3.25 \(\int x^2 \cos ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=94 \[ -\frac {\left (11 a^2-5 a b x+4\right ) \sqrt {1-(a+b x)^2}}{18 b^3}-\frac {a \left (2 a^2+3\right ) \sin ^{-1}(a+b x)}{6 b^3}+\frac {1}{3} x^3 \cos ^{-1}(a+b x)-\frac {x^2 \sqrt {1-(a+b x)^2}}{9 b} \]

[Out]

1/3*x^3*arccos(b*x+a)-1/6*a*(2*a^2+3)*arcsin(b*x+a)/b^3-1/9*x^2*(1-(b*x+a)^2)^(1/2)/b-1/18*(-5*a*b*x+11*a^2+4)
*(1-(b*x+a)^2)^(1/2)/b^3

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Rubi [A]  time = 0.11, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4806, 4744, 743, 780, 216} \[ -\frac {\left (11 a^2-5 a b x+4\right ) \sqrt {1-(a+b x)^2}}{18 b^3}-\frac {a \left (2 a^2+3\right ) \sin ^{-1}(a+b x)}{6 b^3}-\frac {x^2 \sqrt {1-(a+b x)^2}}{9 b}+\frac {1}{3} x^3 \cos ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCos[a + b*x],x]

[Out]

-(x^2*Sqrt[1 - (a + b*x)^2])/(9*b) - ((4 + 11*a^2 - 5*a*b*x)*Sqrt[1 - (a + b*x)^2])/(18*b^3) + (x^3*ArcCos[a +
b*x])/3 - (a*(3 + 2*a^2)*ArcSin[a + b*x])/(6*b^3)

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 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^2 \cos ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^2 \cos ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{3} x^3 \cos ^{-1}(a+b x)+\frac {1}{3} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^3}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {x^2 \sqrt {1-(a+b x)^2}}{9 b}+\frac {1}{3} x^3 \cos ^{-1}(a+b x)-\frac {1}{9} \operatorname {Subst}\left (\int \frac {\left (-\frac {2+3 a^2}{b^2}+\frac {5 a x}{b^2}\right ) \left (-\frac {a}{b}+\frac {x}{b}\right )}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {x^2 \sqrt {1-(a+b x)^2}}{9 b}-\frac {\left (4+11 a^2-5 a b x\right ) \sqrt {1-(a+b x)^2}}{18 b^3}+\frac {1}{3} x^3 \cos ^{-1}(a+b x)-\frac {\left (a \left (3+2 a^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{6 b^3}\\ &=-\frac {x^2 \sqrt {1-(a+b x)^2}}{9 b}-\frac {\left (4+11 a^2-5 a b x\right ) \sqrt {1-(a+b x)^2}}{18 b^3}+\frac {1}{3} x^3 \cos ^{-1}(a+b x)-\frac {a \left (3+2 a^2\right ) \sin ^{-1}(a+b x)}{6 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 83, normalized size = 0.88 \[ -\frac {\sqrt {-a^2-2 a b x-b^2 x^2+1} \left (11 a^2-5 a b x+2 b^2 x^2+4\right )+3 a \left (2 a^2+3\right ) \sin ^{-1}(a+b x)-6 b^3 x^3 \cos ^{-1}(a+b x)}{18 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCos[a + b*x],x]

[Out]

-1/18*(Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*(4 + 11*a^2 - 5*a*b*x + 2*b^2*x^2) - 6*b^3*x^3*ArcCos[a + b*x] + 3*a*
(3 + 2*a^2)*ArcSin[a + b*x])/b^3

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fricas [A]  time = 0.48, size = 75, normalized size = 0.80 \[ \frac {3 \, {\left (2 \, b^{3} x^{3} + 2 \, a^{3} + 3 \, a\right )} \arccos \left (b x + a\right ) - {\left (2 \, b^{2} x^{2} - 5 \, a b x + 11 \, a^{2} + 4\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{18 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccos(b*x+a),x, algorithm="fricas")

[Out]

1/18*(3*(2*b^3*x^3 + 2*a^3 + 3*a)*arccos(b*x + a) - (2*b^2*x^2 - 5*a*b*x + 11*a^2 + 4)*sqrt(-b^2*x^2 - 2*a*b*x
- a^2 + 1))/b^3

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giac [A]  time = 1.93, size = 156, normalized size = 1.66 \[ \frac {{\left (b x + a\right )}^{3} \arccos \left (b x + a\right )}{3 \, b^{3}} - \frac {{\left (b x + a\right )}^{2} a \arccos \left (b x + a\right )}{b^{3}} + \frac {{\left (b x + a\right )} a^{2} \arccos \left (b x + a\right )}{b^{3}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}^{2}}{9 \, b^{3}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a}{2 \, b^{3}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a^{2}}{b^{3}} + \frac {a \arccos \left (b x + a\right )}{2 \, b^{3}} - \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1}}{9 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccos(b*x+a),x, algorithm="giac")

[Out]

1/3*(b*x + a)^3*arccos(b*x + a)/b^3 - (b*x + a)^2*a*arccos(b*x + a)/b^3 + (b*x + a)*a^2*arccos(b*x + a)/b^3 -
1/9*sqrt(-(b*x + a)^2 + 1)*(b*x + a)^2/b^3 + 1/2*sqrt(-(b*x + a)^2 + 1)*(b*x + a)*a/b^3 - sqrt(-(b*x + a)^2 +
1)*a^2/b^3 + 1/2*a*arccos(b*x + a)/b^3 - 2/9*sqrt(-(b*x + a)^2 + 1)/b^3

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maple [A]  time = 0.01, size = 161, normalized size = 1.71 \[ \frac {\frac {\arccos \left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\arccos \left (b x +a \right ) \left (b x +a \right )^{2} a +\arccos \left (b x +a \right ) \left (b x +a \right ) a^{2}-\frac {\arccos \left (b x +a \right ) a^{3}}{3}-\frac {\left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}}{9}-\frac {2 \sqrt {1-\left (b x +a \right )^{2}}}{9}-a \left (-\frac {\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{2}+\frac {\arcsin \left (b x +a \right )}{2}\right )-a^{2} \sqrt {1-\left (b x +a \right )^{2}}-\frac {\arcsin \left (b x +a \right ) a^{3}}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccos(b*x+a),x)

[Out]

1/b^3*(1/3*arccos(b*x+a)*(b*x+a)^3-arccos(b*x+a)*(b*x+a)^2*a+arccos(b*x+a)*(b*x+a)*a^2-1/3*arccos(b*x+a)*a^3-1
/9*(b*x+a)^2*(1-(b*x+a)^2)^(1/2)-2/9*(1-(b*x+a)^2)^(1/2)-a*(-1/2*(b*x+a)*(1-(b*x+a)^2)^(1/2)+1/2*arcsin(b*x+a)
)-a^2*(1-(b*x+a)^2)^(1/2)-1/3*arcsin(b*x+a)*a^3)

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maxima [B]  time = 0.42, size = 220, normalized size = 2.34 \[ \frac {1}{3} \, x^{3} \arccos \left (b x + a\right ) - \frac {1}{18} \, b {\left (\frac {2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x^{2}}{b^{2}} - \frac {15 \, a^{3} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{4}} - \frac {5 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x}{b^{3}} + \frac {9 \, {\left (a^{2} - 1\right )} a \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{4}} + \frac {15 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2}}{b^{4}} - \frac {4 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{b^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccos(b*x+a),x, algorithm="maxima")

[Out]

1/3*x^3*arccos(b*x + a) - 1/18*b*(2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*x^2/b^2 - 15*a^3*arcsin(-(b^2*x + a*b)/
sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^4 - 5*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a*x/b^3 + 9*(a^2 - 1)*a*arcsin(-(b^2
*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^4 + 15*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^2/b^4 - 4*sqrt(-b^2*x^2
- 2*a*b*x - a^2 + 1)*(a^2 - 1)/b^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {acos}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(a + b*x),x)

[Out]

int(x^2*acos(a + b*x), x)

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sympy [A]  time = 0.64, size = 170, normalized size = 1.81 \[ \begin {cases} \frac {a^{3} \operatorname {acos}{\left (a + b x \right )}}{3 b^{3}} - \frac {11 a^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{18 b^{3}} + \frac {5 a x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{18 b^{2}} + \frac {a \operatorname {acos}{\left (a + b x \right )}}{2 b^{3}} + \frac {x^{3} \operatorname {acos}{\left (a + b x \right )}}{3} - \frac {x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{9 b} - \frac {2 \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{9 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {acos}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acos(b*x+a),x)

[Out]

Piecewise((a**3*acos(a + b*x)/(3*b**3) - 11*a**2*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(18*b**3) + 5*a*x*sqrt(
-a**2 - 2*a*b*x - b**2*x**2 + 1)/(18*b**2) + a*acos(a + b*x)/(2*b**3) + x**3*acos(a + b*x)/3 - x**2*sqrt(-a**2
- 2*a*b*x - b**2*x**2 + 1)/(9*b) - 2*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(9*b**3), Ne(b, 0)), (x**3*acos(a)
/3, True))

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