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

Optimal. Leaf size=77 \[ -\frac{1}{3} c^2 d x^3 \left (a+b \sin ^{-1}(c x)\right )+d x \left (a+b \sin ^{-1}(c x)\right )+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{9 c}+\frac{2 b d \sqrt{1-c^2 x^2}}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.0609626, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4645, 12, 444, 43} \[ -\frac{1}{3} c^2 d x^3 \left (a+b \sin ^{-1}(c x)\right )+d x \left (a+b \sin ^{-1}(c x)\right )+\frac{b d \left (1-c^2 x^2\right )^{3/2}}{9 c}+\frac{2 b d \sqrt{1-c^2 x^2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4645

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(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]] /; F
reeQ[{a, b, c, d, e}, 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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

Mathematica [A]  time = 0.0692109, size = 88, normalized size = 1.14 \[ -\frac{1}{3} a c^2 d x^3+a d x-\frac{1}{9} b c d x^2 \sqrt{1-c^2 x^2}+\frac{7 b d \sqrt{1-c^2 x^2}}{9 c}-\frac{1}{3} b c^2 d x^3 \sin ^{-1}(c x)+b d x \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 82, normalized size = 1.1 \begin{align*}{\frac{1}{c} \left ( -da \left ({\frac{{c}^{3}{x}^{3}}{3}}-cx \right ) -db \left ({\frac{{c}^{3}{x}^{3}\arcsin \left ( cx \right ) }{3}}-cx\arcsin \left ( cx \right ) +{\frac{{c}^{2}{x}^{2}}{9}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{7}{9}\sqrt{-{c}^{2}{x}^{2}+1}} \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)),x)

[Out]

1/c*(-d*a*(1/3*c^3*x^3-c*x)-d*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 [A]  time = 1.68806, size = 131, normalized size = 1.7 \begin{align*} -\frac{1}{3} \, a c^{2} d 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 c^{2} d + a d x + \frac{{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.16592, size = 163, normalized size = 2.12 \begin{align*} -\frac{3 \, a c^{3} d x^{3} - 9 \, a c d x + 3 \,{\left (b c^{3} d x^{3} - 3 \, b c d x\right )} \arcsin \left (c x\right ) +{\left (b c^{2} d x^{2} - 7 \, b d\right )} \sqrt{-c^{2} x^{2} + 1}}{9 \, 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.52337, size = 90, normalized size = 1.17 \begin{align*} \begin{cases} - \frac{a c^{2} d x^{3}}{3} + a d x - \frac{b c^{2} d x^{3} \operatorname{asin}{\left (c x \right )}}{3} - \frac{b c d x^{2} \sqrt{- c^{2} x^{2} + 1}}{9} + b d x \operatorname{asin}{\left (c x \right )} + \frac{7 b d \sqrt{- c^{2} x^{2} + 1}}{9 c} & \text{for}\: c \neq 0 \\a 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)),x)

[Out]

Piecewise((-a*c**2*d*x**3/3 + a*d*x - b*c**2*d*x**3*asin(c*x)/3 - b*c*d*x**2*sqrt(-c**2*x**2 + 1)/9 + b*d*x*as
in(c*x) + 7*b*d*sqrt(-c**2*x**2 + 1)/(9*c), Ne(c, 0)), (a*d*x, True))

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Giac [A]  time = 1.32507, size = 108, normalized size = 1.4 \begin{align*} -\frac{1}{3} \, a c^{2} d x^{3} - \frac{1}{3} \,{\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right ) + \frac{2}{3} \, b d x \arcsin \left (c x\right ) + a d x + \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b d}{9 \, c} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1} 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)),x, algorithm="giac")

[Out]

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