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

Optimal. Leaf size=94 \[ -\frac {a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}}-\frac {a \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {1}{3} x^2 \sqrt [3]{a+b x^3} \]

[Out]

1/3*x^2*(b*x^3+a)^(1/3)-1/6*a*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)-1/9*a*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^
(1/3))*3^(1/2))/b^(2/3)*3^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 145, normalized size of antiderivative = 1.54, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {279, 331, 292, 31, 634, 617, 204, 628} \[ -\frac {a \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{2/3}}+\frac {a \log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{18 b^{2/3}}-\frac {a \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {1}{3} x^2 \sqrt [3]{a+b x^3} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^3)^(1/3),x]

[Out]

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

Rule 31

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

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 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 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 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 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 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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.54 \[ \frac {x^2 \sqrt [3]{a+b x^3} \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{2 \sqrt [3]{\frac {b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^3)^(1/3),x]

[Out]

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

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fricas [B]  time = 0.86, size = 193, normalized size = 2.05 \[ \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} x^{2} + 2 \, \sqrt {3} a b \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) - 2 \, \left (-b^{2}\right )^{\frac {2}{3}} a \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + \left (-b^{2}\right )^{\frac {2}{3}} a \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(6*(b*x^3 + a)^(1/3)*b^2*x^2 + 2*sqrt(3)*a*b*sqrt(-(-b^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-b^2)^(1/3)*b*x -
2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b^2)^(2/3))*sqrt(-(-b^2)^(1/3))/(b^2*x)) - 2*(-b^2)^(2/3)*a*log(-((-b^2)^(2/3)*x
- (b*x^3 + a)^(1/3)*b)/x) + (-b^2)^(2/3)*a*log(-((-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))/b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {1}{3}} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(1/3)*x, x)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{\frac {1}{3}} x\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.98, size = 134, normalized size = 1.43 \[ \frac {\sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{9 \, b^{\frac {2}{3}}} + \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, b^{\frac {2}{3}}} - \frac {a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{9 \, b^{\frac {2}{3}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{3 \, {\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(2/3) + 1/18*a*log(b^(2/3) + (b*
x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(2/3) - 1/9*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(2/3)
- 1/3*(b*x^3 + a)^(1/3)*a/((b - (b*x^3 + a)/x^3)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^3)^(1/3),x)

[Out]

int(x*(a + b*x^3)^(1/3), x)

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sympy [C]  time = 1.21, size = 39, normalized size = 0.41 \[ \frac {\sqrt [3]{a} x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/3)*x**2*gamma(2/3)*hyper((-1/3, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3))

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