3.36 \(\int x \cos ^{-1}(a x)^4 \, dx\)

Optimal. Leaf size=112 \[ -\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{a}+\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}-\frac{\cos ^{-1}(a x)^4}{4 a^2}+\frac{3 \cos ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^4-\frac{3}{2} x^2 \cos ^{-1}(a x)^2+\frac{3 x^2}{4} \]

[Out]

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

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Rubi [A]  time = 0.237046, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4628, 4708, 4642, 30} \[ -\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{a}+\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}-\frac{\cos ^{-1}(a x)^4}{4 a^2}+\frac{3 \cos ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^4-\frac{3}{2} x^2 \cos ^{-1}(a x)^2+\frac{3 x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCos[a*x]^4,x]

[Out]

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

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 4708

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcCos[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcCos[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 4642

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp[(a + b*ArcCos[c*x])
^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x \cos ^{-1}(a x)^4 \, dx &=\frac{1}{2} x^2 \cos ^{-1}(a x)^4+(2 a) \int \frac{x^2 \cos ^{-1}(a x)^3}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{a}+\frac{1}{2} x^2 \cos ^{-1}(a x)^4-3 \int x \cos ^{-1}(a x)^2 \, dx+\frac{\int \frac{\cos ^{-1}(a x)^3}{\sqrt{1-a^2 x^2}} \, dx}{a}\\ &=-\frac{3}{2} x^2 \cos ^{-1}(a x)^2-\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{a}-\frac{\cos ^{-1}(a x)^4}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^4-(3 a) \int \frac{x^2 \cos ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}-\frac{3}{2} x^2 \cos ^{-1}(a x)^2-\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{a}-\frac{\cos ^{-1}(a x)^4}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^4+\frac{3 \int x \, dx}{2}-\frac{3 \int \frac{\cos ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{2 a}\\ &=\frac{3 x^2}{4}+\frac{3 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}+\frac{3 \cos ^{-1}(a x)^2}{4 a^2}-\frac{3}{2} x^2 \cos ^{-1}(a x)^2-\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3}{a}-\frac{\cos ^{-1}(a x)^4}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^4\\ \end{align*}

Mathematica [A]  time = 0.0381788, size = 96, normalized size = 0.86 \[ \frac{3 a^2 x^2+\left (2 a^2 x^2-1\right ) \cos ^{-1}(a x)^4-4 a x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3+\left (3-6 a^2 x^2\right ) \cos ^{-1}(a x)^2+6 a x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCos[a*x]^4,x]

[Out]

(3*a^2*x^2 + 6*a*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x] + (3 - 6*a^2*x^2)*ArcCos[a*x]^2 - 4*a*x*Sqrt[1 - a^2*x^2]*Arc
Cos[a*x]^3 + (-1 + 2*a^2*x^2)*ArcCos[a*x]^4)/(4*a^2)

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Maple [A]  time = 0.053, size = 113, normalized size = 1. \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{{a}^{2}{x}^{2} \left ( \arccos \left ( ax \right ) \right ) ^{4}}{2}}- \left ( \arccos \left ( ax \right ) \right ) ^{3} \left ( ax\sqrt{-{a}^{2}{x}^{2}+1}+\arccos \left ( ax \right ) \right ) -{\frac{3\,{a}^{2}{x}^{2} \left ( \arccos \left ( ax \right ) \right ) ^{2}}{2}}+{\frac{3\,\arccos \left ( ax \right ) }{2} \left ( ax\sqrt{-{a}^{2}{x}^{2}+1}+\arccos \left ( ax \right ) \right ) }-{\frac{3\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}{4}}+{\frac{3\,{a}^{2}{x}^{2}}{4}}-{\frac{3}{4}}+{\frac{3\, \left ( \arccos \left ( ax \right ) \right ) ^{4}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccos(a*x)^4,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{4} - 2 \, a \int \frac{\sqrt{a x + 1} \sqrt{-a x + 1} x^{2} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{3}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccos(a*x)^4,x, algorithm="maxima")

[Out]

1/2*x^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^4 - 2*a*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1)*x^2*arctan2(
sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3/(a^2*x^2 - 1), x)

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Fricas [A]  time = 2.32759, size = 205, normalized size = 1.83 \begin{align*} \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \arccos \left (a x\right )^{4} + 3 \, a^{2} x^{2} - 3 \,{\left (2 \, a^{2} x^{2} - 1\right )} \arccos \left (a x\right )^{2} - 2 \,{\left (2 \, a x \arccos \left (a x\right )^{3} - 3 \, a x \arccos \left (a x\right )\right )} \sqrt{-a^{2} x^{2} + 1}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccos(a*x)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.34343, size = 110, normalized size = 0.98 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{acos}^{4}{\left (a x \right )}}{2} - \frac{3 x^{2} \operatorname{acos}^{2}{\left (a x \right )}}{2} + \frac{3 x^{2}}{4} - \frac{x \sqrt{- a^{2} x^{2} + 1} \operatorname{acos}^{3}{\left (a x \right )}}{a} + \frac{3 x \sqrt{- a^{2} x^{2} + 1} \operatorname{acos}{\left (a x \right )}}{2 a} - \frac{\operatorname{acos}^{4}{\left (a x \right )}}{4 a^{2}} + \frac{3 \operatorname{acos}^{2}{\left (a x \right )}}{4 a^{2}} & \text{for}\: a \neq 0 \\\frac{\pi ^{4} x^{2}}{32} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acos(a*x)**4,x)

[Out]

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

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Giac [A]  time = 1.16109, size = 136, normalized size = 1.21 \begin{align*} \frac{1}{2} \, x^{2} \arccos \left (a x\right )^{4} - \frac{3}{2} \, x^{2} \arccos \left (a x\right )^{2} - \frac{\sqrt{-a^{2} x^{2} + 1} x \arccos \left (a x\right )^{3}}{a} + \frac{3}{4} \, x^{2} - \frac{\arccos \left (a x\right )^{4}}{4 \, a^{2}} + \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x \arccos \left (a x\right )}{2 \, a} + \frac{3 \, \arccos \left (a x\right )^{2}}{4 \, a^{2}} - \frac{3}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccos(a*x)^4,x, algorithm="giac")

[Out]

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