3.71 \(\int x^3 \cos ^{-1}(a+b x^4) \, dx\)

Optimal. Leaf size=47 \[ \frac {\left (a+b x^4\right ) \cos ^{-1}\left (a+b x^4\right )}{4 b}-\frac {\sqrt {1-\left (a+b x^4\right )^2}}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6715, 4804, 4620, 261} \[ \frac {\left (a+b x^4\right ) \cos ^{-1}\left (a+b x^4\right )}{4 b}-\frac {\sqrt {1-\left (a+b x^4\right )^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCos[a + b*x^4],x]

[Out]

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

Rule 261

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

Rule 4620

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

Rule 4804

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 43, normalized size = 0.91 \[ \frac {\left (a+b x^4\right ) \cos ^{-1}\left (a+b x^4\right )-\sqrt {1-\left (a+b x^4\right )^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCos[a + b*x^4],x]

[Out]

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

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fricas [A]  time = 0.43, size = 48, normalized size = 1.02 \[ \frac {{\left (b x^{4} + a\right )} \arccos \left (b x^{4} + a\right ) - \sqrt {-b^{2} x^{8} - 2 \, a b x^{4} - a^{2} + 1}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(b*x^4+a),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 39, normalized size = 0.83 \[ \frac {{\left (b x^{4} + a\right )} \arccos \left (b x^{4} + a\right ) - \sqrt {-{\left (b x^{4} + a\right )}^{2} + 1}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(b*x^4+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 40, normalized size = 0.85 \[ \frac {\left (b \,x^{4}+a \right ) \arccos \left (b \,x^{4}+a \right )-\sqrt {1-\left (b \,x^{4}+a \right )^{2}}}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccos(b*x^4+a),x)

[Out]

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

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maxima [A]  time = 0.40, size = 39, normalized size = 0.83 \[ \frac {{\left (b x^{4} + a\right )} \arccos \left (b x^{4} + a\right ) - \sqrt {-{\left (b x^{4} + a\right )}^{2} + 1}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(b*x^4+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.65, size = 99, normalized size = 2.11 \[ \frac {x^4\,\mathrm {acos}\left (b\,x^4+a\right )}{4}-\frac {\sqrt {-a^2-2\,a\,b\,x^4-b^2\,x^8+1}}{4\,b}-\frac {a\,\ln \left (\sqrt {-a^2-2\,a\,b\,x^4-b^2\,x^8+1}-\frac {b^2\,x^4+a\,b}{\sqrt {-b^2}}\right )}{4\,\sqrt {-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(a + b*x^4),x)

[Out]

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

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sympy [A]  time = 0.75, size = 61, normalized size = 1.30 \[ \begin {cases} \frac {a \operatorname {acos}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {acos}{\left (a + b x^{4} \right )}}{4} - \frac {\sqrt {- a^{2} - 2 a b x^{4} - b^{2} x^{8} + 1}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acos}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acos(b*x**4+a),x)

[Out]

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

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