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]

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

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Rubi [A]  time = 0.31, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, 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 8

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

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 2638

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

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.17, 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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fricas [A]  time = 0.42, size = 108, normalized size = 0.51 \[ \frac {2 \, {\left (2 \, b^{2} x^{2} - 2 \, a^{2} - 1\right )} \arcsin \left (b x + a\right )^{3} - 3 \, {\left (2 \, b^{2} x^{2} - 12 \, a b x - 14 \, a^{2} - 1\right )} \arcsin \left (b x + a\right ) + 3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (2 \, {\left (b x - 3 \, a\right )} \arcsin \left (b x + a\right )^{2} - b x + 15 \, a\right )}}{8 \, b^{2}} \]

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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giac [A]  time = 0.41, size = 203, normalized size = 0.96 \[ -\frac {{\left (b x + a\right )} a \arcsin \left (b x + a\right )^{3}}{b^{2}} + \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )^{3}}{2 \, b^{2}} + \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{4 \, b^{2}} - \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} a \arcsin \left (b x + a\right )^{2}}{b^{2}} + \frac {6 \, {\left (b x + a\right )} a \arcsin \left (b x + a\right )}{b^{2}} + \frac {\arcsin \left (b x + a\right )^{3}}{4 \, b^{2}} - \frac {3 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )}{4 \, b^{2}} - \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}}{8 \, b^{2}} + \frac {6 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} a}{b^{2}} - \frac {3 \, \arcsin \left (b x + a\right )}{8 \, b^{2}} \]

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

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maple [A]  time = 0.09, size = 185, normalized size = 0.88 \[ \frac {\frac {\arcsin \left (b x +a \right )^{3} \left (-1+\left (b x +a \right )^{2}\right )}{2}+\frac {3 \arcsin \left (b x +a \right )^{2} \left (\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+\arcsin \left (b x +a \right )\right )}{4}-\frac {3 \arcsin \left (b x +a \right ) \left (-1+\left (b x +a \right )^{2}\right )}{4}-\frac {3 \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{8}-\frac {3 \arcsin \left (b x +a \right )}{8}-\frac {\arcsin \left (b x +a \right )^{3}}{2}-a \left (\arcsin \left (b x +a \right )^{3} \left (b x +a \right )+3 \arcsin \left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}-6 \sqrt {1-\left (b x +a \right )^{2}}-6 \left (b x +a \right ) \arcsin \left (b x +a \right )\right )}{b^{2}} \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x^{2} \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )^{3} + 3 \, b \int \frac {\sqrt {b x + a + 1} \sqrt {-b x - a + 1} x^{2} \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )^{2}}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^3 + 3*b*integrate(1/2*sqrt(b*x + a + 1)*sqrt(-b
*x - a + 1)*x^2*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2/(b^2*x^2 + 2*a*b*x + a^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,{\mathrm {asin}\left (a+b\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asin(a + b*x)^3,x)

[Out]

int(x*asin(a + b*x)^3, x)

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sympy [A]  time = 1.20, size = 248, normalized size = 1.18 \[ \begin {cases} - \frac {a^{2} \operatorname {asin}^{3}{\left (a + b x \right )}}{2 b^{2}} + \frac {21 a^{2} \operatorname {asin}{\left (a + b x \right )}}{4 b^{2}} + \frac {9 a x \operatorname {asin}{\left (a + b x \right )}}{2 b} - \frac {9 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac {45 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{8 b^{2}} + \frac {x^{2} \operatorname {asin}^{3}{\left (a + b x \right )}}{2} - \frac {3 x^{2} \operatorname {asin}{\left (a + b x \right )}}{4} + \frac {3 x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{4 b} - \frac {3 x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{8 b} - \frac {\operatorname {asin}^{3}{\left (a + b x \right )}}{4 b^{2}} + \frac {3 \operatorname {asin}{\left (a + b x \right )}}{8 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {asin}^{3}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

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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