3.201 \(\int (a+b \sin ^{-1}(c+d x))^3 \, dx\)

Optimal. Leaf size=104 \[ -6 a b^2 x+\frac{3 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac{6 b^3 \sqrt{1-(c+d x)^2}}{d}-\frac{6 b^3 (c+d x) \sin ^{-1}(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.114004, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4803, 4619, 4677, 261} \[ -6 a b^2 x+\frac{3 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac{6 b^3 \sqrt{1-(c+d x)^2}}{d}-\frac{6 b^3 (c+d x) \sin ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^3,x]

[Out]

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

Rule 4803

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

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 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]

Rubi steps

\begin{align*} \int \left (a+b \sin ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \sin ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{x \left (a+b \sin ^{-1}(x)\right )^2}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{3 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=-6 a b^2 x+\frac{3 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}-\frac{\left (6 b^3\right ) \operatorname{Subst}\left (\int \sin ^{-1}(x) \, dx,x,c+d x\right )}{d}\\ &=-6 a b^2 x-\frac{6 b^3 (c+d x) \sin ^{-1}(c+d x)}{d}+\frac{3 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}+\frac{\left (6 b^3\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{d}\\ &=-6 a b^2 x-\frac{6 b^3 \sqrt{1-(c+d x)^2}}{d}-\frac{6 b^3 (c+d x) \sin ^{-1}(c+d x)}{d}+\frac{3 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d}\\ \end{align*}

Mathematica [A]  time = 0.0770902, size = 96, normalized size = 0.92 \[ \frac{-6 b^2 \left (a (c+d x)+b \sqrt{1-(c+d x)^2}+b (c+d x) \sin ^{-1}(c+d x)\right )+(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^3+3 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.029, size = 166, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ( \left ( dx+c \right ){a}^{3}+{b}^{3} \left ( \left ( \arcsin \left ( dx+c \right ) \right ) ^{3} \left ( dx+c \right ) +3\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}\sqrt{1- \left ( dx+c \right ) ^{2}}-6\,\sqrt{1- \left ( dx+c \right ) ^{2}}-6\, \left ( dx+c \right ) \arcsin \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( \left ( \arcsin \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) -2\,dx-2\,c+2\,\sqrt{1- \left ( dx+c \right ) ^{2}}\arcsin \left ( dx+c \right ) \right ) +3\,{a}^{2}b \left ( \left ( dx+c \right ) \arcsin \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))^3,x)

[Out]

1/d*((d*x+c)*a^3+b^3*(arcsin(d*x+c)^3*(d*x+c)+3*arcsin(d*x+c)^2*(1-(d*x+c)^2)^(1/2)-6*(1-(d*x+c)^2)^(1/2)-6*(d
*x+c)*arcsin(d*x+c))+3*a*b^2*(arcsin(d*x+c)^2*(d*x+c)-2*d*x-2*c+2*(1-(d*x+c)^2)^(1/2)*arcsin(d*x+c))+3*a^2*b*(
(d*x+c)*arcsin(d*x+c)+(1-(d*x+c)^2)^(1/2)))

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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((a+b*arcsin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.39141, size = 370, normalized size = 3.56 \begin{align*} \frac{{\left (b^{3} d x + b^{3} c\right )} \arcsin \left (d x + c\right )^{3} +{\left (a^{3} - 6 \, a b^{2}\right )} d x + 3 \,{\left (a b^{2} d x + a b^{2} c\right )} \arcsin \left (d x + c\right )^{2} + 3 \,{\left ({\left (a^{2} b - 2 \, b^{3}\right )} d x +{\left (a^{2} b - 2 \, b^{3}\right )} c\right )} \arcsin \left (d x + c\right ) + 3 \,{\left (b^{3} \arcsin \left (d x + c\right )^{2} + 2 \, a b^{2} \arcsin \left (d x + c\right ) + a^{2} b - 2 \, b^{3}\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.16492, size = 282, normalized size = 2.71 \begin{align*} \begin{cases} a^{3} x + \frac{3 a^{2} b c \operatorname{asin}{\left (c + d x \right )}}{d} + 3 a^{2} b x \operatorname{asin}{\left (c + d x \right )} + \frac{3 a^{2} b \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{d} + \frac{3 a b^{2} c \operatorname{asin}^{2}{\left (c + d x \right )}}{d} + 3 a b^{2} x \operatorname{asin}^{2}{\left (c + d x \right )} - 6 a b^{2} x + \frac{6 a b^{2} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname{asin}{\left (c + d x \right )}}{d} + \frac{b^{3} c \operatorname{asin}^{3}{\left (c + d x \right )}}{d} - \frac{6 b^{3} c \operatorname{asin}{\left (c + d x \right )}}{d} + b^{3} x \operatorname{asin}^{3}{\left (c + d x \right )} - 6 b^{3} x \operatorname{asin}{\left (c + d x \right )} + \frac{3 b^{3} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname{asin}^{2}{\left (c + d x \right )}}{d} - \frac{6 b^{3} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \operatorname{asin}{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))**3,x)

[Out]

Piecewise((a**3*x + 3*a**2*b*c*asin(c + d*x)/d + 3*a**2*b*x*asin(c + d*x) + 3*a**2*b*sqrt(-c**2 - 2*c*d*x - d*
*2*x**2 + 1)/d + 3*a*b**2*c*asin(c + d*x)**2/d + 3*a*b**2*x*asin(c + d*x)**2 - 6*a*b**2*x + 6*a*b**2*sqrt(-c**
2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/d + b**3*c*asin(c + d*x)**3/d - 6*b**3*c*asin(c + d*x)/d + b**3*x*a
sin(c + d*x)**3 - 6*b**3*x*asin(c + d*x) + 3*b**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)**2/d - 6
*b**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/d, Ne(d, 0)), (x*(a + b*asin(c))**3, True))

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Giac [B]  time = 1.20051, size = 281, normalized size = 2.7 \begin{align*} \frac{{\left (d x + c\right )} b^{3} \arcsin \left (d x + c\right )^{3}}{d} + \frac{3 \,{\left (d x + c\right )} a b^{2} \arcsin \left (d x + c\right )^{2}}{d} + \frac{3 \, \sqrt{-{\left (d x + c\right )}^{2} + 1} b^{3} \arcsin \left (d x + c\right )^{2}}{d} + \frac{3 \,{\left (d x + c\right )} a^{2} b \arcsin \left (d x + c\right )}{d} - \frac{6 \,{\left (d x + c\right )} b^{3} \arcsin \left (d x + c\right )}{d} + \frac{6 \, \sqrt{-{\left (d x + c\right )}^{2} + 1} a b^{2} \arcsin \left (d x + c\right )}{d} + \frac{{\left (d x + c\right )} a^{3}}{d} - \frac{6 \,{\left (d x + c\right )} a b^{2}}{d} + \frac{3 \, \sqrt{-{\left (d x + c\right )}^{2} + 1} a^{2} b}{d} - \frac{6 \, \sqrt{-{\left (d x + c\right )}^{2} + 1} b^{3}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)*b^3*arcsin(d*x + c)^3/d + 3*(d*x + c)*a*b^2*arcsin(d*x + c)^2/d + 3*sqrt(-(d*x + c)^2 + 1)*b^3*arcsi
n(d*x + c)^2/d + 3*(d*x + c)*a^2*b*arcsin(d*x + c)/d - 6*(d*x + c)*b^3*arcsin(d*x + c)/d + 6*sqrt(-(d*x + c)^2
+ 1)*a*b^2*arcsin(d*x + c)/d + (d*x + c)*a^3/d - 6*(d*x + c)*a*b^2/d + 3*sqrt(-(d*x + c)^2 + 1)*a^2*b/d - 6*s
qrt(-(d*x + c)^2 + 1)*b^3/d