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

Optimal. Leaf size=211 \[ -\frac{a^2 \sin ^{-1}(a+b x)^3}{2 b^2}-\frac{3 (a+b x) \sqrt{1-(a+b x)^2}}{8 b^2}+\frac{6 a \sqrt{1-(a+b x)^2}}{b^2}-\frac{\sin ^{-1}(a+b x)^3}{4 b^2}+\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{4 b^2}-\frac{3 a \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{b^2}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)}{4 b^2}+\frac{6 a (a+b x) \sin ^{-1}(a+b x)}{b^2}+\frac{3 \sin ^{-1}(a+b x)}{8 b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3 \]

[Out]

(6*a*Sqrt[1 - (a + b*x)^2])/b^2 - (3*(a + b*x)*Sqrt[1 - (a + b*x)^2])/(8*b^2) + (3*ArcSin[a + b*x])/(8*b^2) +
(6*a*(a + b*x)*ArcSin[a + b*x])/b^2 - (3*(a + b*x)^2*ArcSin[a + b*x])/(4*b^2) - (3*a*Sqrt[1 - (a + b*x)^2]*Arc
Sin[a + b*x]^2)/b^2 + (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x]^2)/(4*b^2) - ArcSin[a + b*x]^3/(4*b^2
) - (a^2*ArcSin[a + b*x]^3)/(2*b^2) + (x^2*ArcSin[a + b*x]^3)/2

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Rubi [A]  time = 0.308878, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {4805, 4743, 4773, 3317, 3296, 2638, 3311, 30, 2635, 8} \[ -\frac{a^2 \sin ^{-1}(a+b x)^3}{2 b^2}-\frac{3 (a+b x) \sqrt{1-(a+b x)^2}}{8 b^2}+\frac{6 a \sqrt{1-(a+b x)^2}}{b^2}-\frac{\sin ^{-1}(a+b x)^3}{4 b^2}+\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{4 b^2}-\frac{3 a \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{b^2}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)}{4 b^2}+\frac{6 a (a+b x) \sin ^{-1}(a+b x)}{b^2}+\frac{3 \sin ^{-1}(a+b x)}{8 b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSin[a + b*x]^3,x]

[Out]

(6*a*Sqrt[1 - (a + b*x)^2])/b^2 - (3*(a + b*x)*Sqrt[1 - (a + b*x)^2])/(8*b^2) + (3*ArcSin[a + b*x])/(8*b^2) +
(6*a*(a + b*x)*ArcSin[a + b*x])/b^2 - (3*(a + b*x)^2*ArcSin[a + b*x])/(4*b^2) - (3*a*Sqrt[1 - (a + b*x)^2]*Arc
Sin[a + b*x]^2)/b^2 + (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x]^2)/(4*b^2) - ArcSin[a + b*x]^3/(4*b^2
) - (a^2*ArcSin[a + b*x]^3)/(2*b^2) + (x^2*ArcSin[a + b*x]^3)/2

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 4743

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

Rule 4773

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sin[x])^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a,
b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x \sin ^{-1}(a+b x)^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right ) \sin ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3-\frac{3}{2} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sin ^{-1}(x)^2}{\sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3-\frac{3}{2} \operatorname{Subst}\left (\int x^2 \left (-\frac{a}{b}+\frac{\sin (x)}{b}\right )^2 \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3-\frac{3}{2} \operatorname{Subst}\left (\int \left (\frac{a^2 x^2}{b^2}-\frac{2 a x^2 \sin (x)}{b^2}+\frac{x^2 \sin ^2(x)}{b^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac{a^2 \sin ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}+\frac{(3 a) \operatorname{Subst}\left (\int x^2 \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)}{4 b^2}-\frac{3 a \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{b^2}+\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{4 b^2}-\frac{a^2 \sin ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}+\frac{3 \operatorname{Subst}\left (\int \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}+\frac{(6 a) \operatorname{Subst}\left (\int x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{3 (a+b x) \sqrt{1-(a+b x)^2}}{8 b^2}+\frac{6 a (a+b x) \sin ^{-1}(a+b x)}{b^2}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)}{4 b^2}-\frac{3 a \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{b^2}+\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{4 b^2}-\frac{\sin ^{-1}(a+b x)^3}{4 b^2}-\frac{a^2 \sin ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3+\frac{3 \operatorname{Subst}\left (\int 1 \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^2}-\frac{(6 a) \operatorname{Subst}\left (\int \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{6 a \sqrt{1-(a+b x)^2}}{b^2}-\frac{3 (a+b x) \sqrt{1-(a+b x)^2}}{8 b^2}+\frac{3 \sin ^{-1}(a+b x)}{8 b^2}+\frac{6 a (a+b x) \sin ^{-1}(a+b x)}{b^2}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)}{4 b^2}-\frac{3 a \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{b^2}+\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{4 b^2}-\frac{\sin ^{-1}(a+b x)^3}{4 b^2}-\frac{a^2 \sin ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^3\\ \end{align*}

Mathematica [A]  time = 0.141906, size = 135, normalized size = 0.64 \[ \frac{3 (15 a-b x) \sqrt{-a^2-2 a b x-b^2 x^2+1}+\left (-4 a^2+4 b^2 x^2-2\right ) \sin ^{-1}(a+b x)^3-6 (3 a-b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)^2+\left (42 a^2+36 a b x-6 b^2 x^2+3\right ) \sin ^{-1}(a+b x)}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSin[a + b*x]^3,x]

[Out]

(3*(15*a - b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] + (3 + 42*a^2 + 36*a*b*x - 6*b^2*x^2)*ArcSin[a + b*x] - 6*(3
*a - b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x]^2 + (-2 - 4*a^2 + 4*b^2*x^2)*ArcSin[a + b*x]^3)/(8
*b^2)

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Maple [A]  time = 0.047, size = 185, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{3} \left ( -1+ \left ( bx+a \right ) ^{2} \right ) }{2}}+{\frac{3\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}{4} \left ( \left ( bx+a \right ) \sqrt{1- \left ( bx+a \right ) ^{2}}+\arcsin \left ( bx+a \right ) \right ) }-{\frac{3\,\arcsin \left ( bx+a \right ) \left ( -1+ \left ( bx+a \right ) ^{2} \right ) }{4}}-{\frac{3\,bx+3\,a}{8}\sqrt{1- \left ( bx+a \right ) ^{2}}}-{\frac{3\,\arcsin \left ( bx+a \right ) }{8}}-{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}}{2}}-a \left ( \left ( \arcsin \left ( bx+a \right ) \right ) ^{3} \left ( bx+a \right ) +3\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}\sqrt{1- \left ( bx+a \right ) ^{2}}-6\,\sqrt{1- \left ( bx+a \right ) ^{2}}-6\, \left ( bx+a \right ) \arcsin \left ( bx+a \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(1/2*arcsin(b*x+a)^3*(-1+(b*x+a)^2)+3/4*arcsin(b*x+a)^2*((b*x+a)*(1-(b*x+a)^2)^(1/2)+arcsin(b*x+a))-3/4*
arcsin(b*x+a)*(-1+(b*x+a)^2)-3/8*(b*x+a)*(1-(b*x+a)^2)^(1/2)-3/8*arcsin(b*x+a)-1/2*arcsin(b*x+a)^3-a*(arcsin(b
*x+a)^3*(b*x+a)+3*arcsin(b*x+a)^2*(1-(b*x+a)^2)^(1/2)-6*(1-(b*x+a)^2)^(1/2)-6*(b*x+a)*arcsin(b*x+a)))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.90174, size = 266, normalized size = 1.26 \begin{align*} \frac{2 \,{\left (2 \, b^{2} x^{2} - 2 \, a^{2} - 1\right )} \arcsin \left (b x + a\right )^{3} - 3 \,{\left (2 \, b^{2} x^{2} - 12 \, a b x - 14 \, a^{2} - 1\right )} \arcsin \left (b x + a\right ) + 3 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (2 \,{\left (b x - 3 \, a\right )} \arcsin \left (b x + a\right )^{2} - b x + 15 \, a\right )}}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(2*b^2*x^2 - 2*a^2 - 1)*arcsin(b*x + a)^3 - 3*(2*b^2*x^2 - 12*a*b*x - 14*a^2 - 1)*arcsin(b*x + a) + 3*s
qrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(2*(b*x - 3*a)*arcsin(b*x + a)^2 - b*x + 15*a))/b^2

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Sympy [A]  time = 1.59027, size = 248, normalized size = 1.18 \begin{align*} \begin{cases} - \frac{a^{2} \operatorname{asin}^{3}{\left (a + b x \right )}}{2 b^{2}} + \frac{21 a^{2} \operatorname{asin}{\left (a + b x \right )}}{4 b^{2}} + \frac{9 a x \operatorname{asin}{\left (a + b x \right )}}{2 b} - \frac{9 a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac{45 a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{8 b^{2}} + \frac{x^{2} \operatorname{asin}^{3}{\left (a + b x \right )}}{2} - \frac{3 x^{2} \operatorname{asin}{\left (a + b x \right )}}{4} + \frac{3 x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}^{2}{\left (a + b x \right )}}{4 b} - \frac{3 x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{8 b} - \frac{\operatorname{asin}^{3}{\left (a + b x \right )}}{4 b^{2}} + \frac{3 \operatorname{asin}{\left (a + b x \right )}}{8 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{asin}^{3}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*asin(a + b*x)**3/(2*b**2) + 21*a**2*asin(a + b*x)/(4*b**2) + 9*a*x*asin(a + b*x)/(2*b) - 9*a*
sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)**2/(4*b**2) + 45*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(
8*b**2) + x**2*asin(a + b*x)**3/2 - 3*x**2*asin(a + b*x)/4 + 3*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a
+ b*x)**2/(4*b) - 3*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(8*b) - asin(a + b*x)**3/(4*b**2) + 3*asin(a + b*x
)/(8*b**2), Ne(b, 0)), (x**2*asin(a)**3/2, True))

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Giac [A]  time = 1.19873, size = 274, normalized size = 1.3 \begin{align*} -\frac{{\left (b x + a\right )} a \arcsin \left (b x + a\right )^{3}}{b^{2}} + \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )^{3}}{2 \, b^{2}} + \frac{3 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{4 \, b^{2}} - \frac{3 \, \sqrt{-{\left (b x + a\right )}^{2} + 1} a \arcsin \left (b x + a\right )^{2}}{b^{2}} + \frac{6 \,{\left (b x + a\right )} a \arcsin \left (b x + a\right )}{b^{2}} + \frac{\arcsin \left (b x + a\right )^{3}}{4 \, b^{2}} - \frac{3 \,{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )}{4 \, b^{2}} - \frac{3 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )}}{8 \, b^{2}} + \frac{6 \, \sqrt{-{\left (b x + a\right )}^{2} + 1} a}{b^{2}} - \frac{3 \, \arcsin \left (b x + a\right )}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x + a)*a*arcsin(b*x + a)^3/b^2 + 1/2*((b*x + a)^2 - 1)*arcsin(b*x + a)^3/b^2 + 3/4*sqrt(-(b*x + a)^2 + 1)*
(b*x + a)*arcsin(b*x + a)^2/b^2 - 3*sqrt(-(b*x + a)^2 + 1)*a*arcsin(b*x + a)^2/b^2 + 6*(b*x + a)*a*arcsin(b*x
+ a)/b^2 + 1/4*arcsin(b*x + a)^3/b^2 - 3/4*((b*x + a)^2 - 1)*arcsin(b*x + a)/b^2 - 3/8*sqrt(-(b*x + a)^2 + 1)*
(b*x + a)/b^2 + 6*sqrt(-(b*x + a)^2 + 1)*a/b^2 - 3/8*arcsin(b*x + a)/b^2