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

Optimal. Leaf size=76 \[ \frac{1}{4} x^4 \left (a+b \cos ^{-1}(c x)\right )-\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}-\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}+\frac{3 b \sin ^{-1}(c x)}{32 c^4} \]

[Out]

(-3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) - (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) + (x^4*(a + b*ArcCos[c*x]))/4 + (3*b*Ar
cSin[c*x])/(32*c^4)

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Rubi [A]  time = 0.0355778, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4628, 321, 216} \[ \frac{1}{4} x^4 \left (a+b \cos ^{-1}(c x)\right )-\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}-\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}+\frac{3 b \sin ^{-1}(c x)}{32 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcCos[c*x]),x]

[Out]

(-3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) - (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) + (x^4*(a + b*ArcCos[c*x]))/4 + (3*b*Ar
cSin[c*x])/(32*c^4)

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]

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 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 \left (a+b \cos ^{-1}(c x)\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{1}{4} (b c) \int \frac{x^4}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{1}{4} x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{(3 b) \int \frac{x^2}{\sqrt{1-c^2 x^2}} \, dx}{16 c}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}-\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{1}{4} x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{(3 b) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{32 c^3}\\ &=-\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}-\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{1}{4} x^4 \left (a+b \cos ^{-1}(c x)\right )+\frac{3 b \sin ^{-1}(c x)}{32 c^4}\\ \end{align*}

Mathematica [A]  time = 0.0474766, size = 68, normalized size = 0.89 \[ \frac{a x^4}{4}+b \sqrt{1-c^2 x^2} \left (-\frac{3 x}{32 c^3}-\frac{x^3}{16 c}\right )+\frac{3 b \sin ^{-1}(c x)}{32 c^4}+\frac{1}{4} b x^4 \cos ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcCos[c*x]),x]

[Out]

(a*x^4)/4 + b*Sqrt[1 - c^2*x^2]*((-3*x)/(32*c^3) - x^3/(16*c)) + (b*x^4*ArcCos[c*x])/4 + (3*b*ArcSin[c*x])/(32
*c^4)

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Maple [A]  time = 0.006, size = 72, normalized size = 1. \begin{align*}{\frac{1}{{c}^{4}} \left ({\frac{{c}^{4}{x}^{4}a}{4}}+b \left ({\frac{{c}^{4}{x}^{4}\arccos \left ( cx \right ) }{4}}-{\frac{{c}^{3}{x}^{3}}{16}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,cx}{32}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{3\,\arcsin \left ( cx \right ) }{32}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.44846, size = 112, normalized size = 1.47 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{32} \,{\left (8 \, x^{4} \arccos \left (c x\right ) -{\left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac{3 \, \sqrt{-c^{2} x^{2} + 1} x}{c^{4}} - \frac{3 \, \arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4 + 1/32*(8*x^4*arccos(c*x) - (2*sqrt(-c^2*x^2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c^
2*x/sqrt(c^2))/(sqrt(c^2)*c^4))*c)*b

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Fricas [A]  time = 2.45426, size = 139, normalized size = 1.83 \begin{align*} \frac{8 \, a c^{4} x^{4} +{\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \arccos \left (c x\right ) -{\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt{-c^{2} x^{2} + 1}}{32 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*a*c^4*x^4 + (8*b*c^4*x^4 - 3*b)*arccos(c*x) - (2*b*c^3*x^3 + 3*b*c*x)*sqrt(-c^2*x^2 + 1))/c^4

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Sympy [A]  time = 1.28894, size = 85, normalized size = 1.12 \begin{align*} \begin{cases} \frac{a x^{4}}{4} + \frac{b x^{4} \operatorname{acos}{\left (c x \right )}}{4} - \frac{b x^{3} \sqrt{- c^{2} x^{2} + 1}}{16 c} - \frac{3 b x \sqrt{- c^{2} x^{2} + 1}}{32 c^{3}} - \frac{3 b \operatorname{acos}{\left (c x \right )}}{32 c^{4}} & \text{for}\: c \neq 0 \\\frac{x^{4} \left (a + \frac{\pi b}{2}\right )}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**4/4 + b*x**4*acos(c*x)/4 - b*x**3*sqrt(-c**2*x**2 + 1)/(16*c) - 3*b*x*sqrt(-c**2*x**2 + 1)/(32
*c**3) - 3*b*acos(c*x)/(32*c**4), Ne(c, 0)), (x**4*(a + pi*b/2)/4, True))

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Giac [A]  time = 1.1333, size = 90, normalized size = 1.18 \begin{align*} \frac{1}{4} \, b x^{4} \arccos \left (c x\right ) + \frac{1}{4} \, a x^{4} - \frac{\sqrt{-c^{2} x^{2} + 1} b x^{3}}{16 \, c} - \frac{3 \, \sqrt{-c^{2} x^{2} + 1} b x}{32 \, c^{3}} - \frac{3 \, b \arccos \left (c x\right )}{32 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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