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

Optimal. Leaf size=137 \[ -\frac{\left (4 a \left (19 a^2+16\right )-\left (26 a^2+9\right ) (a+b x)\right ) \sqrt{1-(a+b x)^2}}{96 b^4}-\frac{\left (8 a^4+24 a^2+3\right ) \sin ^{-1}(a+b x)}{32 b^4}-\frac{7 a x^2 \sqrt{1-(a+b x)^2}}{48 b^2}+\frac{x^3 \sqrt{1-(a+b x)^2}}{16 b}+\frac{1}{4} x^4 \sin ^{-1}(a+b x) \]

[Out]

(-7*a*x^2*Sqrt[1 - (a + b*x)^2])/(48*b^2) + (x^3*Sqrt[1 - (a + b*x)^2])/(16*b) - ((4*a*(16 + 19*a^2) - (9 + 26
*a^2)*(a + b*x))*Sqrt[1 - (a + b*x)^2])/(96*b^4) - ((3 + 24*a^2 + 8*a^4)*ArcSin[a + b*x])/(32*b^4) + (x^4*ArcS
in[a + b*x])/4

________________________________________________________________________________________

Rubi [A]  time = 0.199654, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4805, 4743, 743, 833, 780, 216} \[ -\frac{\left (4 a \left (19 a^2+16\right )-\left (26 a^2+9\right ) (a+b x)\right ) \sqrt{1-(a+b x)^2}}{96 b^4}-\frac{\left (8 a^4+24 a^2+3\right ) \sin ^{-1}(a+b x)}{32 b^4}-\frac{7 a x^2 \sqrt{1-(a+b x)^2}}{48 b^2}+\frac{x^3 \sqrt{1-(a+b x)^2}}{16 b}+\frac{1}{4} x^4 \sin ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSin[a + b*x],x]

[Out]

(-7*a*x^2*Sqrt[1 - (a + b*x)^2])/(48*b^2) + (x^3*Sqrt[1 - (a + b*x)^2])/(16*b) - ((4*a*(16 + 19*a^2) - (9 + 26
*a^2)*(a + b*x))*Sqrt[1 - (a + b*x)^2])/(96*b^4) - ((3 + 24*a^2 + 8*a^4)*ArcSin[a + b*x])/(32*b^4) + (x^4*ArcS
in[a + b*x])/4

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 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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^3 \sin ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sin ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \sin ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^4}{\sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=\frac{x^3 \sqrt{1-(a+b x)^2}}{16 b}+\frac{1}{4} x^4 \sin ^{-1}(a+b x)+\frac{1}{16} \operatorname{Subst}\left (\int \frac{\left (-\frac{3+4 a^2}{b^2}+\frac{7 a x}{b^2}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )^2}{\sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac{7 a x^2 \sqrt{1-(a+b x)^2}}{48 b^2}+\frac{x^3 \sqrt{1-(a+b x)^2}}{16 b}+\frac{1}{4} x^4 \sin ^{-1}(a+b x)-\frac{1}{48} \operatorname{Subst}\left (\int \frac{\left (-\frac{a \left (23+12 a^2\right )}{b^3}+\frac{\left (9+26 a^2\right ) x}{b^3}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )}{\sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac{7 a x^2 \sqrt{1-(a+b x)^2}}{48 b^2}+\frac{x^3 \sqrt{1-(a+b x)^2}}{16 b}-\frac{\left (4 a \left (16+19 a^2\right )-\left (9+26 a^2\right ) (a+b x)\right ) \sqrt{1-(a+b x)^2}}{96 b^4}+\frac{1}{4} x^4 \sin ^{-1}(a+b x)-\frac{\left (3+24 a^2+8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{32 b^4}\\ &=-\frac{7 a x^2 \sqrt{1-(a+b x)^2}}{48 b^2}+\frac{x^3 \sqrt{1-(a+b x)^2}}{16 b}-\frac{\left (4 a \left (16+19 a^2\right )-\left (9+26 a^2\right ) (a+b x)\right ) \sqrt{1-(a+b x)^2}}{96 b^4}-\frac{\left (3+24 a^2+8 a^4\right ) \sin ^{-1}(a+b x)}{32 b^4}+\frac{1}{4} x^4 \sin ^{-1}(a+b x)\\ \end{align*}

Mathematica [A]  time = 0.0884133, size = 99, normalized size = 0.72 \[ \frac{\sqrt{-a^2-2 a b x-b^2 x^2+1} \left (26 a^2 b x-50 a^3-a \left (14 b^2 x^2+55\right )+6 b^3 x^3+9 b x\right )-3 \left (8 a^4+24 a^2-8 b^4 x^4+3\right ) \sin ^{-1}(a+b x)}{96 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSin[a + b*x],x]

[Out]

(Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*(-50*a^3 + 9*b*x + 26*a^2*b*x + 6*b^3*x^3 - a*(55 + 14*b^2*x^2)) - 3*(3 + 2
4*a^2 + 8*a^4 - 8*b^4*x^4)*ArcSin[a + b*x])/(96*b^4)

________________________________________________________________________________________

Maple [A]  time = 0.016, size = 213, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{\arcsin \left ( bx+a \right ) \left ( bx+a \right ) ^{4}}{4}}-\arcsin \left ( bx+a \right ) \left ( bx+a \right ) ^{3}a+{\frac{3\,\arcsin \left ( bx+a \right ) \left ( bx+a \right ) ^{2}{a}^{2}}{2}}-\arcsin \left ( bx+a \right ) \left ( bx+a \right ){a}^{3}+{\frac{ \left ( bx+a \right ) ^{3}}{16}\sqrt{1- \left ( bx+a \right ) ^{2}}}+{\frac{3\,bx+3\,a}{32}\sqrt{1- \left ( bx+a \right ) ^{2}}}-{\frac{3\,\arcsin \left ( bx+a \right ) }{32}}+a \left ( -{\frac{ \left ( bx+a \right ) ^{2}}{3}\sqrt{1- \left ( bx+a \right ) ^{2}}}-{\frac{2}{3}\sqrt{1- \left ( bx+a \right ) ^{2}}} \right ) -{\frac{3\,{a}^{2}}{2} \left ( -{\frac{bx+a}{2}\sqrt{1- \left ( bx+a \right ) ^{2}}}+{\frac{\arcsin \left ( bx+a \right ) }{2}} \right ) }-{a}^{3}\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.75963, size = 219, normalized size = 1.6 \begin{align*} \frac{3 \,{\left (8 \, b^{4} x^{4} - 8 \, a^{4} - 24 \, a^{2} - 3\right )} \arcsin \left (b x + a\right ) +{\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} +{\left (26 \, a^{2} + 9\right )} b x - 55 \, a\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{96 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(8*b^4*x^4 - 8*a^4 - 24*a^2 - 3)*arcsin(b*x + a) + (6*b^3*x^3 - 14*a*b^2*x^2 - 50*a^3 + (26*a^2 + 9)*b
*x - 55*a)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/b^4

________________________________________________________________________________________

Sympy [A]  time = 1.85865, size = 255, normalized size = 1.86 \begin{align*} \begin{cases} - \frac{a^{4} \operatorname{asin}{\left (a + b x \right )}}{4 b^{4}} - \frac{25 a^{3} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{4}} + \frac{13 a^{2} x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{3}} - \frac{3 a^{2} \operatorname{asin}{\left (a + b x \right )}}{4 b^{4}} - \frac{7 a x^{2} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{2}} - \frac{55 a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{96 b^{4}} + \frac{x^{4} \operatorname{asin}{\left (a + b x \right )}}{4} + \frac{x^{3} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{16 b} + \frac{3 x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{32 b^{3}} - \frac{3 \operatorname{asin}{\left (a + b x \right )}}{32 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{asin}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**4*asin(a + b*x)/(4*b**4) - 25*a**3*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(48*b**4) + 13*a**2*x*
sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(48*b**3) - 3*a**2*asin(a + b*x)/(4*b**4) - 7*a*x**2*sqrt(-a**2 - 2*a*b*
x - b**2*x**2 + 1)/(48*b**2) - 55*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(96*b**4) + x**4*asin(a + b*x)/4 + x
**3*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(16*b) + 3*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(32*b**3) - 3*asi
n(a + b*x)/(32*b**4), Ne(b, 0)), (x**4*asin(a)/4, True))

________________________________________________________________________________________

Giac [B]  time = 1.20477, size = 383, normalized size = 2.8 \begin{align*} -\frac{{\left (b x + a\right )} a^{3} \arcsin \left (b x + a\right )}{b^{4}} - \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )}{\left (b x + a\right )} a \arcsin \left (b x + a\right )}{b^{4}} + \frac{3 \,{\left ({\left (b x + a\right )}^{2} - 1\right )} a^{2} \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac{3 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} a^{2}}{4 \, b^{4}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a^{3}}{b^{4}} + \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )}^{2} \arcsin \left (b x + a\right )}{4 \, b^{4}} - \frac{{\left (b x + a\right )} a \arcsin \left (b x + a\right )}{b^{4}} + \frac{3 \, a^{2} \arcsin \left (b x + a\right )}{4 \, b^{4}} - \frac{{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}{\left (b x + a\right )}}{16 \, b^{4}} + \frac{{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}} a}{3 \, b^{4}} + \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac{5 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )}}{32 \, b^{4}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a}{b^{4}} + \frac{5 \, \arcsin \left (b x + a\right )}{32 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x + a)*a^3*arcsin(b*x + a)/b^4 - ((b*x + a)^2 - 1)*(b*x + a)*a*arcsin(b*x + a)/b^4 + 3/2*((b*x + a)^2 - 1)
*a^2*arcsin(b*x + a)/b^4 + 3/4*sqrt(-(b*x + a)^2 + 1)*(b*x + a)*a^2/b^4 - sqrt(-(b*x + a)^2 + 1)*a^3/b^4 + 1/4
*((b*x + a)^2 - 1)^2*arcsin(b*x + a)/b^4 - (b*x + a)*a*arcsin(b*x + a)/b^4 + 3/4*a^2*arcsin(b*x + a)/b^4 - 1/1
6*(-(b*x + a)^2 + 1)^(3/2)*(b*x + a)/b^4 + 1/3*(-(b*x + a)^2 + 1)^(3/2)*a/b^4 + 1/2*((b*x + a)^2 - 1)*arcsin(b
*x + a)/b^4 + 5/32*sqrt(-(b*x + a)^2 + 1)*(b*x + a)/b^4 - sqrt(-(b*x + a)^2 + 1)*a/b^4 + 5/32*arcsin(b*x + a)/
b^4