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

Optimal. Leaf size=80 \[ \frac {\left (2 a^2+1\right ) \sin ^{-1}(a+b x)}{4 b^2}+\frac {3 a \sqrt {1-(a+b x)^2}}{4 b^2}+\frac {1}{2} x^2 \cos ^{-1}(a+b x)-\frac {x \sqrt {1-(a+b x)^2}}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4806, 4744, 743, 641, 216} \[ \frac {\left (2 a^2+1\right ) \sin ^{-1}(a+b x)}{4 b^2}+\frac {3 a \sqrt {1-(a+b x)^2}}{4 b^2}+\frac {1}{2} x^2 \cos ^{-1}(a+b x)-\frac {x \sqrt {1-(a+b x)^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCos[a + b*x],x]

[Out]

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

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -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 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 \cos ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right ) \cos ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{2} x^2 \cos ^{-1}(a+b x)+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {x \sqrt {1-(a+b x)^2}}{4 b}+\frac {1}{2} x^2 \cos ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \frac {-\frac {1+2 a^2}{b^2}+\frac {3 a x}{b^2}}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=\frac {3 a \sqrt {1-(a+b x)^2}}{4 b^2}-\frac {x \sqrt {1-(a+b x)^2}}{4 b}+\frac {1}{2} x^2 \cos ^{-1}(a+b x)+\frac {\left (1+2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{4 b^2}\\ &=\frac {3 a \sqrt {1-(a+b x)^2}}{4 b^2}-\frac {x \sqrt {1-(a+b x)^2}}{4 b}+\frac {1}{2} x^2 \cos ^{-1}(a+b x)+\frac {\left (1+2 a^2\right ) \sin ^{-1}(a+b x)}{4 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.86 \[ \frac {(3 a-b x) \sqrt {-a^2-2 a b x-b^2 x^2+1}+\left (2 a^2+1\right ) \sin ^{-1}(a+b x)+2 b^2 x^2 \cos ^{-1}(a+b x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCos[a + b*x],x]

[Out]

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

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fricas [A]  time = 0.46, size = 59, normalized size = 0.74 \[ \frac {{\left (2 \, b^{2} x^{2} - 2 \, a^{2} - 1\right )} \arccos \left (b x + a\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x - 3 \, a\right )}}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.94, size = 88, normalized size = 1.10 \[ \frac {{\left (b x + a\right )}^{2} \arccos \left (b x + a\right )}{2 \, b^{2}} - \frac {{\left (b x + a\right )} a \arccos \left (b x + a\right )}{b^{2}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}}{4 \, b^{2}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a}{b^{2}} - \frac {\arccos \left (b x + a\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x + a)^2*arccos(b*x + a)/b^2 - (b*x + a)*a*arccos(b*x + a)/b^2 - 1/4*sqrt(-(b*x + a)^2 + 1)*(b*x + a)/b
^2 + sqrt(-(b*x + a)^2 + 1)*a/b^2 - 1/4*arccos(b*x + a)/b^2

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maple [A]  time = 0.01, size = 78, normalized size = 0.98 \[ \frac {\frac {\arccos \left (b x +a \right ) \left (b x +a \right )^{2}}{2}-\arccos \left (b x +a \right ) a \left (b x +a \right )-\frac {\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{4}+\frac {\arcsin \left (b x +a \right )}{4}+a \sqrt {1-\left (b x +a \right )^{2}}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 153, normalized size = 1.91 \[ \frac {1}{2} \, x^{2} \arccos \left (b x + a\right ) - \frac {1}{4} \, b {\left (\frac {3 \, a^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{3}} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{b^{2}} - \frac {{\left (a^{2} - 1\right )} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{3}} - \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acos(a + b*x),x)

[Out]

int(x*acos(a + b*x), x)

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sympy [A]  time = 0.29, size = 104, normalized size = 1.30 \[ \begin {cases} - \frac {a^{2} \operatorname {acos}{\left (a + b x \right )}}{2 b^{2}} + \frac {3 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{4 b^{2}} + \frac {x^{2} \operatorname {acos}{\left (a + b x \right )}}{2} - \frac {x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{4 b} - \frac {\operatorname {acos}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acos}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*acos(a + b*x)/(2*b**2) + 3*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(4*b**2) + x**2*acos(a + b
*x)/2 - x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/(4*b) - acos(a + b*x)/(4*b**2), Ne(b, 0)), (x**2*acos(a)/2, Tr
ue))

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