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

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

[Out]

1/4*x^4*arccos(b*x+a)+1/32*(8*a^4+24*a^2+3)*arcsin(b*x+a)/b^4+7/48*a*x^2*(1-(b*x+a)^2)^(1/2)/b^2-1/16*x^3*(1-(
b*x+a)^2)^(1/2)/b+1/96*(4*a*(19*a^2+16)-(26*a^2+9)*(b*x+a))*(1-(b*x+a)^2)^(1/2)/b^4

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Rubi [A]  time = 0.19, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.12, size = 104, normalized size = 0.76 \[ \frac {3 \left (8 a^4+24 a^2+3\right ) \sin ^{-1}(a+b x)+\sqrt {-a^2-2 a b x-b^2 x^2+1} \left (50 a^3-26 a^2 b x+14 a b^2 x^2+55 a-6 b^3 x^3-9 b x\right )+24 b^4 x^4 \cos ^{-1}(a+b x)}{96 b^4} \]

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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fricas [A]  time = 0.47, size = 94, normalized size = 0.69 \[ \frac {3 \, {\left (8 \, b^{4} x^{4} - 8 \, a^{4} - 24 \, a^{2} - 3\right )} \arccos \left (b x + a\right ) - {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} + {\left (26 \, a^{2} + 9\right )} b x - 55 \, a\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{96 \, b^{4}} \]

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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giac [B]  time = 0.72, size = 242, normalized size = 1.77 \[ \frac {{\left (b x + a\right )}^{4} \arccos \left (b x + a\right )}{4 \, b^{4}} - \frac {{\left (b x + a\right )}^{3} a \arccos \left (b x + a\right )}{b^{4}} + \frac {3 \, {\left (b x + a\right )}^{2} a^{2} \arccos \left (b x + a\right )}{2 \, b^{4}} - \frac {{\left (b x + a\right )} a^{3} \arccos \left (b x + a\right )}{b^{4}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}^{3}}{16 \, b^{4}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}^{2} a}{3 \, b^{4}} - \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a^{2}}{4 \, b^{4}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{3}}{b^{4}} - \frac {3 \, a^{2} \arccos \left (b x + a\right )}{4 \, b^{4}} - \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}}{32 \, b^{4}} + \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} a}{3 \, b^{4}} - \frac {3 \, \arccos \left (b x + a\right )}{32 \, b^{4}} \]

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

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maple [A]  time = 0.02, size = 235, normalized size = 1.72 \[ \frac {\frac {\arccos \left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\arccos \left (b x +a \right ) \left (b x +a \right )^{3} a +\frac {3 \arccos \left (b x +a \right ) \left (b x +a \right )^{2} a^{2}}{2}-\arccos \left (b x +a \right ) \left (b x +a \right ) a^{3}+\frac {\arccos \left (b x +a \right ) a^{4}}{4}-\frac {\left (b x +a \right )^{3} \sqrt {1-\left (b x +a \right )^{2}}}{16}-\frac {3 \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{32}+\frac {3 \arcsin \left (b x +a \right )}{32}-a \left (-\frac {\left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}}{3}-\frac {2 \sqrt {1-\left (b x +a \right )^{2}}}{3}\right )+\frac {3 a^{2} \left (-\frac {\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{2}+\frac {\arcsin \left (b x +a \right )}{2}\right )}{2}+a^{3} \sqrt {1-\left (b x +a \right )^{2}}+\frac {\arcsin \left (b x +a \right ) a^{4}}{4}}{b^{4}} \]

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 [B]  time = 0.42, size = 333, normalized size = 2.43 \[ \frac {1}{4} \, x^{4} \arccos \left (b x + a\right ) - \frac {1}{96} \, {\left (\frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x^{3}}{b^{2}} - \frac {14 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x^{2}}{b^{3}} + \frac {105 \, a^{4} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{5}} + \frac {35 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2} x}{b^{4}} - \frac {90 \, {\left (a^{2} - 1\right )} a^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{5}} - \frac {105 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{3}}{b^{5}} - \frac {9 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} x}{b^{4}} + \frac {9 \, {\left (a^{2} - 1\right )}^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{5}} + \frac {55 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} a}{b^{5}}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccos(b*x + a) - 1/96*(6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*x^3/b^2 - 14*sqrt(-b^2*x^2 - 2*a*b*x - a^
2 + 1)*a*x^2/b^3 + 105*a^4*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^5 + 35*sqrt(-b^2*x^2 - 2*a*b
*x - a^2 + 1)*a^2*x/b^4 - 90*(a^2 - 1)*a^2*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^5 - 105*sqrt
(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^3/b^5 - 9*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1)*x/b^4 + 9*(a^2 - 1)^2*
arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^5 + 55*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1)*a/b
^5)*b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {acos}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(a + b*x),x)

[Out]

int(x^3*acos(a + b*x), x)

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sympy [A]  time = 1.35, size = 255, normalized size = 1.86 \[ \begin {cases} - \frac {a^{4} \operatorname {acos}{\left (a + b x \right )}}{4 b^{4}} + \frac {25 a^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{4}} - \frac {13 a^{2} x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{3}} - \frac {3 a^{2} \operatorname {acos}{\left (a + b x \right )}}{4 b^{4}} + \frac {7 a x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{48 b^{2}} + \frac {55 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{96 b^{4}} + \frac {x^{4} \operatorname {acos}{\left (a + b x \right )}}{4} - \frac {x^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{16 b} - \frac {3 x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{32 b^{3}} - \frac {3 \operatorname {acos}{\left (a + b x \right )}}{32 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acos}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

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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