3.25 \(\int x \cos ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=99 \[ \frac {3 x \sqrt {1-a^2 x^2}}{8 a}-\frac {3 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2}{4 a}-\frac {3 \sin ^{-1}(a x)}{8 a^2}-\frac {\cos ^{-1}(a x)^3}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^3-\frac {3}{4} x^2 \cos ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4628, 4708, 4642, 321, 216} \[ \frac {3 x \sqrt {1-a^2 x^2}}{8 a}-\frac {3 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2}{4 a}-\frac {3 \sin ^{-1}(a x)}{8 a^2}-\frac {\cos ^{-1}(a x)^3}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^3-\frac {3}{4} x^2 \cos ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCos[a*x]^3,x]

[Out]

(3*x*Sqrt[1 - a^2*x^2])/(8*a) - (3*x^2*ArcCos[a*x])/4 - (3*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2)/(4*a) - ArcCos[a
*x]^3/(4*a^2) + (x^2*ArcCos[a*x]^3)/2 - (3*ArcSin[a*x])/(8*a^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]

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 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]

Rubi steps

\begin {align*} \int x \cos ^{-1}(a x)^3 \, dx &=\frac {1}{2} x^2 \cos ^{-1}(a x)^3+\frac {1}{2} (3 a) \int \frac {x^2 \cos ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {3 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2}{4 a}+\frac {1}{2} x^2 \cos ^{-1}(a x)^3-\frac {3}{2} \int x \cos ^{-1}(a x) \, dx+\frac {3 \int \frac {\cos ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{4 a}\\ &=-\frac {3}{4} x^2 \cos ^{-1}(a x)-\frac {3 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2}{4 a}-\frac {\cos ^{-1}(a x)^3}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^3-\frac {1}{4} (3 a) \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {3 x \sqrt {1-a^2 x^2}}{8 a}-\frac {3}{4} x^2 \cos ^{-1}(a x)-\frac {3 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2}{4 a}-\frac {\cos ^{-1}(a x)^3}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^3-\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a}\\ &=\frac {3 x \sqrt {1-a^2 x^2}}{8 a}-\frac {3}{4} x^2 \cos ^{-1}(a x)-\frac {3 x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2}{4 a}-\frac {\cos ^{-1}(a x)^3}{4 a^2}+\frac {1}{2} x^2 \cos ^{-1}(a x)^3-\frac {3 \sin ^{-1}(a x)}{8 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 85, normalized size = 0.86 \[ \frac {3 a x \sqrt {1-a^2 x^2}+\left (4 a^2 x^2-2\right ) \cos ^{-1}(a x)^3-6 a x \sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2-6 a^2 x^2 \cos ^{-1}(a x)-3 \sin ^{-1}(a x)}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCos[a*x]^3,x]

[Out]

(3*a*x*Sqrt[1 - a^2*x^2] - 6*a^2*x^2*ArcCos[a*x] - 6*a*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2 + (-2 + 4*a^2*x^2)*Ar
cCos[a*x]^3 - 3*ArcSin[a*x])/(8*a^2)

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fricas [A]  time = 0.43, size = 69, normalized size = 0.70 \[ \frac {2 \, {\left (2 \, a^{2} x^{2} - 1\right )} \arccos \left (a x\right )^{3} - 3 \, {\left (2 \, a^{2} x^{2} - 1\right )} \arccos \left (a x\right ) - 3 \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x \arccos \left (a x\right )^{2} - a x\right )}}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 83, normalized size = 0.84 \[ \frac {1}{2} \, x^{2} \arccos \left (a x\right )^{3} - \frac {3}{4} \, x^{2} \arccos \left (a x\right ) - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x \arccos \left (a x\right )^{2}}{4 \, a} - \frac {\arccos \left (a x\right )^{3}}{4 \, a^{2}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{8 \, a} + \frac {3 \, \arccos \left (a x\right )}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 90, normalized size = 0.91 \[ \frac {\frac {a^{2} x^{2} \arccos \left (a x \right )^{3}}{2}-\frac {3 \arccos \left (a x \right )^{2} \left (a x \sqrt {-a^{2} x^{2}+1}+\arccos \left (a x \right )\right )}{4}-\frac {3 a^{2} x^{2} \arccos \left (a x \right )}{4}+\frac {3 a x \sqrt {-a^{2} x^{2}+1}}{8}+\frac {3 \arccos \left (a x \right )}{8}+\frac {\arccos \left (a x \right )^{3}}{2}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccos(a*x)^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x^{2} \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )^{3} - 3 \, a \int \frac {\sqrt {a x + 1} \sqrt {-a x + 1} x^{2} \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )^{2}}{2 \, {\left (a^{2} x^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3 - 3*a*integrate(1/2*sqrt(a*x + 1)*sqrt(-a*x + 1)*x^2*arct
an2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2/(a^2*x^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {acos}\left (a\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acos(a*x)^3,x)

[Out]

int(x*acos(a*x)^3, x)

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sympy [A]  time = 0.87, size = 99, normalized size = 1.00 \[ \begin {cases} \frac {x^{2} \operatorname {acos}^{3}{\left (a x \right )}}{2} - \frac {3 x^{2} \operatorname {acos}{\left (a x \right )}}{4} - \frac {3 x \sqrt {- a^{2} x^{2} + 1} \operatorname {acos}^{2}{\left (a x \right )}}{4 a} + \frac {3 x \sqrt {- a^{2} x^{2} + 1}}{8 a} - \frac {\operatorname {acos}^{3}{\left (a x \right )}}{4 a^{2}} + \frac {3 \operatorname {acos}{\left (a x \right )}}{8 a^{2}} & \text {for}\: a \neq 0 \\\frac {\pi ^{3} x^{2}}{16} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acos(a*x)**3,x)

[Out]

Piecewise((x**2*acos(a*x)**3/2 - 3*x**2*acos(a*x)/4 - 3*x*sqrt(-a**2*x**2 + 1)*acos(a*x)**2/(4*a) + 3*x*sqrt(-
a**2*x**2 + 1)/(8*a) - acos(a*x)**3/(4*a**2) + 3*acos(a*x)/(8*a**2), Ne(a, 0)), (pi**3*x**2/16, True))

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