3.516 \(\int x^4 \sqrt [3]{a+b x^3} \, dx\)

Optimal. Leaf size=120 \[ \frac{a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3}}+\frac{a^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{5/3}}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}+\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b} \]

[Out]

(a*x^2*(a + b*x^3)^(1/3))/(18*b) + (x^5*(a + b*x^3)^(1/3))/6 + (a^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3
))/Sqrt[3]])/(9*Sqrt[3]*b^(5/3)) + (a^2*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(18*b^(5/3))

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Rubi [A]  time = 0.100495, antiderivative size = 173, normalized size of antiderivative = 1.44, number of steps used = 9, number of rules used = 9, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {279, 321, 331, 292, 31, 634, 617, 204, 628} \[ \frac{a^2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}-\frac{a^2 \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{54 b^{5/3}}+\frac{a^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{5/3}}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}+\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^3)^(1/3),x]

[Out]

(a*x^2*(a + b*x^3)^(1/3))/(18*b) + (x^5*(a + b*x^3)^(1/3))/6 + (a^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3
))/Sqrt[3]])/(9*Sqrt[3]*b^(5/3)) + (a^2*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(27*b^(5/3)) - (a^2*Log[1 + (b
^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(54*b^(5/3))

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

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

Rubi steps

\begin{align*} \int x^4 \sqrt [3]{a+b x^3} \, dx &=\frac{1}{6} x^5 \sqrt [3]{a+b x^3}+\frac{1}{6} a \int \frac{x^4}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}-\frac{a^2 \int \frac{x}{\left (a+b x^3\right )^{2/3}} \, dx}{9 b}\\ &=\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}-\frac{a^2 \operatorname{Subst}\left (\int \frac{x}{1-b x^3} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{9 b}\\ &=\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1-\sqrt [3]{b} x} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{4/3}}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1-\sqrt [3]{b} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{4/3}}\\ &=\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}+\frac{a^2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{\sqrt [3]{b}+2 b^{2/3} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{54 b^{5/3}}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{18 b^{4/3}}\\ &=\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}+\frac{a^2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}-\frac{a^2 \log \left (1+\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{54 b^{5/3}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3}}\\ &=\frac{a x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac{1}{6} x^5 \sqrt [3]{a+b x^3}+\frac{a^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{9 \sqrt{3} b^{5/3}}+\frac{a^2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{5/3}}-\frac{a^2 \log \left (1+\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{54 b^{5/3}}\\ \end{align*}

Mathematica [C]  time = 0.0560021, size = 64, normalized size = 0.53 \[ \frac{x^2 \sqrt [3]{a+b x^3} \left (-\frac{a \, _2F_1\left (-\frac{1}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{\frac{b x^3}{a}+1}}+a+b x^3\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^3)^(1/3),x]

[Out]

(x^2*(a + b*x^3)^(1/3)*(a + b*x^3 - (a*Hypergeometric2F1[-1/3, 2/3, 5/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^(1/3))
)/(6*b)

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{x}^{4}\sqrt [3]{b{x}^{3}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^3+a)^(1/3),x)

[Out]

int(x^4*(b*x^3+a)^(1/3),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87821, size = 481, normalized size = 4.01 \begin{align*} -\frac{2 \, \sqrt{3} a^{2}{\left (b^{2}\right )}^{\frac{1}{6}} b \arctan \left (\frac{{\left (\sqrt{3}{\left (b^{2}\right )}^{\frac{1}{3}} b x + 2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b^{2}\right )}^{\frac{2}{3}}\right )}{\left (b^{2}\right )}^{\frac{1}{6}}}{3 \, b^{2} x}\right ) - 2 \, a^{2}{\left (b^{2}\right )}^{\frac{2}{3}} \log \left (-\frac{{\left (b^{2}\right )}^{\frac{2}{3}} x -{\left (b x^{3} + a\right )}^{\frac{1}{3}} b}{x}\right ) + a^{2}{\left (b^{2}\right )}^{\frac{2}{3}} \log \left (\frac{{\left (b^{2}\right )}^{\frac{1}{3}} b x^{2} +{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b^{2}\right )}^{\frac{2}{3}} x +{\left (b x^{3} + a\right )}^{\frac{2}{3}} b}{x^{2}}\right ) - 3 \,{\left (3 \, b^{3} x^{5} + a b^{2} x^{2}\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{54 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

-1/54*(2*sqrt(3)*a^2*(b^2)^(1/6)*b*arctan(1/3*(sqrt(3)*(b^2)^(1/3)*b*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(b^2)^(2/
3))*(b^2)^(1/6)/(b^2*x)) - 2*a^2*(b^2)^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + a^2*(b^2)^(2/3)*l
og(((b^2)^(1/3)*b*x^2 + (b*x^3 + a)^(1/3)*(b^2)^(2/3)*x + (b*x^3 + a)^(2/3)*b)/x^2) - 3*(3*b^3*x^5 + a*b^2*x^2
)*(b*x^3 + a)^(1/3))/b^3

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Sympy [C]  time = 2.06643, size = 39, normalized size = 0.32 \begin{align*} \frac{\sqrt [3]{a} x^{5} \Gamma \left (\frac{5}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, \frac{5}{3} \\ \frac{8}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{8}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**3+a)**(1/3),x)

[Out]

a**(1/3)*x**5*gamma(5/3)*hyper((-1/3, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(8/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)*x^4, x)