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

Optimal. Leaf size=128 \[ \frac{1}{3} d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2+\frac{2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac{4 b d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{2}{3} d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac{2}{27} b^2 c^2 d x^3-\frac{14}{9} b^2 d x \]

[Out]

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

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Rubi [A]  time = 0.137281, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4649, 4619, 4677, 8} \[ \frac{1}{3} d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2+\frac{2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac{4 b d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{2}{3} d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac{2}{27} b^2 c^2 d x^3-\frac{14}{9} b^2 d x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4649

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.203339, size = 137, normalized size = 1.07 \[ -\frac{d \left (9 a^2 c x \left (c^2 x^2-3\right )+6 a b \sqrt{1-c^2 x^2} \left (c^2 x^2-7\right )+6 b \sin ^{-1}(c x) \left (3 a c x \left (c^2 x^2-3\right )+b \sqrt{1-c^2 x^2} \left (c^2 x^2-7\right )\right )-2 b^2 c x \left (c^2 x^2-21\right )+9 b^2 c x \left (c^2 x^2-3\right ) \sin ^{-1}(c x)^2\right )}{27 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(-2*b^2*c*x*(-21 + c^2*x^2) + 6*a*b*Sqrt[1 - c^2*x^2]*(-7 + c^2*x^2) + 9*a^2*c*x*(-3 + c^2*x^2) + 6*b*(b*S
qrt[1 - c^2*x^2]*(-7 + c^2*x^2) + 3*a*c*x*(-3 + c^2*x^2))*ArcSin[c*x] + 9*b^2*c*x*(-3 + c^2*x^2)*ArcSin[c*x]^2
))/(27*c)

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Maple [A]  time = 0.033, size = 173, normalized size = 1.4 \begin{align*}{\frac{1}{c} \left ( -d{a}^{2} \left ({\frac{{c}^{3}{x}^{3}}{3}}-cx \right ) -d{b}^{2} \left ({\frac{ \left ( \arcsin \left ( cx \right ) \right ) ^{2} \left ({c}^{2}{x}^{2}-3 \right ) cx}{3}}+{\frac{4\,cx}{3}}-{\frac{4\,\arcsin \left ( cx \right ) }{3}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{2\, \left ({c}^{2}{x}^{2}-1 \right ) \arcsin \left ( cx \right ) }{9}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{ \left ( 2\,{c}^{2}{x}^{2}-6 \right ) cx}{27}} \right ) -2\,dab \left ( 1/3\,{c}^{3}{x}^{3}\arcsin \left ( cx \right ) -cx\arcsin \left ( cx \right ) +1/9\,{c}^{2}{x}^{2}\sqrt{-{c}^{2}{x}^{2}+1}-{\frac{7\,\sqrt{-{c}^{2}{x}^{2}+1}}{9}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-d*a^2*(1/3*c^3*x^3-c*x)-d*b^2*(1/3*arcsin(c*x)^2*(c^2*x^2-3)*c*x+4/3*c*x-4/3*arcsin(c*x)*(-c^2*x^2+1)^(1
/2)+2/9*arcsin(c*x)*(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-2/27*(c^2*x^2-3)*c*x)-2*d*a*b*(1/3*c^3*x^3*arcsin(c*x)-c*x*
arcsin(c*x)+1/9*c^2*x^2*(-c^2*x^2+1)^(1/2)-7/9*(-c^2*x^2+1)^(1/2)))

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Maxima [B]  time = 1.6771, size = 315, normalized size = 2.46 \begin{align*} -\frac{1}{3} \, b^{2} c^{2} d x^{3} \arcsin \left (c x\right )^{2} - \frac{1}{3} \, a^{2} c^{2} d 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 c^{2} d - \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} c^{2} d + b^{2} d x \arcsin \left (c x\right )^{2} - 2 \, b^{2} d{\left (x - \frac{\sqrt{-c^{2} x^{2} + 1} \arcsin \left (c x\right )}{c}\right )} + a^{2} d x + \frac{2 \,{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} a b d}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.81685, size = 335, normalized size = 2.62 \begin{align*} -\frac{{\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{3} d x^{3} - 3 \,{\left (9 \, a^{2} - 14 \, b^{2}\right )} c d x + 9 \,{\left (b^{2} c^{3} d x^{3} - 3 \, b^{2} c d x\right )} \arcsin \left (c x\right )^{2} + 18 \,{\left (a b c^{3} d x^{3} - 3 \, a b c d x\right )} \arcsin \left (c x\right ) + 6 \,{\left (a b c^{2} d x^{2} - 7 \, a b d +{\left (b^{2} c^{2} d x^{2} - 7 \, b^{2} d\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} x^{2} + 1}}{27 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*((9*a^2 - 2*b^2)*c^3*d*x^3 - 3*(9*a^2 - 14*b^2)*c*d*x + 9*(b^2*c^3*d*x^3 - 3*b^2*c*d*x)*arcsin(c*x)^2 +
18*(a*b*c^3*d*x^3 - 3*a*b*c*d*x)*arcsin(c*x) + 6*(a*b*c^2*d*x^2 - 7*a*b*d + (b^2*c^2*d*x^2 - 7*b^2*d)*arcsin(c
*x))*sqrt(-c^2*x^2 + 1))/c

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Sympy [A]  time = 1.74528, size = 224, normalized size = 1.75 \begin{align*} \begin{cases} - \frac{a^{2} c^{2} d x^{3}}{3} + a^{2} d x - \frac{2 a b c^{2} d x^{3} \operatorname{asin}{\left (c x \right )}}{3} - \frac{2 a b c d x^{2} \sqrt{- c^{2} x^{2} + 1}}{9} + 2 a b d x \operatorname{asin}{\left (c x \right )} + \frac{14 a b d \sqrt{- c^{2} x^{2} + 1}}{9 c} - \frac{b^{2} c^{2} d x^{3} \operatorname{asin}^{2}{\left (c x \right )}}{3} + \frac{2 b^{2} c^{2} d x^{3}}{27} - \frac{2 b^{2} c d x^{2} \sqrt{- c^{2} x^{2} + 1} \operatorname{asin}{\left (c x \right )}}{9} + b^{2} d x \operatorname{asin}^{2}{\left (c x \right )} - \frac{14 b^{2} d x}{9} + \frac{14 b^{2} d \sqrt{- c^{2} x^{2} + 1} \operatorname{asin}{\left (c x \right )}}{9 c} & \text{for}\: c \neq 0 \\a^{2} d x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*c**2*d*x**3/3 + a**2*d*x - 2*a*b*c**2*d*x**3*asin(c*x)/3 - 2*a*b*c*d*x**2*sqrt(-c**2*x**2 + 1
)/9 + 2*a*b*d*x*asin(c*x) + 14*a*b*d*sqrt(-c**2*x**2 + 1)/(9*c) - b**2*c**2*d*x**3*asin(c*x)**2/3 + 2*b**2*c**
2*d*x**3/27 - 2*b**2*c*d*x**2*sqrt(-c**2*x**2 + 1)*asin(c*x)/9 + b**2*d*x*asin(c*x)**2 - 14*b**2*d*x/9 + 14*b*
*2*d*sqrt(-c**2*x**2 + 1)*asin(c*x)/(9*c), Ne(c, 0)), (a**2*d*x, True))

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Giac [A]  time = 1.45402, size = 265, normalized size = 2.07 \begin{align*} -\frac{1}{3} \, a^{2} c^{2} d x^{3} - \frac{1}{3} \,{\left (c^{2} x^{2} - 1\right )} b^{2} d x \arcsin \left (c x\right )^{2} - \frac{2}{3} \,{\left (c^{2} x^{2} - 1\right )} a b d x \arcsin \left (c x\right ) + \frac{2}{3} \, b^{2} d x \arcsin \left (c x\right )^{2} + \frac{2}{27} \,{\left (c^{2} x^{2} - 1\right )} b^{2} d x + \frac{4}{3} \, a b d x \arcsin \left (c x\right ) + \frac{2 \,{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b^{2} d \arcsin \left (c x\right )}{9 \, c} + a^{2} d x - \frac{40}{27} \, b^{2} d x + \frac{2 \,{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} a b d}{9 \, c} + \frac{4 \, \sqrt{-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{3 \, c} + \frac{4 \, \sqrt{-c^{2} x^{2} + 1} a b d}{3 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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