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

Optimal. Leaf size=90 \[ -\frac{d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2}+\frac{b d x \left (1-c^2 x^2\right )^{3/2}}{16 c}+\frac{3 b d x \sqrt{1-c^2 x^2}}{32 c}+\frac{3 b d \sin ^{-1}(c x)}{32 c^2} \]

[Out]

(3*b*d*x*Sqrt[1 - c^2*x^2])/(32*c) + (b*d*x*(1 - c^2*x^2)^(3/2))/(16*c) + (3*b*d*ArcSin[c*x])/(32*c^2) - (d*(1
- c^2*x^2)^2*(a + b*ArcSin[c*x]))/(4*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.041917, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4677, 195, 216} \[ -\frac{d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2}+\frac{b d x \left (1-c^2 x^2\right )^{3/2}}{16 c}+\frac{3 b d x \sqrt{1-c^2 x^2}}{32 c}+\frac{3 b d \sin ^{-1}(c x)}{32 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*d*x*Sqrt[1 - c^2*x^2])/(32*c) + (b*d*x*(1 - c^2*x^2)^(3/2))/(16*c) + (3*b*d*ArcSin[c*x])/(32*c^2) - (d*(1
- c^2*x^2)^2*(a + b*ArcSin[c*x]))/(4*c^2)

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

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

Rubi steps

\begin{align*} \int x \left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=-\frac{d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2}+\frac{(b d) \int \left (1-c^2 x^2\right )^{3/2} \, dx}{4 c}\\ &=\frac{b d x \left (1-c^2 x^2\right )^{3/2}}{16 c}-\frac{d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2}+\frac{(3 b d) \int \sqrt{1-c^2 x^2} \, dx}{16 c}\\ &=\frac{3 b d x \sqrt{1-c^2 x^2}}{32 c}+\frac{b d x \left (1-c^2 x^2\right )^{3/2}}{16 c}-\frac{d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2}+\frac{(3 b d) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{32 c}\\ &=\frac{3 b d x \sqrt{1-c^2 x^2}}{32 c}+\frac{b d x \left (1-c^2 x^2\right )^{3/2}}{16 c}+\frac{3 b d \sin ^{-1}(c x)}{32 c^2}-\frac{d \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0864117, size = 77, normalized size = 0.86 \[ -\frac{d \left (c x \left (8 a c x \left (c^2 x^2-2\right )+b \sqrt{1-c^2 x^2} \left (2 c^2 x^2-5\right )\right )+b \left (8 c^4 x^4-16 c^2 x^2+5\right ) \sin ^{-1}(c x)\right )}{32 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(c*x*(8*a*c*x*(-2 + c^2*x^2) + b*Sqrt[1 - c^2*x^2]*(-5 + 2*c^2*x^2)) + b*(5 - 16*c^2*x^2 + 8*c^4*x^4)*ArcS
in[c*x]))/(32*c^2)

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 98, normalized size = 1.1 \begin{align*}{\frac{1}{{c}^{2}} \left ( -da \left ({\frac{{c}^{4}{x}^{4}}{4}}-{\frac{{c}^{2}{x}^{2}}{2}} \right ) -db \left ({\frac{{c}^{4}{x}^{4}\arcsin \left ( cx \right ) }{4}}-{\frac{{c}^{2}{x}^{2}\arcsin \left ( cx \right ) }{2}}+{\frac{{c}^{3}{x}^{3}}{16}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{5\,cx}{32}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{5\,\arcsin \left ( cx \right ) }{32}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-d*a*(1/4*c^4*x^4-1/2*c^2*x^2)-d*b*(1/4*c^4*x^4*arcsin(c*x)-1/2*c^2*x^2*arcsin(c*x)+1/16*c^3*x^3*(-c^2*
x^2+1)^(1/2)-5/32*c*x*(-c^2*x^2+1)^(1/2)+5/32*arcsin(c*x)))

________________________________________________________________________________________

Maxima [A]  time = 1.58258, size = 205, normalized size = 2.28 \begin{align*} -\frac{1}{4} \, a c^{2} d x^{4} - \frac{1}{32} \,{\left (8 \, x^{4} \arcsin \left (c x\right ) +{\left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac{3 \, \sqrt{-c^{2} x^{2} + 1} x}{c^{4}} - \frac{3 \, \arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b c^{2} d + \frac{1}{2} \, a d x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \arcsin \left (c x\right ) + c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x}{c^{2}} - \frac{\arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*c^2*d*x^4 - 1/32*(8*x^4*arcsin(c*x) + (2*sqrt(-c^2*x^2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*ar
csin(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^4))*c)*b*c^2*d + 1/2*a*d*x^2 + 1/4*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 +
1)*x/c^2 - arcsin(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^2)))*b*d

________________________________________________________________________________________

Fricas [A]  time = 2.17125, size = 200, normalized size = 2.22 \begin{align*} -\frac{8 \, a c^{4} d x^{4} - 16 \, a c^{2} d x^{2} +{\left (8 \, b c^{4} d x^{4} - 16 \, b c^{2} d x^{2} + 5 \, b d\right )} \arcsin \left (c x\right ) +{\left (2 \, b c^{3} d x^{3} - 5 \, b c d x\right )} \sqrt{-c^{2} x^{2} + 1}}{32 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(8*a*c^4*d*x^4 - 16*a*c^2*d*x^2 + (8*b*c^4*d*x^4 - 16*b*c^2*d*x^2 + 5*b*d)*arcsin(c*x) + (2*b*c^3*d*x^3
- 5*b*c*d*x)*sqrt(-c^2*x^2 + 1))/c^2

________________________________________________________________________________________

Sympy [A]  time = 7.07325, size = 117, normalized size = 1.3 \begin{align*} \begin{cases} - \frac{a c^{2} d x^{4}}{4} + \frac{a d x^{2}}{2} - \frac{b c^{2} d x^{4} \operatorname{asin}{\left (c x \right )}}{4} - \frac{b c d x^{3} \sqrt{- c^{2} x^{2} + 1}}{16} + \frac{b d x^{2} \operatorname{asin}{\left (c x \right )}}{2} + \frac{5 b d x \sqrt{- c^{2} x^{2} + 1}}{32 c} - \frac{5 b d \operatorname{asin}{\left (c x \right )}}{32 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*c**2*d*x**4/4 + a*d*x**2/2 - b*c**2*d*x**4*asin(c*x)/4 - b*c*d*x**3*sqrt(-c**2*x**2 + 1)/16 + b*
d*x**2*asin(c*x)/2 + 5*b*d*x*sqrt(-c**2*x**2 + 1)/(32*c) - 5*b*d*asin(c*x)/(32*c**2), Ne(c, 0)), (a*d*x**2/2,
True))

________________________________________________________________________________________

Giac [A]  time = 1.35259, size = 124, normalized size = 1.38 \begin{align*} \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b d x}{16 \, c} - \frac{{\left (c^{2} x^{2} - 1\right )}^{2} b d \arcsin \left (c x\right )}{4 \, c^{2}} + \frac{3 \, \sqrt{-c^{2} x^{2} + 1} b d x}{32 \, c} - \frac{{\left (c^{2} x^{2} - 1\right )}^{2} a d}{4 \, c^{2}} + \frac{3 \, b d \arcsin \left (c x\right )}{32 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(-c^2*x^2 + 1)^(3/2)*b*d*x/c - 1/4*(c^2*x^2 - 1)^2*b*d*arcsin(c*x)/c^2 + 3/32*sqrt(-c^2*x^2 + 1)*b*d*x/c
- 1/4*(c^2*x^2 - 1)^2*a*d/c^2 + 3/32*b*d*arcsin(c*x)/c^2