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

Optimal. Leaf size=51 \[ \frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {b \sin ^{-1}(c x)}{4 c^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4628, 321, 216} \[ \frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {b \sin ^{-1}(c x)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcCos[c*x]),x]

[Out]

-(b*x*Sqrt[1 - c^2*x^2])/(4*c) + (x^2*(a + b*ArcCos[c*x]))/2 + (b*ArcSin[c*x])/(4*c^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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4628

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
s[c*x])^n)/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCos[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \left (a+b \cos ^{-1}(c x)\right ) \, dx &=\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )+\frac {1}{2} (b c) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )+\frac {b \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c}\\ &=-\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )+\frac {b \sin ^{-1}(c x)}{4 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 1.10 \[ \frac {a x^2}{2}-\frac {b x \sqrt {1-c^2 x^2}}{4 c}+\frac {b \sin ^{-1}(c x)}{4 c^2}+\frac {1}{2} b x^2 \cos ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcCos[c*x]),x]

[Out]

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

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fricas [A]  time = 0.47, size = 50, normalized size = 0.98 \[ \frac {2 \, a c^{2} x^{2} - \sqrt {-c^{2} x^{2} + 1} b c x + {\left (2 \, b c^{2} x^{2} - b\right )} \arccos \left (c x\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 46, normalized size = 0.90 \[ \frac {1}{2} \, b x^{2} \arccos \left (c x\right ) + \frac {1}{2} \, a x^{2} - \frac {\sqrt {-c^{2} x^{2} + 1} b x}{4 \, c} - \frac {b \arccos \left (c x\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 52, normalized size = 1.02 \[ \frac {\frac {c^{2} x^{2} a}{2}+b \left (\frac {c^{2} x^{2} \arccos \left (c x \right )}{2}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{4}+\frac {\arcsin \left (c x \right )}{4}\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 50, normalized size = 0.98 \[ \frac {1}{2} \, a x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arccos \left (c x\right ) - c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.30, size = 45, normalized size = 0.88 \[ \frac {a\,x^2}{2}+\frac {b\,\left (\frac {\mathrm {acos}\left (c\,x\right )\,\left (2\,c^2\,x^2-1\right )}{4}-\frac {c\,x\,\sqrt {1-c^2\,x^2}}{4}\right )}{c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*acos(c*x)),x)

[Out]

(a*x^2)/2 + (b*((acos(c*x)*(2*c^2*x^2 - 1))/4 - (c*x*(1 - c^2*x^2)^(1/2))/4))/c^2

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sympy [A]  time = 0.24, size = 60, normalized size = 1.18 \[ \begin {cases} \frac {a x^{2}}{2} + \frac {b x^{2} \operatorname {acos}{\left (c x \right )}}{2} - \frac {b x \sqrt {- c^{2} x^{2} + 1}}{4 c} - \frac {b \operatorname {acos}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {x^{2} \left (a + \frac {\pi b}{2}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**2/2 + b*x**2*acos(c*x)/2 - b*x*sqrt(-c**2*x**2 + 1)/(4*c) - b*acos(c*x)/(4*c**2), Ne(c, 0)), (
x**2*(a + pi*b/2)/2, True))

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