^2}\right ) \int \sin ^{-1}(c x) \, dx}{3 c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{4 a b x \sqrt{1-c^2 x^2}}{3 c^3 \sqrt{d-c^2 d x^2}}+\frac{4 b^2 x \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{3 c^3 \sqrt{d-c^2 d x^2}}+\frac{2 b x^3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c \sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d}-\frac{x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{\left (b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \sqrt{1-c^2 x}}-\frac{\sqrt{1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{9 \sqrt{d-c^2 d x^2}}-\frac{\left (4 b^2 \sqrt{1-c^2 x^2}\right ) \int \frac{x}{\sqrt{1-c^2 x^2}} \, dx}{3 c^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{4 a b x \sqrt{1-c^2 x^2}}{3 c^3 \sqrt{d-c^2 d x^2}}+\frac{14 b^2 \left (1-c^2 x^2\right )}{9 c^4 \sqrt{d-c^2 d x^2}}-\frac{2 b^2 \left (1-c^2 x^2\right )^2}{27 c^4 \sqrt{d-c^2 d x^2}}+\frac{4 b^2 x \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{3 c^3 \sqrt{d-c^2 d x^2}}+\frac{2 b x^3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c \sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d}-\frac{x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}\\ \end{align*}

Mathematica [A]  time = 0.123847, size = 176, normalized size = 0.64 \[ \frac{9 a^2 \left (c^4 x^4+c^2 x^2-2\right )+6 a b c x \sqrt{1-c^2 x^2} \left (c^2 x^2+6\right )+6 b \sin ^{-1}(c x) \left (3 a \left (c^4 x^4+c^2 x^2-2\right )+b c x \sqrt{1-c^2 x^2} \left (c^2 x^2+6\right )\right )-2 b^2 \left (c^4 x^4+19 c^2 x^2-20\right )+9 b^2 \left (c^4 x^4+c^2 x^2-2\right ) \sin ^{-1}(c x)^2}{27 c^4 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSin[c*x])^2)/Sqrt[d - c^2*d*x^2],x]

[Out]

(6*a*b*c*x*Sqrt[1 - c^2*x^2]*(6 + c^2*x^2) + 9*a^2*(-2 + c^2*x^2 + c^4*x^4) - 2*b^2*(-20 + 19*c^2*x^2 + c^4*x^
4) + 6*b*(b*c*x*Sqrt[1 - c^2*x^2]*(6 + c^2*x^2) + 3*a*(-2 + c^2*x^2 + c^4*x^4))*ArcSin[c*x] + 9*b^2*(-2 + c^2*
x^2 + c^4*x^4)*ArcSin[c*x]^2)/(27*c^4*Sqrt[d - c^2*d*x^2])

________________________________________________________________________________________

Maple [C]  time = 0.394, size = 750, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x)

[Out]

a^2*(-1/3*x^2/c^2/d*(-c^2*d*x^2+d)^(1/2)-2/3/d/c^4*(-c^2*d*x^2+d)^(1/2))+b^2*(-1/216*(-d*(c^2*x^2-1))^(1/2)*(4
*c^4*x^4-5*c^2*x^2-4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+3*I*(-c^2*x^2+1)^(1/2)*x*c+1)*(6*I*arcsin(c*x)+9*arcsin(c*x)
^2-2)/c^4/d/(c^2*x^2-1)-3/8*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)^2-2+2*I*a
rcsin(c*x))/c^4/d/(c^2*x^2-1)-3/8*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(arcsin(c*x)^2-2
-2*I*arcsin(c*x))/c^4/d/(c^2*x^2-1)-1/216*(-d*(c^2*x^2-1))^(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+4*c^4*x^4-3*I
*(-c^2*x^2+1)^(1/2)*x*c-5*c^2*x^2+1)*(-6*I*arcsin(c*x)+9*arcsin(c*x)^2-2)/c^4/d/(c^2*x^2-1))+2*a*b*(-1/72*(-d*
(c^2*x^2-1))^(1/2)*(4*c^4*x^4-5*c^2*x^2-4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+3*I*(-c^2*x^2+1)^(1/2)*x*c+1)*(I+3*arcs
in(c*x))/c^4/d/(c^2*x^2-1)-3/8*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)+I)/c^4
/d/(c^2*x^2-1)-3/8*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(arcsin(c*x)-I)/c^4/d/(c^2*x^2-
1)-1/72*(-d*(c^2*x^2-1))^(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+4*c^4*x^4-3*I*(-c^2*x^2+1)^(1/2)*x*c-5*c^2*x^2+
1)*(-I+3*arcsin(c*x))/c^4/d/(c^2*x^2-1))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.01907, size = 456, normalized size = 1.65 \begin{align*} -\frac{6 \,{\left (a b c^{3} x^{3} + 6 \, a b c x +{\left (b^{2} c^{3} x^{3} + 6 \, b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} +{\left ({\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{4} x^{4} +{\left (9 \, a^{2} - 38 \, b^{2}\right )} c^{2} x^{2} + 9 \,{\left (b^{2} c^{4} x^{4} + b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \arcsin \left (c x\right )^{2} - 18 \, a^{2} + 40 \, b^{2} + 18 \,{\left (a b c^{4} x^{4} + a b c^{2} x^{2} - 2 \, a b\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{27 \,{\left (c^{6} d x^{2} - c^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

-1/27*(6*(a*b*c^3*x^3 + 6*a*b*c*x + (b^2*c^3*x^3 + 6*b^2*c*x)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2
+ 1) + ((9*a^2 - 2*b^2)*c^4*x^4 + (9*a^2 - 38*b^2)*c^2*x^2 + 9*(b^2*c^4*x^4 + b^2*c^2*x^2 - 2*b^2)*arcsin(c*x)
^2 - 18*a^2 + 40*b^2 + 18*(a*b*c^4*x^4 + a*b*c^2*x^2 - 2*a*b)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^6*d*x^2 -
c^4*d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (a + b \operatorname{asin}{\left (c x \right )}\right )^{2}}{\sqrt{- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asin(c*x))**2/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**3*(a + b*asin(c*x))**2/sqrt(-d*(c*x - 1)*(c*x + 1)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{3}}{\sqrt{-c^{2} d x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^2*x^3/sqrt(-c^2*d*x^2 + d), x)