3.47 \(\int x^3 \cos ^{-1}(a x^2) \, dx\)

Optimal. Leaf size=51 \[ \frac {\sin ^{-1}\left (a x^2\right )}{8 a^2}-\frac {x^2 \sqrt {1-a^2 x^4}}{8 a}+\frac {1}{4} x^4 \cos ^{-1}\left (a x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4843, 12, 275, 321, 216} \[ -\frac {x^2 \sqrt {1-a^2 x^4}}{8 a}+\frac {\sin ^{-1}\left (a x^2\right )}{8 a^2}+\frac {1}{4} x^4 \cos ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCos[a*x^2],x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 4843

Int[((a_.) + ArcCos[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcCos[
u]))/(d*(m + 1)), x] + Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int x^3 \cos ^{-1}\left (a x^2\right ) \, dx &=\frac {1}{4} x^4 \cos ^{-1}\left (a x^2\right )+\frac {1}{4} \int \frac {2 a x^5}{\sqrt {1-a^2 x^4}} \, dx\\ &=\frac {1}{4} x^4 \cos ^{-1}\left (a x^2\right )+\frac {1}{2} a \int \frac {x^5}{\sqrt {1-a^2 x^4}} \, dx\\ &=\frac {1}{4} x^4 \cos ^{-1}\left (a x^2\right )+\frac {1}{4} a \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^2 \sqrt {1-a^2 x^4}}{8 a}+\frac {1}{4} x^4 \cos ^{-1}\left (a x^2\right )+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx,x,x^2\right )}{8 a}\\ &=-\frac {x^2 \sqrt {1-a^2 x^4}}{8 a}+\frac {1}{4} x^4 \cos ^{-1}\left (a x^2\right )+\frac {\sin ^{-1}\left (a x^2\right )}{8 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 0.94 \[ \frac {-a x^2 \sqrt {1-a^2 x^4}+2 a^2 x^4 \cos ^{-1}\left (a x^2\right )+\sin ^{-1}\left (a x^2\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCos[a*x^2],x]

[Out]

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

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fricas [A]  time = 0.45, size = 41, normalized size = 0.80 \[ -\frac {\sqrt {-a^{2} x^{4} + 1} a x^{2} - {\left (2 \, a^{2} x^{4} - 1\right )} \arccos \left (a x^{2}\right )}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.02, size = 46, normalized size = 0.90 \[ \frac {2 \, a^{2} x^{4} \arccos \left (a x^{2}\right ) - \sqrt {-a^{2} x^{4} + 1} a x^{2} - \arccos \left (a x^{2}\right )}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 65, normalized size = 1.27 \[ \frac {x^{4} \arccos \left (a \,x^{2}\right )}{4}-\frac {x^{2} \sqrt {-a^{2} x^{4}+1}}{8 a}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x^{2}}{\sqrt {-a^{2} x^{4}+1}}\right )}{8 a \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccos(a*x^2),x)

[Out]

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

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maxima [A]  time = 0.41, size = 79, normalized size = 1.55 \[ \frac {1}{4} \, x^{4} \arccos \left (a x^{2}\right ) - \frac {1}{8} \, a {\left (\frac {\arctan \left (\frac {\sqrt {-a^{2} x^{4} + 1}}{a x^{2}}\right )}{a^{3}} + \frac {\sqrt {-a^{2} x^{4} + 1}}{{\left (a^{4} - \frac {{\left (a^{2} x^{4} - 1\right )} a^{2}}{x^{4}}\right )} x^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.30, size = 42, normalized size = 0.82 \[ \frac {\mathrm {acos}\left (a\,x^2\right )\,\left (2\,a^2\,x^4-1\right )}{8\,a^2}-\frac {x^2\,\sqrt {1-a^2\,x^4}}{8\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(a*x^2),x)

[Out]

(acos(a*x^2)*(2*a^2*x^4 - 1))/(8*a^2) - (x^2*(1 - a^2*x^4)^(1/2))/(8*a)

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sympy [A]  time = 0.75, size = 48, normalized size = 0.94 \[ \begin {cases} \frac {x^{4} \operatorname {acos}{\left (a x^{2} \right )}}{4} - \frac {x^{2} \sqrt {- a^{2} x^{4} + 1}}{8 a} - \frac {\operatorname {acos}{\left (a x^{2} \right )}}{8 a^{2}} & \text {for}\: a \neq 0 \\\frac {\pi x^{4}}{8} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acos(a*x**2),x)

[Out]

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

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