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

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

[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) + (x^4*ArcCos[a + b*x])/4 + ((3 + 24*a^2 + 8*a^4)*ArcSin[a + b
*x])/(32*b^4)

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Rubi [A]  time = 0.194006, 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 = {4806, 4744, 743, 833, 780, 216} \[ \frac{\left (4 a \left (19 a^2+16\right )-\left (26 a^2+9\right ) (a+b x)\right ) \sqrt{1-(a+b x)^2}}{96 b^4}+\frac{\left (8 a^4+24 a^2+3\right ) \sin ^{-1}(a+b x)}{32 b^4}+\frac{7 a x^2 \sqrt{1-(a+b x)^2}}{48 b^2}-\frac{x^3 \sqrt{1-(a+b x)^2}}{16 b}+\frac{1}{4} x^4 \cos ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCos[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) + (x^4*ArcCos[a + b*x])/4 + ((3 + 24*a^2 + 8*a^4)*ArcSin[a + b
*x])/(32*b^4)

Rule 4806

Int[((a_.) + ArcCos[(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*ArcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 4744

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*ArcCos[c*x])^n)/(e*(m + 1)), x] + Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcCos[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 \cos ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \cos ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \cos ^{-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 \cos ^{-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 \cos ^{-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 \cos ^{-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{1}{4} x^4 \cos ^{-1}(a+b x)+\frac{\left (3+24 a^2+8 a^4\right ) \sin ^{-1}(a+b x)}{32 b^4}\\ \end{align*}

Mathematica [A]  time = 0.100011, size = 104, normalized size = 0.76 \[ \frac{\sqrt{-a^2-2 a b x-b^2 x^2+1} \left (-26 a^2 b x+50 a^3+14 a b^2 x^2+55 a-6 b^3 x^3-9 b x\right )+3 \left (8 a^4+24 a^2+3\right ) \sin ^{-1}(a+b x)+24 b^4 x^4 \cos ^{-1}(a+b x)}{96 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCos[a + b*x],x]

[Out]

(Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*(55*a + 50*a^3 - 9*b*x - 26*a^2*b*x + 14*a*b^2*x^2 - 6*b^3*x^3) + 24*b^4*x^
4*ArcCos[a + b*x] + 3*(3 + 24*a^2 + 8*a^4)*ArcSin[a + b*x])/(96*b^4)

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Maple [A]  time = 0.02, size = 235, normalized size = 1.7 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{\arccos \left ( bx+a \right ) \left ( bx+a \right ) ^{4}}{4}}-\arccos \left ( bx+a \right ) \left ( bx+a \right ) ^{3}a+{\frac{3\,\arccos \left ( bx+a \right ) \left ( bx+a \right ) ^{2}{a}^{2}}{2}}-\arccos \left ( bx+a \right ) \left ( bx+a \right ){a}^{3}+{\frac{\arccos \left ( bx+a \right ){a}^{4}}{4}}-{\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}}+{\frac{\arcsin \left ( bx+a \right ){a}^{4}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.47944, size = 219, normalized size = 1.6 \begin{align*} \frac{3 \,{\left (8 \, b^{4} x^{4} - 8 \, a^{4} - 24 \, a^{2} - 3\right )} \arccos \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*arccos(b*x+a),x, algorithm="fricas")

[Out]

1/96*(3*(8*b^4*x^4 - 8*a^4 - 24*a^2 - 3)*arccos(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

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Sympy [A]  time = 1.75661, size = 255, normalized size = 1.86 \begin{align*} \begin{cases} - \frac{a^{4} \operatorname{acos}{\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{acos}{\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{acos}{\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{acos}{\left (a + b x \right )}}{32 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{acos}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**4*acos(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*acos(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*acos(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*aco
s(a + b*x)/(32*b**4), Ne(b, 0)), (x**4*acos(a)/4, True))

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Giac [B]  time = 1.31574, size = 327, normalized size = 2.39 \begin{align*} \frac{{\left (b x + a\right )}^{4} \arccos \left (b x + a\right )}{4 \, b^{4}} - \frac{{\left (b x + a\right )}^{3} a \arccos \left (b x + a\right )}{b^{4}} + \frac{3 \,{\left (b x + a\right )}^{2} a^{2} \arccos \left (b x + a\right )}{2 \, b^{4}} - \frac{{\left (b x + a\right )} a^{3} \arccos \left (b x + a\right )}{b^{4}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )}^{3}}{16 \, b^{4}} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )}^{2} a}{3 \, 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{3 \, a^{2} \arccos \left (b x + a\right )}{4 \, b^{4}} - \frac{3 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )}}{32 \, b^{4}} + \frac{2 \, \sqrt{-{\left (b x + a\right )}^{2} + 1} a}{3 \, b^{4}} - \frac{3 \, \arccos \left (b x + a\right )}{32 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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